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

…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).

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

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

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.