I am a computer scientist also interested in the physical sciences, pseudoscience, and the borders of religion and science. More about me here. More about this blog here. You can contact me here.
TOP POST
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.
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.
1. Adal
50-50, it is always 50-50: heads or tails!
2. Adal
I mean regardless of what the first coin is…
3. Jess
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
4. Anonymous
… 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!
5. Stavros
…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%!
6. Stavros
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!
7. dave
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.
9. Mirk
Absolutely it is not a Paradox
10. Sezgin
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.
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.
11. Stavros
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%
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%
12. Laz
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.
13. Stavros
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.
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.
14. Anonymous
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
…or
- 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).
It seems to me there are two possible ways of looking at the question:
- I know that there is a coin which is heads
…or
- 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).
15. Laz
Just to let you know who sent the previous post - I’m not totally anonymous.
16. David
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.
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
TH
HH
HT
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
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
TH
HH
HT
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
18. Stavros
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.
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.
19. Gianna
Hi guys!!! I have a “nice question”: The probability that 4 coins will come head up when flipped simlutaneously is…
20. Stavros
1/2 x 1/2 x 1/2 x 1/2 = (1/2)^4 = 1/16
21. Gianna
I’m sorry but I don’t get it!
22. Stavros
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.
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.
23. Gianna
ok…thank you!
24. ajay
whats the parodox . no confusion here it is 1/2
25. Chris
ajay, you are confused. The probability is 1/3 not a half. Read the question. You have answered a question that wasn’t posted.
26. davidh
“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.
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.
27. Chris
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.
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.
28. Chris
… 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.
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.
29. davidh
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
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
30. Chris
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 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
31. davidh
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.
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.
32. Chris
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.
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.
33. Chris
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.
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.
34. davidh
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
might
wind up as opposed to how it
does
finally wind up.
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
might
wind up as opposed to how it
does
finally wind up.
35. Chris
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.
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.
36. Stavros
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”
Cheers
Cheers
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..!!
(http://ragsoverriches.blogspot.com/2009/10/un-lightenment.html)
And great that you’ve now provided an “Experiment”—in the true Spirit of Science..as Feynman himself would have loved..!
Talk of Probability (or Serendipity..!!)..!
Cheers..!!
“raghu”
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..!!
(http://ragsoverriches.blogspot.com/2009/10/un-lightenment.html)
And great that you’ve now provided an “Experiment”—in the true Spirit of Science..as Feynman himself would have loved..!
Talk of Probability (or Serendipity..!!)..!
Cheers..!!
“raghu”
38. Lucas
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
39. davidh
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.
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.
40. Stavros
If not convinced, please look at the experimental observations here
41. Chris
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.
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.
42. davidh
Chris,
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.
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.
43. Stavros
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.
44. Chris
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.
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.
45. Chris
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.
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.
46. PassingThru
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.
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.
47. Chris
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.
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.
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