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If we throw two fair coins then there are 4 equally probable possibilities: HH, TT, HT, TH.

Suppose we can't see the result, but we can check one of those two coins. (doesn't matter which one) Suppose the checked one is H. Then we know that both of them cannot be TT. So there are now only 3 possibilities: HH, HT, TH.

******And here is my mistake******

Therefore, the probability that the other coin in the pair is T (when we know that one of them is H) is 2/3.

Actually the probability is 1/2, because by checking one coin at random we remove only half of the cases in which they are different, so we are left with equal probabilities that both are same, and that they are different.

So there is no paradox.

Similarly, if we get T when we check one of those two coins, then the possibilities are: TT, HT, TH.

So, in this case the probability that the other coin in the pair is H is 2/3.

Let say that we have N such random pairs, and for each pair we check one coin and than place a bet on the state of the other. We would naturally always choose the opposite value for the other coin because, as demonstrated above, the probability that the other one will be opposite of the one we checked is always 2/3.

However, if those N pairs are truly random, then for large N there will be about N/2 pairs with the coins of same value (TT or HH), and about N/2 pairs with the coins of opposite values (TH or HT). And since we are effectively always betting that the coins have the opposite values, then this implies that we would win about half of the time.

How can that be? We are constantly betting on the results which have 2/3 probability, but we win only half of the times. What is wrong with my reasoning?

cod3r
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3 Answers3

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When you specify which of the two coins is heads (here the coin is specified as 'the one you looked at'), the other coin has a .5 chance of being tails.

For your situation to work, you need someone to tell you 'at least one of the coins is heads', whenever this is indeed the case. Then there is a $\frac23$ probability that one of the coins is tails.

paw88789
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  • So the other person in fact says "the configuration is not TT". Ok, this is the same thing I find out when I check one coin and find it to be H. What's the difference. (Why do I feel that this has something to do with quantum mechanics?) – cod3r Oct 02 '14 at 16:35
  • @cod3r I don't think hearing 'not TT ' is the same as you looking and seeing a head. In the first case, the remaining sample space has size 3 (HT, TH, HH); while in the latter case, the remaining sample space has size 2 (the state of the unseen coin). – paw88789 Oct 02 '14 at 16:44
  • Just a small addition. The other person must answer our question if she sees at least one type. That's important so we can 'measure' her 'measurement' in basis of our choice. Then we can discard the negative answer, and for positive answers we than have a $2/3$ probability that the coins are different. The other person may not decide what to say by herself because then either the probability is $1/2$, or she is biased in choosing what she is going to say. (will she prefer to say 'at least one head', or 'at least one tail', when able to) – cod3r Aug 30 '15 at 16:03
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Let say that we have N such random pairs, and for each pair we check one coin and than place a bet on the state of the other. We would naturally always choose the opposite value for the other coin because, as demonstrated above, the probability that the other one will be opposite of the one we checked is always 2/3.

$2/3$ is the probability that the coins are different when you know "at least one coin is a head".

However that is not the same thing as knowing "the coin looked at is a head", which is what you are really being told.

The coin is being tossed, and the person looks at either the left or the right coin, so there are actually eight equally probable outcomes. Underlining the coin looked at, these are: $$\Large \{\mathsf {\underbrace{\overbrace{\underline{H}H, H\underline{H}, \underline{H}T, T\underline{H}}^{\text{the coin looked at is a head}}, H\underline{T}, \underline{T}H}_{\text{at least one coin is a head}}, \underline{T}T, T\underline{T}}\}$$

So the coins will be different on only $1/2$ of the times the person tells you they looked at a head.

Now, if we change the set up so that instead of looking at just one coin the other person looks at both coins and tells you "at least one coin is $\underline{\quad}$".

Suppose further that when the coins are different the other person guarantees to say "heads" or "tails" without bias.   So when you hear "at least one coin is a heads" you know you will know the probability that the coins are different is only $1/2$ -- because on the $1/4$ chance they are the same you will always hear the words, but on the $1/2$ probability that they are different you will only hear those words $1/2$ the time.   So your expected chance of being right by guessing opposite of what you're told is: $(1/2)(1/2)+(1/2)(1/2)=1/2$

Suppose instead that when the coins are different the other person is biased to always say "at least one coin is a head".   Then there will be a $2/3$ probability that the coins are different when you hear those words, and you will hear those words $3/4$ of the time.   However, there will also be a $0$ probability that they are different when you hear "at least one coin is a tail", and you will hear that $1/4$ of the time.   So your expected chance to be right by guessing opposite of what you're told is: $(3/4)(2/3)+(1/4)(0) = 1/2$

You could improve your odds by noticing such a bias and changing your strategy accordingly -- if the other person didn't change their bias in response.

Thus the apparent paradox is avoided.

Graham Kemp
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  • But is it possible to know that "at least one coin is a head" without looking? I don't see how this 2/3 probability can ever be used. – cod3r Oct 08 '14 at 21:20
  • @cod3r That's the very point. Unless the other person looks at both coins and tells you "at least one coin is a head/tail" you don't have that information. The way the trials are set up the other person looks at just one coin and tells what it is, so all you know is "one coin is a head/tail" and that is not the same thing. – Graham Kemp Oct 08 '14 at 23:09
  • I still don't see how my odds can depend on the way the other person makes her "measurement" since the information she gives me is exactly the same, and I bet solely on the information passed. In my original question I indeed misunderstood what I was doing by looking at one coin myself, but now that the other person is involved I see a different problem. (continued in my next comment... sorry this will be several comments long because of the limits) – cod3r Oct 09 '14 at 06:50
  • I understand that if she looks at only one coin and tells me the info (at least one is...), then this can't give me the edge because then it would give her the edge too (she could bet exactly as I), and as was explained by you and others the odds are not better after looking at one coin. However, I'm afraid that I don't see how I can get better odds if other person looks at both coins before passing the info. I'll explain in my next two comments... – cod3r Oct 09 '14 at 06:50
  • Other person looks at both coins in $n$ experiments, $n$ is very large, so about $n/4$ HH, $n/4$ TT, $n/2$ HT or TH. For those $n/2$ cases she must choose what to tell me. If she chooses to tell half of the times at least one is H, and other half of the times T, then I can never surpass $1/2$ probability of winning. As before (according to $2/3$ probability) I should always bet to the opposite, and again, I will win about $n/2$ times. – cod3r Oct 09 '14 at 06:51
  • However, if I suspect that for HT or TH she always chooses to tell the same, for example at least one is H, then I can experiment. For some time I can try to bet opposite for H and same for T. After some time I'll notice that I win with $3/4$ probability. (or if she always chooses to say T then with $1/4$ probability) So my point is, unless she is biased I can't make the use of the info she passes. Moreover, how can clear $2/3$ probability end up in $1/2$ wins? What am I missing? – cod3r Oct 09 '14 at 06:54
  • @cod3r Nothing much. You are making good points. I've edited my answer to include a change of set up. – Graham Kemp Oct 09 '14 at 21:00
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I think regardless if you view the first coin or not, the 2nd coin has a 50% chance of being a tail. Why would that ever change?

Also, how would your experiment be any different than just tossing one coin first, seeing that it is heads, and then tossing the 2nd coin? Would you say if the first coin in the individual coin toss experiment is heads that the 2nd toss will then have a 75% chance of being tails? Those two coin tosses, whether done at the same time or sequentially are independent of each other.

There are only 4 heads you can see, namely HH, HT, and TH. In the first case, if either of those heads are seen, then other coin is also a H. In the HT and TH cases, if an H is seen, the other coin flip is a T. So if you see an H for one coin, there is a 2 out of 4 chance the other coin will be an H and a 2 out of 4 chance the other coin will be a T.

Here are the possible cases where we see a H on the one coin checked:

HH (we see the left H)
HH (we see the right H)
TH
HT

Upon seeing a H for one coin, there are 2 cases where the other coin is an H and 2 cases where the other coin is a T.

Your initial statement that if a H is seen then there are only 3 possibilities (HH, HT, and TH) is correct but they are not equiprobable because you could see the H in the HH case twice as often as the others, thus making the effective outcomes HH, HH, HT, and TH. The HH case is "doubly weighted" since no matter what H we see, the other one is also an H and there are 2 ways to do that.

I don't see any paradox.

David
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  • I think it is different because we are checking random coin from a pair, as you suspected. I must say that I don't quite understand why in that case HT and TH would count only once. – cod3r Oct 02 '14 at 16:30
  • Thanks, I get it. Btw, your "HT and TH would count only once" which you later deleted from your answer, reminded me of Bose and his famous "mistake". (http://www.isical.ac.in/~econophys/bose.html) – cod3r Oct 02 '14 at 17:07
  • I was thinking "relative" to HH but I deleted that comment because it was confusing and I instead worded my answer to "double weight" HH so it was like 2 ways to say something similar. That is, if we count HH twice (which I did), then we can now count both TH and HT as possible outcomes to make things "fair". – David Oct 02 '14 at 17:57