Mathematical Paradoxes: Twin coin flipping
I’ve recently come across some fascinating probability and statistical paradoxes that made me think that our possibly distorted perception of reality and our intuition can lead us to the wrong conclusions quite often.

I will not post any relevant links yet, but I will post some of the paradoxes, some are very common, others are very frustrating, and others very difficult to understand and explain (these are the worst ones!). So here we go, an easy one for warm up, coin flipping:

Flip two coins simultaneously. If one coin is heads, what is the probability that the other coin is also heads?

UPDATE: There is a new post “Twin coin flipping ‘paradox’ revisited” which explains this scenario and includes a javascript that demonstrates the correct results.
118 Responses to this post
50-50, it is always 50-50: heads or tails!
I mean regardless of what the first coin is…
Agree, coins do not have memory so it doesn't matter what the first one came out to be. It is still 50-50 that it is the same as the other
… provided that the events are independent … the combination H(ead) and H(ead) is 1/4. Mathematically, P(H) * P(H) = 1/2 * 1/2 = 1/4 … It is not a paradox! Note that Independence does matter!
…and of course is a paradox in the sense that it contradicts the common response of an average person who would say it is 50%!
OK, the initial probability is indeed 1/4, but after you know that one (not any one specifically) is heads, you can safely eliminate the T - T case, so the probability is 1/3 actually!
Starvos is absolutely correct. 50% of the time a h-t combo comes up. the h can occur for either coin and you note the other is t. since a t-t combo is thrown out, the remaining possiblilities are: h-t; t-h; and h-h. the only combo that results in a 2nd h is h-h. 1/3 of the time.
Nonsense. If you already know that one of the coins is showing "heads" you cannot assume that it is showing "tails" with probability 0,5. (The only probability for that is 0). Knowing that one of the coins shows "heads" doesn't give you any information at all about the other coin because of the independence. (As we are assuming that.) If you don't want to believe that the probability for the other coin showing "heads" as well is 0,5 I recommend you to make a simulation of the coin toss. I think that some of you may have confused this problem with the likes with the so called Monty Hall -problem, where the initial probabilities affect the final probability, but that is just because the events are not independent.
Absolutely it is not a Paradox
When the first coin is heads two possibilities remain in the sample space. H-T and H-H
Therefore this is not a paradox. H-H outcome has probability 0.5. You can also use bayes theorem to calculate this probability, with the added information that first coin is heads, the probability that second coin will be heads is 0.5. This is not a paradox but rather a mathematical mistake.
Nobody said the FIRST coin is heads! The problem says if ONE coin is heads, anyone and not the first or the second! There lies the difference.

If any is heads, there are three possibilities: H-H, H-T, T-H. So for both to be heads it is 1/3 or 33%
Suppose the 2 coins are A and B. What does “If one coin is heads” mean? Does it mean “If coin A is heads”, “If coin B is heads”, or “If at least A or B is heads”? It makes a difference.
Hi Laz,

it means “if at least one of the coins is heads” *without* specifying which one! And it is exactly therein that lies the difference. If someone said “coin A” or “the first coin” is heads then the probability is 50%. Also, the coins are thrown simultaneously, which also makes a difference.
Stavros, I should have read your previous post more carefully before sending my own, because I think I only repeated your comment, which clearly I agree with.

It seems to me there are two possible ways of looking at the question:

- I know that there is a coin which is heads


- I know that a particular coin is heads.

As you say, it makes a difference (I think - I’m not totally sure because I’ve just put a bottle of wine inside me).
Just to let you know who sent the previous post - I’m not totally anonymous.
Stavros is exactly correct. If both coins are flipped at the same time, there are four ways they can land:

1. tail - tail
2. tail - head
3. head - tail
4. head - head

“If one coin is heads” eliminated possibility 1 since neither is heads. Therefore, if one coin is heads (landings 2, 3 or 4), the other can be tails (landings 2 or 3) or heads (landing 4).

Clearly then if ONE coin lands heads, the possibility of the other one also being heads is 1 out of 3.
Assume that one coin is heads, what are the odds that it will come up heads again.

As it been stated we already know that one of the coins is (H)
So therefore the solution will start with 50% of a chance
Outcome: 1 coin (H) 3 possibilities


If one coin we know is H –» HT = TH will produce 50% chance

I think 50% chance only will work correctly if we tell exact event of the coin in the beginning, else the probability of 1/3 is not working

Is like having different idea by assuming a traffic light labelled with following [TT, TH, HT, HH]. Assume we know one of the coins, same as knowing which light it’s have been lit; If we already know which coin, it’s almost knowing that two of the lights can’t have been lit.

Therefore this leads to: if the coin is (H) it will follow to one of the lights going off.. .

Regards Smith
Smith, thanks for taking a ride on the blog and commenting :-)

we must be careful to distinguish from the case where the “first (or second) coin is heads” with the case where “one of the coins” is heads. These cases produce different results.
Hi guys!!! I have a “nice question”: The probability that 4 coins will come head up when flipped simlutaneously is…
1/2 x 1/2 x 1/2 x 1/2 = (1/2)^4 = 1/16
I’m sorry but I don’t get it!
Four independent events (coin flippings) each one with a probability of landing heads being 1/2 (or 0.5).

To calculate the probability that all four coins come heads you just multiply their individual probabilities: 1/2 x 1/2 x 1/2 x 1/2 which equals 1/2 at the power of 4, which equals 1/16 (or 0.0625).

So the probability that all four coins come heads is 1/16.
ok…thank you!
whats the parodox . no confusion here it is 1/2
ajay, you are confused. The probability is 1/3 not a half. Read the question. You have answered a question that wasn’t posted.
“Flip two coins simultaneously. If one coin is heads, what is the probability that the other coin is also heads?”

The question here is not the probability that both coins land heads up, nor is the question whether either coin matches the other.

The question is what is the probability that one of the coins lands heads up. The statement that one of the coins lands heads up has no bearing on the question asking that another coin lands heads up.

You can flip 1000 coins at once. If 999 land heads up, the probability that the 1000th coin also lands heads up is 1/2.
davidh, the question is the one that you quoted. It is not even remotely similar to the one in your fourth paragraph.

You are correct in believing that the other coins don’t influence the 1000th coin. The posters who have given the correct answer (1/3) are also fully aware of that. Posts 6,7,11,16 and 17 make it clear why it’s 1/3 and that the law’s of nature are not being broken.

If you flip 1000 coins and (at least) 999 come up heads, then the probability that the 1000th coin is a heads is 1/1001. That’s because the only admissible arrangements are: 1000H and 1000 variations on 999H+1T - a total of 1001 equally likely events. Any time you thrown 2 or more T, disgard the result as it doesn’t have at least 999 H.

The probability of actually throwing at least 999 heads is incomprehensibly small, being 1001/(2^1000) ≈ 9.34*10^-299.
… oops, I meant your third paragraph.

As I’m posting again: I believe that you are misinterpreting the question as “if you throw 999 heads in a row, then what is the probability that the next throw will be a head?” For that, the answer is 1/2.
OK, let’s look at it this way -

Obviously, the outcome can’t be determined until after the action has been completed. The original postulation is that both coins are flipped simultaneously. The results can be determined only after the two coins have landed on the floor.

If one coin shows tails it doesn’t matter what the second one is because it can’t also be heads and the premise of the problem isn’t met.

However, if one coin is heads then we can continue and look at the other coin. The second coin can only be heads or tails, or one out of two options.

The question isn’t asking about all possible outcomes of the two coins being flipped; it is asking about the condition of coin B only if the coin A is heads.

The second coin can only be heads or tails or odds of 1/2
Hi davidh. If you change the description of the problem then of course you will gets different solutions. So introducing that one coin is definitely showing tails is irrelevant to the original problem. I’m sure we are agreed on that.

The question doesn’t say that coin A is showing heads. If it had, then the answer would be 1/2. It only says, admittedly only implicitly, that A or B or both are showing heads. Presumably someone else has provided this information after they have been asked “is either of the coins showing a head?”. If the answer was “no” then both coins must be showing tails, so we discard this trial and have another go.

Rather than repeating my previous argument, I’ll offer the other way of viewing it (which I don’t like as introduces probability too early and relatively advanced probability at that) is that P(HH) = P(HT) = P(TH) = P(TT) = 1/4, then using Bayesian probability to say P(HH given at least one H) is P(HH)/(P(HH)+P(HT)+P(TH)) = 1/3. Or if you prefer P(HH)/(1-P(TT)).

At the risk of appearing to be condescending, I assure you that I didn’t interpret the question properly when I first came across it and also believed that the answer was 1/2. I think that the argument I made in my last post with the 1000 coins and the corresponding result of 1/1001 should be enough to force the penny to drop (groan) - it hasn’t done for you yet - but that’s all. I’m sure it will.

There is a danger that the second half of your argument could be interpreted as follows: if I throw a die it can land showing a 6 or something else. That’s two possible results, so that means there’s a 50-50 chance of throwing a 6. I’m not sure that I’m being fair in that interpretation. I mention it as it’s, to me, precisely the same “logic” that people use when discussing the Monty Hall problem and I just wanted to sneak that reference in ;)
The big problem is in the interpretation of the question - that is, after the flip, do we look at the coins one at a time, or both together?

If, when we look at them one at a time, the first is heads, then the other can only be either heads or tails (HH or HT) - a possibility of 1/2 that it’s heads.

However, if we look at them together, then the possibilities are HH, HT, TH or TT. In the first three cases one is heads, but in only one of these three outcomes is the other also heads - a possibility of 1/3.

It’s pretty hard to look at two things at the same time, so as a gedankenexperiment this question yields a final answer of 1/3; however as a practical reality I would argue an answer of 1/2.
Hi davidh. I see you’re beginning to get it. i.e. that’s the first time that you’ve acknowledged that 1/3 is possibly the answer; although you call it a possibility rather than a probability. Something is possible, or it isn’t; it may not be known, or even knowable, which it is.

To me, the question is crystal clear in that no particular coin is being specified as showing heads. It explicitly states “one of them” is heads and unambiguously implies that the other may be heads or tails. There are 3 possibilities. They are “obviously” equally likely so the probability of each possibility occurring is 1/3. All three satisfy that one of the coins is heads, only one of them corresponds to the other coin being a head.

I don’t see how, after all that’s been said, that you can still find any ambiguity in the question. Your second paragraph is not the one that the question is about.

Why do you need to see both coins at the same time? They aren’t likely to flip over just because you look away. However, I don’t find it especially difficult to see two coins at the same time, although I’ll admit it was harder to do than I thought it would be.

The last 1/3 of fourth paragraph is a denial of reality - the answer is 1/3 - that’s an inescapable fact, you can’t philosophise into 1/2 - not even that scumbag fraud Uri Geller can do that.
davidh. On re-reading your last post, I realise that I have misinterpreted what you meant when you were talking about looking at two things at the same time - I now realise you meant looking at the cases corresponding to two different questions, at the same time. I now don’t understand why you would want to do that.

Just look at the case that corresponds to the question being asked, it’s stands alone, it doesn’t communicate with another coining tossing experiment and change it’s behaviour accordingly. This is not a multiverse quantum-entanglement problem. It is an extremely simple coin tossing problem.
Although you ask how I might find any ambiguity in the question, there is some in how the whole scenario is perceived. Are we looking at how the coins react to being flipped (A), or how we see them after the event has happened (B)?

In the first case (before the flip), there are four ways the coins can wind up, with only one way being heads on one coin AND heads on the other coin too. (1/4)

In the second case (after the flip) (looking at the coins one at a time and seeing one head first), if we see one head showing, then we know the other one can only be heads (HH) or tails (HT or TH). (1/3)

Or (after the flip), (looking at the coins at the same time with at least one head), we’ll see either two heads or not. (1/2)

There is a difference in how it

wind up as opposed to how it

finally wind up.
Your doing metaphysics again. The reaction by a coin to being flipped is that it lands with a head or a tail showing with probability 1/2 for either case. Your observation of the result doesn’t affect the result - just your knowledge of it. These are just ordinary coins. You can throw the coins one at a time. You can look at the first one before throwing the second one if you wish. At some point you throw the second coin (or even the first coin a second time), it doesn’t matter one jot.

You are also changing the question yet again. You keep making the first coin you look at be a heads. The question does not say the first coin or the second coin is a heads - it has deliberately not done so. Why do you refuse to see that that is the only question that I’m arguing for P(HH│H? or ?H) = 1/3. It’s pretty obvious that I understand that when you throw two coins, that the probability of them both being heads is 1/4 and that if you throw two coins and the first one you look at being a heads has no influence on whether or not the second one is a heads - the probability of the second one being a heads is 1/2.

You seem to have no difficulty in realising that the probability of throwing two heads is 1/4. I wonder if I could persuade you that the probability is “in practice” a half. After all, we’ve already thrown a head, there is no memory effect etc, so the chance of getting a second head is a half so the chance of getting two heads is a 1/2. Or I perhaps I could argue that either you get two heads or you don’t, that’s two possibilities, so the probability of getting two heads is a half. The latter is unashamedly using your logic. Please don’t try to correct me on that, it was hard enough coming up with such nonsense.

An exactly equivalent phrasing of the question is: “Flip two coins simultaneously. Discarding all cases where two tails are thrown, of the remaining possibilities, what is the probability that both coins are showing heads?”. The advantage to that version is that the psychological reaction is different, the symmetry is perfect. No-one (I’m an eternal optimist) is going to start seeing ambiguities about a first coin etc., which they seem to do with the original form of the question, probably because it reminds them of another slightly similar sounding problem where the answer is 1/2.
Because a little experimentation can go a long way, for those still unable to grasp why the probability is 0.33 and NOT 0.5 I have hacked up a little javascript to calculate it. Find it here: “Twin coin flipping ‘paradox’ revisited

Hi Stavros..!

Indeed a very beautiful “paradox”…! :)

The solution hinges totally on that tiny part where you say “one of the coins”…& not “the first-coin…” :))

BTW, I was pleasantly surprised that I too posted an article on the EXACT portion of Feynman’s lecture on Gravitation as you did TODAY..!!

And great that you’ve now provided an “Experiment”—in the true Spirit of Feynman himself would have loved..!

Talk of Probability (or Serendipity..!!)..!

the probability of getting two heads is a half but it happens only in half of cases. the other half of cases is when first coin is tail
Still not quite convinced…

You have two coins. You throw them both into the next room where they land on the floor. Someone in the room picks them both up and, without altering whichever face is showing, places them on a table next to each other. You are then called into the room and you look at the two coins laying next to each other on the table.

Suppose that you can plainly see that one coin shows heads. The other one either shows heads also, or tails - a 1 in 2 chance that they are both heads.

If, on the other hand, when you first see the two coins, you see one coin shows tails, then the experiment is over as the conditions of the original question are not met.

After the flip (toss), there is no coin A nor coin B, nor is there any First or Second coin. There are simply two coins, either both heads, or not both heads.

I agree that there are four different ways the coins can land, but we’re only interested in the final condition where one of the coins is heads and the other one either is, or it is not heads too.

Calculating the possibilities of final outcomes is not the same as observing the reality of the final outcome.

If you spend a few weeks in a sensory depravation room where you have no view of the outside and no clock to keep time, your days and nights will eventually diverge from the actual conditions outside. When it’s finally time to exit this room, the predictive odds of it being either day or night outside are 50-50. However, when you open the door, observation changes the odds to 100% that it is what it is.
If not convinced, please look at the experimental observations here
Hi davidh. Throwing the coins into another room etc., doesn’t help because YOU still ended up being the observer. If you do that then there are two possibilities:
1 - you see both coins at the same time - then no probabilities as you see what you see. However, the probability of two heads is 1/4.
2 - you see one coin first (head or tail), then the probability that the other coin is a head is 1/2.
In neither of the above cases have we rejected the two tails result.

It is crucial that you use a third party (or equivalent) to observe the result. Let the third party throw the coins (out of your sight), he will throw (with equally probability) HH, HT, TH and TT. He will simply throw again if he gets TT because TT is not at least one head. We are left with three equally likely outcomes: HH, HT and TH.

You may (or may not) find the related question and answer helpful: “flip two coins simultaneously. If one coin is heads, what is the probability that the other coin is tails?” Answer: 2/3.

I agree that the probability of two coins flipped together both landing heads up is 1/4.

However, what the question is postulating is that after the coins have landed, at least one of the coins is heads. Thus, the TT condition be ignored. Only the condition of at least one coin being heads need be considered.

Since coins are thrown and land as a single pair, when they should be viewed only as a single pair. Ignoring any TT landings, we are left with either a TH or HH pair.

As I mentioned before, there is no coin A or B, nor is there a First or Second coin; there is only a single pair of coins. HT and TH are not different, they are the same thing viewed from opposite sides of the table.

The confusion here seems to be in the wording of the question, “If one coin is heads, what is the probability that the other coin is also heads?” This seems to require that the coins be viewed separately, one after the other. I say that if both coins are viewed together as a single connected pair, and if one of the pair is Heads, then the other half of the pair is either a Heads or it is a Tails.

I can only repeat that the odds of an event happening are not the same as stating the results after the event has taken place.

The odds of correctly guessing a three digit lottery number before a drawing are 1000 to one. However, after the drawing, you have either won or lost. You may have to play 1000 times before you win, but after each drawing, you have either won or you have lost.

The chances of being hit by a car while crossing the street in the middle of a block may be 1 in 10,000; but if you reach the other side unscathed, the odds that you got hit are zero.
davidh, no the two coins are not a single entity, they are separate -we just do not know which of the two coins is heads. So there are still three possibilities (HH, HT, TH) as explained numerous times in the comments. You might still have trouble understanding it, but this does not change the correct answer, which is of course 0.33 -see the link I posted above for a demo.
Hi davidh. You must distinguish between the probability of an event ocurring (whether or not it has actually ocurred) and any actual particular result. The probability that a coin has been thrown and is showing heads implies that the probability of the coin showing heads is 1. You seem to be blurring between what “could happen” with what “has happened” in a way that is going to cause total confusion. The probability of what has happened is always 1.

I’m saying that I know before or after the experiment, that the probability of throwing two heads (after disregarding all two tails results) is 1/3.

As to your HT = TH on the other side of the table - that’s crazy. If you throw two coins, the probability of throwing HT (in that order) is 1/4 and the probability of throwing TH (in that order) is 1/4. So the probability of throwing an H and a T disregarding the order is 1/2 and not 1/4 as your logic would inevitably lead to.

I can only suggest that you actually try (the original problem) experimentally either using Stavros’s applet or try it with real coins - I doubt that you’d have to do it many times before you see the experimental confirmation.

As Stavros says, the correct answer (a proven experimental result) is that the probabilty (original question) is 1/3. Any argument you make to the contrary is necessarily fallacious.
Hi davidh. Sorry, on re-reading your and my last posts, I see my one wasn’t a good response.

There is no confusion in the wording of the problem - the only confusion is in the mind(s) of the reader(s) of the question who “translate” it into something that it isn’t - the question deliberately doesn’t suggest/imply/infer a first or second coin. That neither particular coin is mentioned is the point, the whole point and nothing but the point. Until you fully accept that, you will be not be able to understand the result (1/3).

Forget all the cunning plans of throwing the coins into other rooms etc., they are irrelevant. The only important thing is that there is no first or second coin - they are either both tails, in which case we disregard the result, or they are not - it’s that simple.
The result is 1/2 as Chris states. When the coins are tossed, there is no way to tell which of the coins landed on HEADS only that at least one coin did. By this, you are applying a “magic effect” on the results, setting the probability back to 1/2 and not 1/3.

Now, if the question was applied to a switch gate, where two lights are either on or off but both lights can never be off at the same time (due to how the switch gates were wired), then you get the 1/3 result that you are trying to apply to the coins.

These lights do not require the magic effect to produce a result of 1/3 like the coins do. Think about it.
Hi PassingThru. I assume your first staement is a typo - I said 1/3 not 1/2. The phrase “magic effect” is extremely unhelpful.

Your two lights scenario has a two serious flaws. The first is that you haven’t defined what drives the input side of the gates to the lights. Assuming that it was driven by a four state signal with each state being eqaully likely, then the gating would turn the one that corresponded to both lights being off, into one of the other three output states. That would give two input states that corresponded to one output state, which would then occur with a probability 1/2 and the other two with a probability 1/4.

To get the correct result, when getting the input state that corresponded to both lights being off, some additional gating would have to prematurely trigger the next random input state. In effect the unwanted input state itself would have to be suppressed, leaving the three valid input states only - each now with a probability 1/3 (in the time domain) of occurring.
The fallicy in your logic lies in the way the problem is postulated; “Flip two coins simultaneously. If one coin is heads, what is the probability that the other coin is also heads?”

THe first sentence gives one condition; the second sentence gives another.

To say that you flip two coins simultaneoisly means that a single event is taking place - the two coins being flipped.

But then the second sentence speaks of two events - the viewing of two coins separately from one another.

To put this in its simplest terms, there is only one event - the two coins being flipped and landing one of three ways: H-H, T-T, or H-T.

There is no differece between H-T and T-H. This is simply one of the three ways that two coins can land.

Saying that H-T is different from T-H is only relevent when the coins are flipped, and land, separately.

If you say that my reasoning is faulty, I challenge you to explain to me exactly what is the difference between the H-T and T-H outcome in a simgle event.
davidh, after almost a year, you still cannot get this??

No, this is not “faulty reasoning”, it isn’t even reasoning! You say: “there is only one event - the two coins being flipped and landing one of three ways: H-H, T-T, or H-T” and “explain to me exactly what is the difference between the H-T and T-H outcome in a simgle event“. Are you telling me that then the answer to the question “what is the probability the two coins landed same side” is 66%?!?

Of course there is difference between H-T and T-H! If someone asks “the leftmost coin landed HEADS, what is the probability that …” then it is obvious why. And no, this is not a separate event: the probabilities we assign depend on our level of knowledge.

If we know nothing at all, the chances they landed the same is 50%. If we still know nothing at all, the chances they both landed HEADS is 25%. If we know the LEFT ONE landed HEADS the chances they landed the same is again 50%. If we know ONE of them landed HEADS (but not which one) the chances they landed the same is 33% as discussed.
davidh. Let one coin have sides H and T, and the other have sides H’ and T’. The four possible equally likely throws are HH’, HT’, TH’ and TT’. That’s one way to get two heads, one way to get two tails and two ways to get a head and a tails.
Stavros - there are absolutely two different and separate events. One - flipping the coins and Two - observing the result.

If the flipper holds his arm into a room, flips the coins and leaves before the coins hit the floor - that’s one event. If then an observer enters the room and sees the coins on the floor - that’s the second event.

The conditions of the second event are that either both coins show the same face, or they don’t. If they’re both Tails then the outcome can be ignored. On the other hand, if one is Heads then there is only a 50% chance that the other one is also heads.

If the coins had been flipped one at a time, then there are four ways they can land (one at a time). Since they were flipped at the same time (and presumably land at the same time) there can only be three ways that they will have landed - both sides the same (two Heads or two Tails), or with different sides showing (one Head and one Tail). Of the coins which have landed with at least one side showing a Head, there is only one remaining coin which could possible also have a head showing.

Forget the flipping part - if you are walking down the street and see two pennies on the ground, ignore how they may have gotten there, just focus on the probability that if one is Heads, the other one will be too. Obviously 50% - if one is Heads the other one either is or is not also Heads.
So you *are* indeed saying that the chance of two coins flipped simultaneously landing same face up is 66%!! Goodness me…

This is now an exercise in futility. You can continue missing the logic and stick to your non-reasoning. I have no more patience to repeat the same things. Perhaps someone braver than me will keep going…
davidh. I can’t imagine why you make so much of these separate events. You are taking a perfectly ordinary idea (flipping two coins) and turning it into some sort of mystical event.

All you do is flip two coins. If they are both tails, try again. Of the remaining 3/4 of all the possible throws, the HT+TH combination is twice as likely as the HH combination. So that means HH has a probability of 1/3 and HT+TH a probability of 2/3.

No secret rooms, no holding your breath, no shutting your eyes, no (un)lucky rabbit’s foot etc., are required - just two ordinary coins.
Chris - at least you are admitting that there are two separate events taking place. But flipping the coins is a red herring and is immaterial to the results. All that is needed to solve the problem is to observe the coins after they have landed.

I’ll try again.

The first event is flipping the two coins. before this is done, we know that each coin can land in two ways, for a total of four possibilities (I’m agreeing with you on that point).

However, the answer to the question is only the second event - observing the two coins on the ground after they have already landed.

At that point, it doesn’t matter how the coins got there, they may have been flipped, or fallen out of someones pocket, or been placed there deliberately. The only thing that matters is that there are two coins there.

When they are seen, they can either both be tails, or they can both be heads, or there can be one heads and one tails. Just three possibilities with only two of the possibilities meeting the problem as presented (at least one being heads). Therefore, the posssibility of both being heads is one in two.

When you talk of H-T or T-H, you are introducing the concept of a first and second coin, but the question is only about one coin in relation to the other. This is not the same as the first coin in relation to the second one.

It’s apparent that you won’t see the truth about this until you can free your mind from thinking that setting up the problem is the same as seeing the results. They absolutely are two different things.

As before: “… if you are walking down the street and see two pennies on the ground … if one is Heads the other one either is or is not also Heads.”

If you wish to discuss this further, please do so in reference to the above quote.
Hi davidh. Referring to your coins opn the ground; in that situation (one is heads), there is a 1/3 probability that the other one is also heads.

You say, “flipping the coins is a red herring and is immaterial to the results” - that statement beaggars belief. If you don’t flip the coins, then we don’t have heads or tails to discuss. However, you then go on to discuss two coins that have been pre-flipped, perhaps before the universe was created.

You then go on to show incorrectly assign probabilities to possibilities. I repeat that the logic you use is identical to saying that if you throw a dice, that it shows a 6 or it does not, and assert that throwing a 6 has a probability of a 1/2.

I see nothing in what I said that implies a first or second coin. Which coin do you deduce was the first one? (whichever you choose, I’ll say that it was actually the other one).
Hi again davidh. I’ve just re-read your post, 54. In your third paragraph you acknowledge that there are four possibilities. You don’t say what probability you assign to those four possibilities. I shall assume that you, correctly, assign 1/4 to each. i.e. I shall assume that you have accepted that P(HH) = P(HT) = P(TH) = P(TT) = 0.25. Then you have to, IMHO, accept that the relative probability of throwing HH (i.e. after ignoring trials which resulted in TT) is P(HH)/(P(HH) + P(HT) + P(TH)) = 0.25/(0.25 + 0.25 + 0.25) = 1/3.

Bear in mind that the question implies that we are only considering HH, HT and TH (and discarding TT).
Chris - in the paragraph you refer to where I acknowledge a four-way possibility of how the coins might land, I am referring to the event which I assert is immaterial to the answer. Read the paragraph immediately following that one.

Perhaps an experiment would help you to understand my reasoning.

What I propose is a double-blind experiment where the participants doing the experiment are ignorant of the purpose and are therefore not able to influence the outcome.

To do this, you will need, in addition to yourself, two other persons, call them A and B.

Person A will, outside the view of person B and yourself, take two coins out of his pocket and, without manipulation, lay them on a table.

Person B will then approach the table and view the coins. You, without seeing the coins, will ask person B whether (at least) one of the coins is showing Heads. If he answers “no” then there must be two Tails showing and you simply re-do the test until he answers “yes, at least one coin is Heads”.

Your task is then to predict whether there is just one Heads up coin, or two.

In this test, there is no way that you can logically say that there is a 2 in 3 probability (given there is at least one coin showing Heads) that the other coin is also Heads. The probability is 1 in 2 - 50%.

Of course, you don’t really need two other persons to do this test, just yourself and one other who would flip, drop or lay the two coins out and then tell you if one is Heads. A third person just separates the flipper from the you and reduces the chance of manipulation.
davidh. The “flipping” is just a way of asserting the the coins have a random face showing an that they haven’t been carefully chosen. In principle, someone could be choosing which way up to display the coins, and make a distribution which is not uniformly random. To say that flipping the coins is immaterial, demonstrates your complete failure to understand the problem. How did the coins get to be available for observation? If they came of a production line at the minting factory, they will all be the same way up.

You continue to fail to realise that the problem is merely dealing with a subset of four equally likely events: HH, HT, TH and TT. For that case, the probability of throwing HH is 1/4. By disregarding the case where TT has been thrown, we do not change the fact that HH, HT and TH are equally likely and the probability of throwing any one of them is 1 (i.e. you must throw HH, HT or TH). So the probability of throwing a particular one is 1/3. In particular the probability of throwing HH is 1/3.

Your penultimate paragraph is simply wrong (in two ways) - I say 1/3 for HH (not 2/3), and I have shown several times why I can say that.

How do you calculate 1/2? - you never show how you deduce that, you mereley assert it. I’m quite sure that you are saying to yourself that “we can have two heads or a head and a tail, so that’s two possibilities, so 1/2″. You spectacularly fail to understand that there is only one way to get two heads and that there are two ways to get a head and a tails, and that each way is equally likely.
Chris: you are a very patient man

davidh: what you describe in #57 is *exactly* the javascript demonstration here. I assume you haven’t read that?
Hi Stavros. I kind of commended you on your patience a little over a year ago, when some dick**** was failing to follow your argument about the spread of diseases (or similar) and presenting fallacious and biased arguments.

But my patience is geting rather thin. If only davidh actually tried the experiment (as I have done, to demonsrate it to one of my brothers), he’d know for sure that he is wrong. But sadly he’d rather philosophise.
You mean Chuck here? :-)

Ah, funny times…
Your mistake is in thinking that, after the flip that the coins still have the power to be either heads up or tails up. While that is true before filpping them, the fact is that after the flip, and the coins have landed they cannot change. What you will be presented with is two coins, both heads, both tails or one head and one tail.

If a friend picks two marbles from a box containing an equal number of white marbles and black marbles, and then tells you that at least one of the marbles he is holding is white, what is the probability that the other one is aslo white?

Before the pick you know that, since each marble can be either white or black, he can pick two white ones; or two black ones; or a white one and a black one; or a black one and a white one. In other words, before he picks, there are four combinations that he has equal posssibilities of picking from.

However, after he picks, he will be holding two marbles - either two white ones; or two black ones; or one black one and one white one.

If he tells you that one is white, it must be one of the pairs white-white or white-black. Since one is white, there is a 50% chance that the other one is white also.

Four ways to pick; three ways to present the results.

Back to the coins -
Before the flip, there are four ways that the coins

After the flip there is only one of three ways that they


Before the flip, there are probabilities and possibilities. After the flip there are only actualalities.
Hi Stavros. That’s the fella.

Hi davidh. I have no idea what your point is in your first paragraph. You seem to be saying that I’m in error in believing that the coins will land heads or tails, because they can only actually land heads or tails. The distinction is too subtle for me.

You said it: “After the flip there is only one of three ways that they have landed.” Well that’s not quite true, there’s still four ways that they could have landed; we simply ignore one of those ways (TT). However, now observe that those three remaining ways of interest (HH, HT and TH) are equally likely, and voila, HH is 1 in 3 as claimed.

Now what mistake do you think the coins make when I did the experiment with my brother? Perhaps they hadn’t been taught about davidh’s statistics.

The related marbles question cannot be answered without knowing how many marbles there are. If I knew the bag contained N black and N white marbles, then the probability of drawing two whites, given that at least one is white is (N-1)/(3N-1). e.g. if N = 1, then P(WW) = 0. e.g. if N is infinite, P(WW) = 1/3.

“If he tells you that one is white, it must be one of the pairs white-white or white-black. Since one is white, there is a 50% chance that the other one is white also”. Your fallacy is that he could be holding WW, WB or BW. You seem to be meaning that the first marble is white, and not allowing that the second could be the white one. Maybe that’s what you’re doing with the coins (despite being told that’s wrong several times).

That makes me think of another point: even though the usual probability of flipping two coins and getting 2 heads is 1/4 (do you accept that?), you feel that by only excluding only 1 in 4 of the possible outcomes, that the probability is doubled to 1/2. (It is actually increased by a factor of 4/3 to get 1/3).
Chris - the marble problem can be directly related to the coin problem given the understanding that there are an infinite number of black and white marbles equally divided in number between the two.

Now, we go through picking two marbles from the group. The two marbles can be both white (W-W); they can be both black (B-B); or they can be one of each (B-W). [Or, you say that the mixed combination could also be (W-B).]

If it is stated that one of the marbles is white we can eliminate the black-black combination and concentrate on the other two (or three, in your mind) combinations. These other two are white-white or white-black (or, if you insist, a third combintion of black-white)

Since one of the marbles is white, I contend that the other one can be either white or black, for a 50% possibility that both are white.

You are contending that if one marble is one color, there is only a 1 in 3 possibility that the other marble is also that same color. That would mean that you think it’s twice as likely that the second marble would be the opposite color.

By what logic to you explain how, when selecting two marbles from a universe of marbles of two colors, the second marble would be twice as likely to be the opposite color from the first?
Hi davidh. Marbles (infinite bag case). Let a helper pick the marbles one at a time (out of your sight). Your helper picks a first and

second marble. He could pick WW, WB, BW or BB (in that order); all being equally likely. On average, 3 out of 4 times he’ll be able

to say to you that he has at least one white marble. When that happens, you’ll know that he must have picked WW, WB or BW (in

that order), and that each is equally likely. So each case has a probability of 1/3. Therefore the probability that he has picked WW

is 1/3.

Now back to the coins. Let a helper flip the coins one at a time. He could have thrown HH, HT, TH or TT. Each being equally likely. 3

out of 4 times he’ll say that at least one of the cons is a heads (he’ll not mention TT, and will keep on flipping until he has at least

one head). So you know that he must have flipped HH, HT or TH, and each is equally likely. So both are heads 1/3 of the time.

Both of the above are equivalent (which I knew as soon as I saw your problem). Note that even though the helper has a sense of

first and second coin (or marble), you do not.

Anticipating some of your objections. It is impossible to flip two coins at exactly the same time (it is extremely unlikely that the

result of the flips have been determined within 1 nanosecond of each other, say). Even if the coins have been thrown separately,

both are looked at to determine if either (or both) are heads. It doesn’t matter if the first is a heads, the second is a heads or both

are heads - in each case all that matters is that at least one is a heads.

Nearly missed it. I am contending that if at least one of the marbles is W, the probability that the other is B is 2/3, because in WW,

WB, BW, 2 out of 3 of the combinations is of mixed colour.

What is a 50% possibility? (possibilities are alternatives, not percentages). That foggy thinking shows that you do not distinguish

betweeen possibilities and probabilities. I say there are 3 EQUALLY LIKELY possibilities (WW, WB, BW). You say there are only two

possibilities, WW and B+W, but fail to note that B+W is twice as likely as WW.

Some of the following may not be appropriate, but that’s because you don’t clearly define what you mean when you say WB is a

combination; I’m forced to try and guess what you mean.

You take exception to the BW possibility - what mechanism do you believe is preventing it? Why don’t you say that the

combination is BW and reject the WB combination with equal enthusiasm?

I don’t insist that BW is a combination, nature does. You are the one who’s insisting (without justification) that BW isn’t a

combination. I suspect that you mean WB to imply BW, but without appreciating the consequence that that makes your WB (which

I shall endeavaour to write as W+B from now on) be my WB and BW combined.

I also note that you have never used the phrase “equally likely” in these posts - that is a serious omission on your part and

demonstrates that you don’t usefully understand what probability is.

I note that you have never responded to my challenge regarding throwing a 6 with a die. There are two possibilities, you throw a 6 or you don’t. Do you believe that is a 50% “possibility” of throwing a 6? Whatever, I assume you’ll accept my belief that the probability of throwing a 6 is 1/6. I’d argue that for a fair die, that there are 6 equally likely numbers that could come up, (1,2,3,4,5, and 6), so there is a 1 in 6 probability of throwing a 6. If you used the same logic that you have been with the coins (and now the marbles), you would conclude that the probabilty (or is that possibility) is 50%.
A clarification: when I used the phrase “in that order” above, I was refererring to the marbles (or coins) as a pair (e.g. W first, B second), not to the sequence of four pairs.
Chris -

Believe it or not, I don’t disagree with your logic, or the rules of probability, in regards to how two coins being flipped might land (or, one coin being flipped twice) as long as they are flipped one at a time, but that is not the problem as postulated, which is that two coins are flipped simultaneously, as a pair. It’s true that each coin can land in one of two states but my point is that once they have landed, the ways that they might land in the future (probability) is no longer relevant since they have already landed (reality) and there are only three possible outcomes - a pair of heads, a pair of tails, or a pair consisting of one of each. Your confusion seems to be in talking about a first and a second coin when in fact you should be concentrating on a single pair of coins. That is why I switched to the marbles. They can be selected and displayed together and, once selected are fixed, no longer subject to the rules of probability.

If you reach into a bag of equally distributed black and white marbles and select two together (the same as flipping two coins and letting them land) then their fate, so to speak, is sealed. The pair of marbles you selected consists of two marbles, either both white, both black, or one of each color, with each combination being EQUALLY LIKELY (happy now?). There are no other combinations possibile. Of those three combinations, if one half of the pair is white (from the white pair or from the one-of-each pair) then there is only a 1 in 2 chance that the other one is also white (the other half of the white pair). You seem to be saying that if you picked two marbles one at a time then the second marble would more likely be the opposite color. This might be slightly true if there were only a small number of each - say ten white and ten black. In that insatance if you first picked a white one then there would be ten black and nine white left, giving you a slight advantage in picking a black one. However, in the infinite bag (remember, coin flip probabilities are based on infinite flips) then picking one color first give the opposite color no advantage of being picked second. Anyway, picking both together negates any advantage one color has over the other one being depleted by one (infinity-1 still equals infinity).

We could eliminate the problem of seeing two things separately as opposed to being a pair, and simplify this a bit more by using an infinite bag of marbles, 1/3 of which are black, 1/3 are white, and 1/3 are half white and half black. [The 1/3 all black represent the black-black pair; the 1/3 all white represent the white-white pair; and the 1/3 half black, half white represent the mixed color pair]. Before picking, the odds of picking an all white marble are one in three. Now, pick ONE marble and ask an independent observer if he sees any white. If he answers yes, then what is the chance that the marble is all white? Obviously 50%.

In addition to your thinking that two coins are separate where I say they are a single pair (or don’t you really see the difference in this?), our major point of dissention seems to be that you are conflating what might happen with what has already happened. I see these as two different states. For instance, in your die rolling statement, yes indeed, before you roll the die the chances of it landing six up are one in six. However, once it has been rolled and has settled. the odds of it landing six up no longer apply - it either is a six or it is not (and you don’t get 83% of your money back if you’re wrong). Whether it’s coins, marbles or dice, the odds apply only prior to the throw, but once the throw has been completed, it is what it is, probability no longer applies and reality takes over.

I’ve tried my best to try to see this from your point of view; I don’t think I’ve been given the same consideration. You haven’t convinced me that I’m wrong and I doubt that I’ll win you over to my side so maybe we should just end this useless back and forth.
Hi davidh. As you seriously believe that the probablity of drawing two white marbles from the bag is 1/3 (and equivalently that flipping two heads is 1/3), and you mean this to be the case before discarding the BB (or TT) result, I have no hope of convincing you of the result for the originally posted problem. In both case the probability is 1/4.

It has been suggested to you several times that you actually try the experiment. I did the experiment a few days ago (with real coins) after discussing the problem with one of my brothers. He also thought that the probability (for the original problem) was 1/2. He now accepts that it’s 1/3. It took less time to do than writing one of these posts.
“We could eliminate the problem of seeing two things separately as opposed to being a pair, and simplify this a bit more by using an infinite bag of marbles, 1/3 of which are black, 1/3 are white, and 1/3 are half white and half black. [The 1/3 all black represent the black-black pair; the 1/3 all white represent the white-white pair; and the 1/3 half black, half white represent the mixed color pair]. Before picking, the odds of picking an all white marble are one in three. Now, pick ONE marble and ask an independent observer if he sees any white. If he answers yes, then what is the chance that the marble is all white? Obviously 50%.”
davidh, seriously, just do the experiment or try the javascript code as suggested, and stop the armchair non-arguments…
I was about to post something that should deal with the exactly simultaneous drawing flipping that seems to be of significance to you, but saw your last post.

I’m stunned, I actually agree with your answer. My reasoning is essentially the same as for the original coin tossing experiment.

Using wb to represent the mixed white black marbles: As each type of ball can be selected with equal likelihood (and one of them must be selected) we have P(W) = P(B) = P(wb) = 1/3.

We know that some white has been seen, so we are left with the two equally likely possibilites that a W or a wb was selected, so the probability of selecting a W (given that we haven’t selected B) is 1/2.

How did you calculate 50%? Saying it is obvious is not good enough (I’m sorry, but my guts tell me that you got it right for the wrong reason).

I’ll be at least half an hour before dealing with the simultaneous draw scenario - clarifying that does make my brain hurt a little. But note that I have realistically dealt with it by appealing to physics, by saying that it isn’t possible to flip two coins (or pick two marbles) at EXACTLY the same time (in real life). i.e. there will be a time difference, no matter how small. I hope to show that that is an unnecessary appeal.
Hi davidh. I will start with some circumlocution. Consider a bag with N marbles in it. Focus attention on two particular ones, A and B. The probability of drawing A then B is 1/N * 1/(N-1) = P, say. The probability of drawing B then A is also 1/N * 1/(N-1) = P also. So for those situations, the order of drawing is immaterial.

Now let the time interval between the two picks become smaller and smaller. Let it be that you were about to pick two white balls, Wa and Wb. You are 1 femtosecond away from touching ball Wa and 2 femtoseconds away from touching marble Wb, the probability of having chosen two Ws is 1/4 - then out of the blue, the fart of a passing butterfly makes one of your fingers actually race ahead by 1 femtosecond and you’ve now selected both balls simultaneously. It’s pretty obvious to me that the probability of having chosen those balls hasn’t changed one iota. So why do you say it’s jumped from 1/4 to 1/3?

That wasn’t as good as I hoped for, but I’m sure that buiried in there must be something that should cause you to rethink. I also ended up rushing it as my dinner is overcooked.
Chris - first of all, maybe my terminology isn’t the same as yours but I’m pretty sure you know exactly what I mean when I say 50% - if there are two equally likely events then the probability, or likelihood, or chances, or odds of either occuring is 50%, or one in two, or 1/2, or 50-50, or however you might like to represent it. These words may have subtile nuance differences but they’re all interchangable to me.

Now that we have the one marble experiment done, let’s expand it to two marbles. In this instance, when you reach into the bag and close your hand around two marbles (forgive the metaphysics here) let’s say they fuse into a single dumbbell-shaped marble that is the combination of colors they exhibited separately. That is, either both ends white, both ends black, or ends of opposite colors. It doesn’t really matter which one touched your palm first, only that there were two that, at some point, touched both your palm and each other simultaneously. There are only three states that this dumbbell can exhibit - both ends white, both ends black, or one end white and the other end black. You can turn the black-white dumbbell over and around to switch your perception of the order of the colors, but that doesn’t change its reality of being a single unit that has two differently colored ends. (Even if you now cut the dumbbell in half and wind up with one balck and one white marble, there is no meaningful distinction as to whether it is black-white or white-black; it is simply a pair of two differently colored marbles.) We now have a two marble equilivant of the single marble trial. The same reasoning and logic that goes into the chances of a single marble pick ending in one of three states goes into the chances of a pair of marbles ending up in one of three states.

This all goes to my statement that when you are dealing with a simultaneous action on two things, you are actually dealing with a single action on a single, two component, pair, and the components of the pair no longer act independently of each other.

Picoseconds between grasping two marbles doesn’t matter. Once the marbles have been selected, the action is over and the only thing that matters is what you have in your hand - either two black, or two white, or one black and one white.
Hi David. I have no problem with you saying 50%, 1/2, 50-50 etc.; I agree that they are all equivalent. I’m not quite sure why you have mentioned it? The reason that I asked you to justify your calculation is because you simply said it is “obvious”. That phrase cause me severe worries, it usually means that the author cannot explain how the conclusion was obtained. Unfortunately I suspect that you concluded 50% because of the symmetry in the bag of marbles, and in the question proposed. I accept that because of the symmetry, concluding 50% is in fact reasonable - but it has the disadvantage of not using the “usual” probabilistic methods. You’ll notice that my calculation didn’t appeal to the 50-50 symmetry.

I kind of suspected where you were going with the wb marbles. I accept that there are only three types of dumbbell (just as I accept that you could draw WW, W+B, or BB). But for your analogy to be good (although tortuous), you must change your W, wb, BB bag to 25% white, 25% black and 50% white/black. I don’t expect that you’ll agree about that yet.

I can see that you really are being quite ingenious at coming up with these different attempts to illustrate your point. But, each time you do, you cause more and more things to have to be discussed.

In your last paragraph you mention that picoseconds don’t matter - that’s a good start. I know perfectly well that they don’t, and I wanted to make sure that you accepted that. The reason is that you are making a big thing about the simultaneous selection and the order of selection being relevant when they are manifestly irrelevant.

If I had been a bit more careful, I should have said call the coins A and B rather than the first and second coin. The A and B coins are totally peers to each other and could have been thrown A first, B first or simultaneously - it’s irrelevant, it doesn’t affect the probabilities one jot. As the example I gave was rather specialised, I’ll generalise it a bit:

Let there be m marbles of type M and n marbles of type N. The probability of selecting an M then an N type marble is m/(m+n) * n/(m+n-1). The probability of selecting an N type marble followed by an M type marble is n/(m+n) * m/(m+n-1) => the order of selecting doesn’t matter. I contend that if both were selected together, that the probability of soing so is also mn/((m+n)(m+n-1)).

I therefore say the simultaneous flipping shouldn’t be taken too literally.

Back to the original problem. But first, let’s be clear about what the probabilities are when simply flipping a couple of coins. Let the coins be A and B. The probability of flipping an H on A and an H on B is 1/2 * 1/2 = 1/4 (whether you flip A then B or B then A is irrelevant. The same for the TT case. To flip an H on A and a T on B (regardless of order of doing so) is also 1/4. The same for T on A and H on B. Taking XY refer to coin A then coin B, I say P(HH) = P(HT) = P(TH) = P(TT) = 1/4. As having a mixed HT (I’ll write this as H+T => HT or TH) P(H+T) = 1/4 + 1/4 = 1/2. I can make a logically identical argument for the infinite marbles case; but, as you already know, the two situations are equivalent, so I won’t bother.

Just in case (i.e. another way): I hope that you have no problem accepting that P(HH) = P(BB) = 1/4. As the only possible flips are HH, TT, H+T, and P(HH) + P(TT) + P(H+T) = 1, and P(HH) = P(TT) = 1/4, we must have that P(H+T) = 1 - (1/4 + 1/4) = 1/2. As P(HT) = P(TH), and P(H+T) = P(HT) + P(TH), we also deduce that P(HT) = P(TH) = 1/4.

I now hope that I’ve justified that P(HH) = P(HT) = P(TH) = P(TT) = 1/4 for the elementary coin flipping problem.

Finally, the original posed problem. We simply remove from the observed results of the above coin flip scenario, the TT case. The other case HH, HT and TH are not affected in any manner whatsoever by ignoring the TT flips. In particular, their relative probabilities are unaffected by ditching the TT flips. I’ll use P’ to imply the probabilities after discarding the TT flips. We have P’(HH) = P’(HT) = P’(TH), and their sum must = 1, so P’(whatever) = 1/3.

Note that although the question asks what is the probability of the other coin being an H given that one of them H, this is identical to saying what is the probability that both are heads given that at least one is.

I’m going for a fag and some coffee.
… Of course (assuming that you accept my last set of arguments); if one coin is a heads, then the probability of the other being a tails is P’(HT) + P’(TH) = 2/3. So it is twice as likely to be a tails rather than being a heads. Remember that the notation HT etc., is for coins A and B, and is independent of the order in which A and B are flipped. I’m really annoyed that I hadn’t defined it that way umpten posts back - but I never said or believed that I was perfect.

I believe that Richard Feynman made the follow remark in one of his public lectures: It doesn’t matter how smart a man is, or what his name is; if his theory is contradicted by experimental evidence, then he is wrong.

I’ve got a feeling that you may be coming round to seeing this problem my way (the correct and scientific/logical way). But don’t take my word for it, do the experiment and know for sure what the truth is. It really doesn’t take long to do a few hundred flips.
davidh. Let one coin be brown and the other be silver. I hope that you accept that that doesn’t materially alter the problem. I now use e.g. HT to indicate that the brown coin is a head, and that the silver is a tail. In this notation, the first letter always refers to the brown coin, and the second always refers to the silver coin. The four equally likely flips are HH, HT, TH, TT. The notation says nothing about where you make the observation from, or when the particular coin was flipped.

If you have thrown HT, walking to the other side of the table doesn’t convert that into a TH; the colour of the coins don’t change because you have moved. So in this notation, HT ≠ TH. Obviously I claim that P(HH) = P(HT) = P(TH) = P(TT) = 1/4.

You should be able to take it from there.
Chris, I think we’ve gotten a little too far into minutea here, what with coins, marbles, heads, tails, dumbbells, statistics, notations, A, B, HH, TT, etc.

Let’s go back to the original question:

“Flip two coins simultaneously. If one coin is heads, what is the probability that the other coin is also heads?”

Now, I hope that you can agree with me that the concept of “Heads” and “Tails” is relatively meaningless while the coins are still in the air. It isn’t until the two coins land that their final orientation can be observed.

I will fully agree with you that before the flip, and indeed, while the coins are still in the air, there are four ways that they can land IN RELATION TO EACH OTHRER.

However, the condition that they exhibit after having landed is fixed, and they will each display a face that is beyond further probability, possibility, chance, manipulation or any other factors.

So, if, of the two coins that have landed, we are told that one (without regard to which one it is) has landed heads up, then there is only one other coin to consider. The other coin (not the one we are told has landed heads up) can only be either heads, or tails.

Therefore, of the TWO possibile actualities (no longer probabilities) for the “other” coin (not the one we were told had landed heads up), only ONE possibile actuality is that it is heads, matching the coin we were first told was showing heads.


Two coins have been flipped.

Two coins have landed.

Two coins lie there, each showing a particular face.

We are told that one coin is heads up.

The other coin can not possibly be anything other than heads up or tails up.

Of those two equally likely possibilities, only if it is heads up does it match the coin we were told was heads up already.

ONE condition out of TWO possibilities: or “what is the probability that the other coin is also heads?”


Hi davidh. Although you have changed your mind a few times, I’ll refer you to your posts #34 and #42, where you acknowledge that the probability of throwing two heads is 1/4. If you believe that at the moment, then I expect that you’ll also accept that the probability of throwing two tails is also 1/4 (after all in Martian, fortuitously, “heads” is “tails” and “tails” is “heads”).

Consider an idealised experiment where we flip the coins 100 times. We wouldn’t be overly shocked to find that (approximately) 25 times we got HH, 25 times we got TT and for the remaining 50 times we got H+T. If you are currently of the belief that P(HH) = 1/3 (or whatever other value you randomly deem it to be), then I suggest that you do a Google search using the phrase “probability independent events”. You’ll find that P(HH) = P(H) * P(H) = 1/2 * 1/2 = 1/4.

Out of the 100 flips, we have about 75 which have at least one heads. For about 25 of them (HH), the other coin was also a heads, and for about 50 of them (H+T), the other coin was a tails. So, still confining ourselves to the 75 relevant flips: for the 25 HH we deduce that P(HH)= 25/75 = 1/3, and for the 50 T+H flips, we deduce that P(H+T) = 50/75 = 2/3. I’ve used one of the axioms/definitions of probability to do that (Google: probability relative frequency).

You might object to my example and say that such a perfect case, isn’t realsitic. I will counter that by saying that if you repeated the experiment many thousands of times and took the average result, or alternatively if you considered tossing the coins millions of times, that you’d find that the percentages would approximate very well to those ideal values. Those ideas are central to probability theory: they cannot be proven, but they are nearly self-evident, and so far, no experiment has proven them to be wrong.

Your call.
… having had a fag and a think, I’ll counter some of your anticipated response and bring this farce to a conclusion: Please forgive me Stavros. Obviously I’ll understand that you may delete it.

davidh, if you are a young person, then I apologise wholeheartedly for what follows as you haven’t had a chance to learn things properly. I respect your attempts at trying to see the problem another way.

I know that the coins haven’t landed until they’ve landed. I know that the state that they land in isn’t known until they have landed. I know that once they have landed they don’t spontaneously change state. I know that they must be considered as a pair. I know that they are independent and that they have no idea that they are involved in a pair flipping trial. I know that once they have landed, the state that they are actually in has probability 1. I assume that practically everybody knows those things; they are so obvious that almost nobody sees the need to mention them (yet alone several times, and attach such tremendous importance to them as you do) except perhaps in a text book for the absolute beginner.

You have been repeatedly advised to try an experimental determination of the result. This was in part to stop you from continuing to demonstrate what a colossal idiot you are. I can’t readily understand why you think you have the faintest clue about probability or logical reasoning. You are the paradigm of a fool (in particular you have no clue about your own limitations).

I have told you several times that all possibilities aren’t necessarily equally likely; you ignore me. You have asked me to give logical explanations of my statements; that’s what I’ve been doing all along: you simply aren’t capable of recognising it (or you haven’t bothered to read them). I don’t think you have even once given a logical explanation of your assertions.

I was half expecting that your response won’t take any notice of my post #78, but in view of my tone I don’t expect a response to it.

I’ll stick my neck out and assume that you might finally accept the result (fat chance, you brain drag factor is too high). Your punishment will be to re-read this page, pretend that you are me, and try to imagine what I may have been thinking about your comments.
Hi Stavros and everyone else who didn’t like my last post, I’m sorry about that, but I can only tolerate a fool for so long. If it’s deleted, I won’t be surprised or upset. But I hope that it’ll have made some of you laugh a bit. Let’s face it, it was long overdue.

In mitigation, I did a lot of editing before I posted it. It was MUCH stronger at one point.
Well Chris, since you have descended into name calling, I’ll assume that you have run out of intelligent ideas to present, so I’ll let this be the end.

To give you a small bit of my bio (not that it makes much difference now, or before, but just to let you know who and what I am), I am 71 years old, a graduate of Carnegie Institute of Technology (’61) major EE, minor Mathematics; and later Carnegie-Mellon University (’64) English and BA. I have been a member of both Mensa and the 99’s since 1957. None of this means I am any smarter than you or anyone else but I am a few steps above the fool you take me for.

You seem fixated on asking me to toss coins but you seem incapable of understanding that the question, as posed, asks about the coins AFTER the flip has been completed.

I don’t need to do that experiment since I already know that there are four ways they can land.

However, you are totally ignoring what happens next - that the coins are laying on the ground and someone determines that one of them shows heads and the other one either does, or does not.

Since you seem to be a fan of experiments, try this one - flip two coins; ask someone if one of them has landed heads up. If the answer is in the affirmative, ask what the other one is. I’ll bet that the answer will be heads approximately 50% of the time.


By the way, you’ve nicely sidestepped the single marble experiment I proposed. That was designed to produce the final result without the bothersome flipping. I guess it didn’t fit your model of probabilities.

By the way again, I did waste time setting up an Excel program to do a two coin flip 10,000 times and got the expected results of 1/4 H+H, 1/4 T+T and 1/2 mixed H+T/T+H. I also set up another program to examine two coins which had already been flipped 10,000 times and got the expected results of 1/3 T+T, 1/3 H+H and 1/3 mixed H+T.
Chris, I don’t censor anything unless it is ONLY abusive. You have tried…

davidh, oh dear:

“However, you are totally ignoring what happens next - that the coins are laying on the ground and someone determines that one of them shows heads and the other one either does, or does not”

No. If that makes you any happier, noone determines that the other “does, or does not”. All we know and *care* about is that at least one is.

In regards to your Excel programs, I don’t see how anyone could comprehend what you mean. Could you please share your code with us?

And for the grand finale:

“flip two coins; ask someone if one of them has landed heads up. If the answer is in the affirmative, ask what the other one is. I’ll bet that the answer will be heads approximately 50% of the time”

If you have actually done this experiment, you would see that you are on the wrong. And instead of “betting” you would *know* the answer, and save yourself from embarrassment: the other one is indeed heads approximately 33% of the time. It is never too late though. Go do the experiment yourself and get back to us. or try the code here, which I have repeatedly suggested to you.
Reposting (after severe pruning) as my previous attempts didn’t appear. Stavros, I assume that I’m being filtered. Please delete my last two posts (nominally numbers 83 and 84), and thanks for not deleting post #79.

Hi davidh. I see you finally did the experiment (post #81). OK so it was a program - that’s fine by me. It establishes the result that I claim and that you reject.

“By the way again, I did waste time setting up an Excel program to do a two coin flip 10,000 times and got the expected results of 1/4 H+H, 1/4 T+T and 1/2 mixed H+T/T+H.”

That’s what I expect (see my post #78). I immediately see that you are twice as likely to throw a mixed H+T as throwing a HH (as I claim and you scoff) and that the probability of throwing HH given that you have at least one head is 1/4/(1/4 + 1/2) = 1/3 (again as I claim, with multiple proofs, and that you assert should be 1/2).

Your second Excel program results are inconsistent with the first program’s result. Half of the H+T flips have simply disappeared. I think you have botched an attempt at fabricating results to support your beliefs (in this second case).
… I see that one got through, perhaps there was an illegal character buried in my last two tries, and the web software is goofing up. Stavros, I’m sorry that I implied that you were filtering me.
Chris, indeed your two previous attempts ended up directly on my spam folder (not even the moderation queue!). I don’t know why, but my software decided so. I can revive them but I think you have covered the same issues in your last comment.

Sorry, I may have to revisit my filtering software :-)
The true answer is found in the question. “Flip two coins simultaneously. If ONE COIN IS HEADS, what is the probability that the other coin is also heads?”

Since, as is stated, ONE coin is heads, then logically the other one is not.

The answer is ZERO.
Hehe, you have been quite low so far, but this is very amusingly lame :-)
Good one davidh, that’s the most intelligent observation that you’ve made on this site.

So much for your qualifications and supreme intelligence. I expect that they are as ethereal as your grasp on reality.
Having done a few probability problems on another website, I’ve just realised that some people might be getting this wrong for a fairly understandable reason. I wonder if they think that one of the coins being heads up is just an example of what might have happened, rather than being a condition used to reject certain actual outcomes.

Obviously with this particular problem, some are getting it wrong because they don’t read it properly.

And of course one in particular has cognitive issues ;)
And of course one in particular has cognitive issues

Chris, Chris, Chris, please don’t be so hard on yourself. I know it’s difficult to separate ideas that seem so like each other, but once you can get past the concept of being different from you’ll see the error of your ways.

When the coins are lying on the ground, the possibile ways that they might land is no longer relevant.

Before the toss, seen as two separate coins being tossed, they can land in one of four different ways: H-H; H-T; T-H or T-T.

After the toss, seen as a pair of coins on the ground (HH; HT or TT), with one showing heads(either the HT or the HH pair) the other one can only be either heads, or tails.

It simply no longer matters that the HT pair might be twice as likely to appear than the HH pair. All that matters is that one of the coins shows heads and the other coin of the pair is either heads, or it os tails. No other scenario exissts.


Since you and Stavros are fond of computer models, try a model where you have two coins, one being heads and the other being randomly either heads or tails. This, obviously, will describe the conditions given where two coins have been tossed and one shows heads. Logic and common sense will tell you that the odds of both being heads, if one is heads, will be 1/2. But if you want to model it for a million tries, be my guest and let me know the results.
For some reason the program here edited my first paragraph comment where I used brackets to set off part of the statement. Here is how it should have read:

Chris, Chris, Chris, please don’t be so hard on yourself. I know it’s difficult to separate ideas that seem so like each other, but once you can get past the concept of what might be being different from what is, you’ll see the error of your ways.
davidh, Merry Christmas to you. I see that you again (as in some of your post 34) have actually got some calculations right, ie. that you are twice as likey to have mixed HT as you are to have HH. Let the probability of flipping HH be P. Then the probability of flipping HT (either order) is 2P, so the relative probability of getting HH is P/(P + 2P) = 1/3. But you ignore that and assert that the probabilities don’t matter, I have no idea why you do so as the question is all about probabilities, and “calculate” 1/(1+1) = 1/2, using possibilities. Or are you “calculating” P/(P+P) = 1/2; it’s hard to tell because you (almost) never show how you calculate anything, you merely assert results, under the delusion that you have some deep understanding of this stuff.

Re your “computer challenge”. If one coin is double headed, then you are right; the probability is 1/2 that they will both be showing heads (given that at least one is showing heads) and also the probability is 1/2 that one will be showing tails. That should trouble you enormously; you have changed one of the coins significantly, so you should expect, a priori, to get a different probability from the one that you have been pushing for over a year, when both coins are regular HT ones. It’s manifestly obvious that both probabilities cannot be the same. Perhaps you could go on to argue that if changing one of the coins has no effect (and it doesn’t matter which one you changed), then you could change both of them to double-headed and still get a probability of 1/2 of getting a tails.

Now, using those HH and HT coins, try the following. You jumble them up in a bag and pull one out (no peeking) and place it on a table. Now look at the exposed face. If you see a head, what is the probability that the other side is a tail? No doubt you’ll assert that it’s 1/2 (with some nonsense about possibilities to “explain” it), even though it’s 1/3. I don’t expect you to understand the following explanation.

Let the double-headed coin have sides H and H’, and let the other coin have sides h and t (so each letter unambiguously defines a face). Then using the notation AB to imply exposed and hidden face, we can pull HH’, H’H, ht or th each being equally likely. But we cannot have pulled th, so that leave HH’, H’H and ht. So 1 out of 3 times the other side is a tail.

If I extend that problem to a bag with 99 double-headed coins and 1 normal coin. The probability that the other side is a tail, given that I saw a head is 1/199. No doubt you will continue to assert that the other side is a head or a tail and so the probability is 1/2 or 1/100 or even 1/200 (because you may throw in some ad-hoc probability effect) - again I can’t tell which you’ll deem to be the correct answer because you never show how you do your calculations. But whichever wrong value you state, don’t forget to add QED at the end of your argument (to make it become true).

Notes: You might say 1/2 because there’s only 2 possibilities. You might say 1/100 because that’s the probability of picking the HT coin (but you’ll ignire the fact that it could be showing the tail face). You kight say 1/200 bbecause that’s the probability of drawing the HT coin and it showing the head face.

I don’t expect that you’ll agree with any of my arguments because you have endowed yourself with godlike intelligence and being infallible, you can’t have made a mistake, so you can’t admit to having made a mistake.
Hi again davidh. A few randomly selected points for you to ponder.

Your post 67, para 3. Here you have 3 types of marble, B, W and 1/2 B + 1/2 W. You then correctly assert that if you pick one of them and see (at least some) white, that the probability that marble is completely white is 1/2. You didn’t say why that is so (I suspect that’s because you are incapable of actually calculating it as you only use intuition, badly). I then showed why it actually is 1/2 by using exactly the same type of argument that I used for the original problem - you didn’t complain that I had made a fallacious argument. Amazingly, you later said that I had side-stepped your argument - I still haven’t the faintest clue what you meant by that, as I had completely analysed the situation. What more did you want, what did I side-step?

You associated that marbles case with the original coins case. I’ll arbitrarily assume that you e.g. associated B with HH, W with TT and 1/2B + 1/2W with the mixed H+T case. There is now a problem: P(B) = P(W) = P(1/2B+1/2W) = 1/3, whereas P(HH) = P(TT) = 1/4 and P(H+T mixed) = 1/2. So, the marbles case isn’t the same as the coins case (despite yyour beliefs). That difference alone should be enough to let you know that you could be making a mistake somewhere. You cannot possibly be right in both of these these scenarios - so which one is wrong? (hint: the marbles one is right).

Several times you have suggested that I don’t understand that after the coins have actually landed they are in a definite state and the probability that they are actually in that state is 1. Amongst other things, what is being discussed is what is the probability that they got to be in that state in the first place. Sadly, you are aware of the difference, but not of its meaning.

Despite your pains in trying to convince Stavros and me that H+T and T+H are the same thing and that the coins are in some sort of Vulcan coin melded state, I see that in your post 90 that you now recognise that HT and TH are in fact different, whereas previously you insisted that they were the same thing viewed from the other side of the table. (I do like consistency, why don’t you?) On that topic, you didn’t respond to my post 76. How do you deal with the brown and silver coins not transposing their colours when you view them from the other side of the table? I’m not sure whether or not you were arguing that you are just as likely to throw a mixed H+T as an HH as you have a habit of writing ambiguous sentences. But you seemed to be implying that P(HH) = P(TT) = P(H+T mixed); if not, I don’t understand what your point was. Make your mind up - which is it to be (but not just for today, make it so forever). Your post 90 implies that you recognise that there are 4 equally likely outcomes (before discarding the TT case).

I’ll try to phrase the way of seeing the result differently. Using your post 90, where you seem to accept that (after discarding the TT results) that we have 3 possible outcomes (note that’s 3, not 2 possibilities), H-H, H-T and T-H, and all you know is that one of them has ocurred. As each ocurrence is equally likely and exactly one of them has ocurred, the probability for each case is 1/3. Therefore the probability of it being H-H is 1/3 (QED ;)). If you say that you can ignore the probabilities because the H-T + T-H is in some kind of melded state, then please tell me the probability of that state (sometimes you suggest it’s 1/2 and sometimes 1/3 - but it can only be one of them). Or if you say that’s irrelevant, then please provide a link to a (decent) website to confirm that that is the case (don’t try too hard on that, as you cannot possibly succeed).

My calculation is based on the relative frequency axiomatic definition of probability - Googling with that phrase will provide you with about 40,000 web pages. I can’t do that with your assertion as I don’t know what search terms I should use.

That’s enough for now.
Hi again davidh. I note that you didn’t (perhaps that should be “couldn’t”) explain why the probability for the double-headed coin version of the original problem is 1/2. You spluttered some waffle about logic and common sense - that’s no substitute for actual reasoning. Here’s a way to do it using the method that I’ve demonstrated to you several times:

As before, label the coin faces as H,H’ and h,t (so we can keep track of what’s going on). After (not to be confused with “before”) you flip the coins you will see one of: H&h, H’&h, H&t, H’&t. Each satisfies that at least one is heads, so we don’t disregard any results. In 2 of the 4 cases, we get that the other coin is a heads. In 2 of the 4 cases we get that the other coin is a tails. As each case is equally likely, the probability that the other coin is a heads is 2/4 = 1/2 and the probability of the other coin being a tails is, also, 2/4 = 1/2.

Why is a highly educated super-smart fellow, like yourself, unable to do this simple mathematical modelling?

Again, in case you forgot, that probability (1/2) is what you maintain that you get when both coins are regular head-tails coins. Perhaps that doesn’t trouble you, but if you could say something so incongruous existed with any of my conclusions, I would certainly look for where I had goofed.
Original problem:

“Flip two coins simultaneously. If one coin is heads, what is the probability that the other coin is also heads?”

Pardon my ignorance, but just exactly what does this mean? Does it mean that one of the coins you are flipping is already heads? Does it mean that one of the coins is pre-determined to land heads? Of course not.

From #89: “Obviously with this particular problem, some are getting it wrong because they don’t read it properly.”

Unless you have some other explanation, I read it to mean that two coins have been flipped and, while they are laying on the ground, at least one shows heads, with the problem being to determine the odds of the other one also showing heads.

Note that this is speaking of the coins after being flipped and landing into a settled state.

You say, “Amongst other things, what is being discussed is what is the probability that they got to be in that state in the first place. Sadly, you are aware of the difference, but not of its meaning.” In fact, this is not correct. The original question says nothing of how the coins might land, but only of how they have already landed.

You have attempted to “discuss” this using probabilty notation and a holier-that-thou snarkey attitude. The fact is that this whole thing can be solved with logic and a friendly discussion.

I am capable of doing a computer model, and have done so. Are you? If so, do this: model two coins on the ground, one showing heads and the other randomly either heads or tails. After a million or so trials, what results do you come up with?

Per the original question, all that matters is that one of the coins shows heads and the other coin of the pair is either heads, or it is tails. No other scenario exists.

If you can’t comment on the above paragraph, don’t bother responding.

I could, and would, admit it if I were wrong; it’s too bad you can’t do the same.

Happy Holidays to you.
Hi davidh, as I expected, you are quite unnable/unwilling to address any of the important points that I’ve made. You have now resorted to quibbling about the wording of the question; what is actually being asked has been made more than adequately clear. It’s a pathetic attempt to wriggle out of the fact that you have got the problem wrong and accused me (incorrectly) of being in error. Perhaps you have seen that your logic is inconsistent, and just aren’t man enough to admit it.

The meaning of flipping and landing is accessible to most people; you’re the only person I’ve come across who doesn’t understand this everyday idea. At any time whatsoever, I know what the probabilities of the two coins landing in any given configuration is - it’s very simple, and no philosophy is required.

I note that you did a computer model. The before discarding the TT cases you got it right (and were unnabe to see that’s all that you needed). Then you did another where you were supposed to discard the TT cases; but you retained them and discarded half of the H+T flips - that was hilarious. The computer modelling stuff was suggested to you as you the evidence was that you were unnable to solve the problem using your brain. But it also turns out you can’t even do the computer model right. I didn’t do a computer model, I did it with real coins to show my brother the facts.

Clearly you seem to believe that one of the coins is being constrained to land heads up - well that’s just being silly and a blatant disregard of reality; especially as you have shown that you accept that P(HH)= 1/4 etc.

I have also disposed of several of your side problems e.g. “simultaneous” flipping. I showed why that it doesn’t matter if one coin is flipped before the other - you act as if you hadn’t doubted it for a moment (of course a few picoseconds don’t matter” - quote from memory). Yet it was you who made a big deal of it - not me.

I have run rings around you and your pathetic reasoning. You are so stupid that you can’t recognise that I am far more intelligent than you. And more importantly, I have demonstrated the truth of the main result, and of several other problems that you have posed. You at best resort to arm-flailing, claiming some results are obvious, but not once have you actually shown the results using mathematical modelling.

Holier-than-thou - that’s the pot calling the kettle black.

Despite that lot, of course I hope you have an enjoyable Christmas and new year.
Come on guys give it a rest. Chris there’s no way you’ll ever convince davidh.

Davidh, such a simple probability teaser, you’re still unable to grasp after thorough explanations and experimental verification! What else you need is beyond me…

For crying out loud take a sample space of 100 flips, then on average 25 would be HH, 25 would be TT and the remaining 50 would be HT/TH. Discard the 25 TT. From the 75 flips left of which at least one is heads, how many are actually two heads?!?
Hi Stavros. davidh has already used his computer - but he is unable to understand the result.

I formally concede that he is beyond my ability to reason with and I shan’t try to help him to understand the problem any more. Besides I’ve finally become bored with talking to a brick wall - but I did enjoy making the attempt for the most part.

Merry Christmas.
Chris - you feel I have ignored the points you have made; I feel, also, that you have ignored the points that I have made.

You want me to do computer modeling and use probability notation. I contend that simple logic and reasoning are sufficient.

I have suggested simple experiments that would produce the answer to this whole thing and you have totally ignored them.

At no point have you offered a simple answer, in the English language, to the simple question - if one coin shows heads, what are the three states that you claim the second coin can have landed? I say (1/2) either heads or tails, you say (1/3) heads, tails or ? Heads, tails or tails? This is patently absurd.

For the final time, once the coins have landed, the possible ways that they could have landed is no longer relevant; the only definitive thing is how they are now laying on the ground. If one is heads, the other one either is, or it is not also heads.

Your descent into name calling and insults show that you have given up on the ability to use logical reasoning and are taking your ball and going home. So be it.
davidh. I have analysed and responded to nearly every point you have made and I have shown you several different ways of obtaining the solution.

I have also shown you how to calculate the solution to every problem that you have added in trying to explain your viewpoint to me. I’ve broadly used the same approach each time, you don’t seem to have registered that I get results that agree with your simpler (”obvious” result problems), and not see that my logic is impeccable.

You almost never seem to be actually responding to any point that I’ve made (hence the brick wall reference). I’ve repeatedly asked you to show how you calculate any of your results - you have not done so once. You accuse me of not responding to your posts - I can only conclude that you haven’t actually studied them (or unable to understand them). I’ll give you that you may well have run a computer simulation.

Yes, your result of 1/2 is patently absurd. You have yourself run a computer simulation that shows that (see your post 81 and my response 83). Again if the coins a brown and silver you can get brown heads + silver heads, brown heads + silver tails, brown tails + silver heads - that’s 3 states with equal probability. If you see it as 2 states, then I can’t help you; I’m not a psychiatrist. FYI there is no way to Vulcan meld those coins together.

You have suceeded in demonstrating to me that you do not understand probability, simple arguments or simple logic. So yes, I give up - you are a hopeless cause.
Hi davidh. I’m sorry, I rushed that last post. I’ll accept that it is reasonable to say that there are two states. But I won’t accept that you can ignore the fact that that state is twice as likely as the non-mixed HH state.

As I’m here again, we have a die, 5 of the sides are labelled with 1 dot, the last is labelled with 6 dots. I’ll side with you and say that there are two states that you can roll, a 1 or a 6. So using your assertion that the probabilities don’t matter, and your method of calculating I get the probability of throwing a 6 is 1/2 QED.

Can you see that I’ve used precisely the same argument that you’ve used with the coins? i.e. ignore the probabilities for each state and simply divide the number of states we’re interested in by the total possible number of states.

I would argue (using similar notation to that used with the coins) that there are 6 (equally likely) states, 1, 1′, 1″, 1′”, 1″”, 1′”" and 6. So now the probability of throwing a 6 is 1/(1+1+1+1+1+1) = 1/6. Or I might economise and simply say that you are 5 times as likely to throw a 1 as a 6 and say 1/(1+5) = 1/6.

That really is my last word (unless I change my mind).
Merrry Christmas everyone.

davidh. Re your post 99. You say I haven’t evene once explained why I say there are 3 states. Well check my posts 30,41,50,56,58,63,74 and 76. Also much discussion of that very topic permeates this page. I have also defended and commented throughout. You yourself have written these states down several times (so you have acknowledged them as being correct). I used H, H’ and brown and silver coins in order to establish the states and two reject you coin melding stuff (HT is TH from the other side of the table).

You have tried to argue that TH and HT are the same, well I’ve exploded that one. But I have allowed that description, but only if you accept that mixed state is twice as likely as the non-mixed states.

You say I haven’t explained why the probability is 1/3 - well that’s essentially all I’ve been doing - you must have overlooked most of my posts.

In your post 99, you mention that if one of the coins is heads, then the other is or isn’t heads. I’ve never contradicted you on that point - why do you imply that I have? In fact I have proven that the probability of the other one being heads is 1/3 many times.

As for doing the computer modelling. I think you suggested that once (post 90). Why do you want me to do that, that comment seems to imply that I was disputing your claim (even thoough I had agreed with it). Perhaps you neglected to read my post that agreed with you. On the other hand, Stavros and I had recommended that you try a computer model many times over an extended period before you did it. And we only did this because you kept on failing to obtain the correct result. So in this you are being unreasonable and childish.

The hilarious thing is that you did get it right (for a brief moment) in your post 31.
Once again (why do I torture myself like this?) - the following may be stupid in your mind, but the concept is central to my reasoning.

If you roll a die, the probability of a 6 coming up is 1 in 6. After the die has been rolled, if a 6 is showing, it is a 100% 6, not 1/6 of a 6. There is no longer a 5/6 probability that it is not a 6.

If you are jay-walking in a large city you may have a 1 in 1000 probability of getting hit by a car. However, if you successfully make the crossing, it’s not 999/1000 of you standing on the curb, you have beat the odds and made the 100% cross.

If you play the 3-digit lottery your chances of picking the winning number are 1 in 1000. After the number has been selected, if you have the winning number they pay you 100% of your winnings, not 1 one-thousandth.

The point of all this is what you do not want to see; the odds of what may or can happen are not the same as what did happen. Odds exist only before the event; certainty exists after the completion of the event.

There are four ways that two coins might land if filpped, but after the flip there is only one way that they have already landed. If one head is showing, it may be a head coin of the head-head pair, or it may be the head coin of one of the two head-tail pairs. But no matter which of these it may be, if one of the coins IS heads, the other one can only BE either heads, or tails.


You want to tout your intelligence and your superior understanding of probabilities, while denegrating my use of logic and reasoning. I have several Mensa-level puzzle books full of question and puzzles, each of which have a single answer. I’d gladly let you use your probability knowledge to “slove” each of the questions, while I use logic and reasoning. Which of us do you think would win that race?
That should be “solve”, not “slove” in the last paragraph.
Hi davidh. You didn’t show how you calculated 1/6 - you simply used a well known result. That’s not logic or reasoning (although it is sensible).

Here’s a simple question for you. Do you agree that if you flip two normal and fair coins, that the probability that they will land HH is 1/4, TT is 1/4 and mixed HT is 1/2? Obviously discard the cases where the coins land on their sides or any other such ducking and diving.

On the asumption that you agree, then you must accept that if we were to flip a pair of coins 4 times, then on average, we will get 1 case where we got HH, 1 where we got TT and 2 where we got a mixed HT.

Now back to the original problem. We know we haven’t got a TT. So we must have HH or mixed HT. In 1 case out of the remaining cases, we get HH and in 2 we get mixed HT. So the probability (relative frequency) of getting HH is 1/(1+2) = 1/3 QED.
davidh. Here’s one for you to try.
Someone throws a perfectly ordinary die, out of you sight. They tell you that it’s not showing 5. What is the probability that it’s showing a 6?
Hii davidh. To expedite the proceedings, my next set of questions is for the die scenario, but the roller will announce that it’s not a 5 or a 4. I’ll then go on to say he announces that it’s not a 5,4 or 3; and then in another scenario that it’s not a 5,4,3 or 2 and finally it’s not a 5,4,3,2 or 1. Obviously the roller might have to roll many times before he can make those announcements. In each case, what is the probability that a 6 has been rolled? Altogether, that’s 5 separate scenarios.

Please play with me a little longer and answer those 5 questions, this whole business can be put to rest in a few more posts.
Hi again davidh. Just in case you think I’ve ignored you, I agree with what you say in your last post (103) i.e. everything before your —-. I’ll ignore everything after the —-.

I don’t understand why you think that I disagree with what you say. To be clear, any individual trial ends with a particular result. I have never had any doubt about that. In the case of the coins problem, you aren’t given all of the information. You only know that one of the coins is heads, so the best you can do (to the best
of my knowledge) is to use probabilistic concepts to determine the state of the other coin.

In my mind the problem as given is a little false - after all if you flipped the two coins, you wouldn’t somehow only see that at least one of them is a heads, you’ll see both of them. But I understand the gist of what the question was about, and to make it more sensible, I semi-unconsciously reinterpret it as, “out of your sight, someone flips two coins and announces to you that at least one of them is a heads. What is the probability that the other one is a head also?”. At least that has become a plausible situation. I automatically consider what the outcome would be if the trial was repeated many times, with the understanding that the other guy will keep on flipping the coins until at least one is a heads.

While I may have got rather silly in some of my responses to you, at heart I’m really a nice bloke and just love some of these seemingly counter-intuitive problems. I also love the fact that such simple problems have so many different ways of being analysed.

The fact that I keep on corresponding with you is that I love the challenge and because I’m a sad git.

OK now I’ve been completely straightforward with you, please answer my die problems, or acknowledge the 1/3 result for my suggested interpretation of the original problem.
Hi davidh. I’ve been away for a week, and see you have been unable to answer my questions. Perhaps you’ve been away too.

I’ve just being refreshing my memory of where we had got to and noticed the following from your post 99: “At no point have you offered a simple answer, in the English language, to the simple question - if one coin shows heads, what are the three states that you claim the second coin can have landed? I say (1/2) either heads or tails, you say (1/3) heads, tails or ? Heads, tails or tails? This is patently absurd.” - I have never said that one coin can have three states. Is this a deliberate misrepresentation of what I’ve said? Whatever, the three states that I have been referring to are for the pair of coins, after discarding the TT case. To wit: HH, HT and TH. You also imply that you had asked me to provide that as an answer to a question of yours - where did you ask it?

The initial 4 states, HH,HT,TH,TT are known to anyone who has done more than an hour or two of a probability/statistics course, and it’s also evident that you are fully aware of it (see your own post 31).
OK, I’ll answer your question.

First, I’ve been playing with you all along. I’m a practical joker by heart and have been known to drag out a joke for several years.

I know full well that the probability of a second head appearing if one coin is heads is 33% (rounded). The paradox isn’t that the other coin can only be heads or tails but that the other coin is twice as likely to be tails than heads.

As for the die; if a thrown die is announced to not be a 5, the probability of it being a 6 is 33%. Not a 4 or 5, the probability is 50%. Not a 3, 4 or 5, 67%. Not a 2, 3, 4 or 5, 83%. Finally, not a 1, 2, 3, 4 or 5, 100%. (All % rounded to whole numbers).

For the record, in some of my previous responses, I used terms and expressed results for which you demanded an explanation. I speak English, as do you. The words I use have common meanings which are well known to the majority of speakers. You prefer to use probability notation; I prefer not to. The descriptive words I use convey the same meaning as the notations you use. I feel no compunction to have to explain what I say when I know that you fully inderstand what I am saying.

What’s your opinion of the Bertrand paradox?
Hi davidh. The die: the correct answers are 1/5 (for not a 5), 1/4 (for not a 4 or 5), 1/3 (for not a 3,4 or 5), 1/2 (for not a 2,3,4 or 5) and 1 (for not a 1,2,3,4 or 5). That’s 20% (not 33%), 25% (not 50%), 33% (not 67%), 50% (not 83%) and 100% (you got that one right).

e.g. for not a 4 or 5. We are left with 4 equally likely possibilities: 1,2,3 and 6. We must have one of them, so that’s a 25% probability for each case, and 25% for the 6 in particular.

I was going to introduce a tetrahedral die and associate 1 with HH, 2 with HT, 3 with TH and 4 with TT. I can see that I’d have got out of the frying pan and fallen into the fire.

I don’t know why you are so adverse to what you term my “probability notation”. It is extremely easy to understand, and far more succinct than long-winded descriptions.

Although I’d heard of Bertrand’s Paradox, I didn’t know what it was, so I looked it up. I’m surprised that it is famous; it’s manifestly obvious that the probability distribution must be specified. I have been fully aware of that since my teenage years (over 40 years ago).

Practical joker! Nice try, LOL. Anyway, you finally understood the original problem, so there’s a result.
Hi davidh, you’ll like this one.

You have 1025 coins in a bag. One of them is double-headed, the rest are normal. You pick one at random and flip it 10 times and get 10 heads. What is the probability that you had selected the double-headed coin?

Hint: it’s what you thought the answer was to the posted problem.

PS In the original problem, I had assumed that the coins weren’t biased. With the right bias, you would have been right. The bias would have been such that P(H) = 2/3 and P(T) = 1/3 (for identical coins). I expect that the coins would look very strange though.
…hmmm. Post 110 was sufficiently well written that I suspect that it may be from an impostor.

But #1. The strange sentence “I feel no compunction to have to explain what I say when I know that you fully inderstand what I am saying” is consistent with the real davidh. i.e. I can’t imagine why it was written. I have only asked for calculations to be done, not for the words to be explained. I’ll admit that I have said that a lot of what davidh has written is ambiguous and vague; thus causing me difficulty in nailing him down.

But #2: The incorrect results for the die are consistent with the real davidh, as is the failure to say what magical ritual was used to summon them up.

I’ll go for a probability of 1/3 that post 110 is from a fake davidh.
davidh has a good point in his post 86.

The question is badly written. It should be more like, “if you flip two coins simultaneously and at least one of the coins shows heads, what is the probability that both coins show heads?”

The original question implies (because it refers to “the other coin”) that you are looking at a particular coin and seeing that it’s showing a head. The other coin can be showing a head or a tail with a probability of 0.5 for each case.

Nice problem though.
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Hello everyone. I am a retired physics teacher from Kroatien and, after a brief ponderation, I must say I see no paradox at all. Postulating that one of the two thrown coins “IS” - therefore MUST be - Heads, is esentially EXCLUDING from the overall statistical results ALL Tails-Tails hits. Which, in turn, CUTS precisely a fourth off the Heads’ natural random probability count. It is tantamount to playing roulette with, say, Orphelines covered with some sort of reverberating belt that each time before touchdown hurls the ball back in the channel. It does increase the chances for the rest of the series, while supposedly, rhetorically, retaining its original 1:37 probability. And it is not a fair game! .)
CORRECTION: Cuts off NOT a fourth but a THIRD, of course!
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