A friend of mine tossed a fair coin twice. Suppose I ask him whether he got a head in the two tosses, and he says yes. What is the probability that one toss is tail?

Now suppose instead that I happen to see the result of one of his tosses, and it is a head. What is the probability that the other toss is tail?

  • $\begingroup$ Well, you know his sequence of tosses was one of $HH$, $HT$, $TH$ and that these are equally likely outcomes. $\endgroup$ – David Mitra Feb 10 '13 at 17:18
  • $\begingroup$ Well, isn't the prior probability of two heads or tails 0.25 and one H and one T 0.5? $\endgroup$ – John Feb 10 '13 at 17:21
  • $\begingroup$ @ZevChonoles This question is not the same as the other question by the same OP. The answer to this question is 2/3, whereas the answer to the other question is 1/2. $\endgroup$ – Per Manne Feb 10 '13 at 18:05
  • $\begingroup$ @PerManne: Thank you for catching that, I was responding to a flag and the questions seemed quite similar at first glance. I've edited the newer question into this one, since they are related closely enough that it is better for them to appear in the same question anyway. $\endgroup$ – Zev Chonoles Feb 10 '13 at 18:16
  • $\begingroup$ @John: Sorry for assuming that you'd simply reposted the same question. I've edited your new question into this post. $\endgroup$ – Zev Chonoles Feb 10 '13 at 18:16

The friend says yes if he got one head or two.

When a coin is tossed twice, the following $4$ outcomes are equally likely: HH, HT, TH, TT. For $3$ of these outcomes, the friend says yes. In $2$ of these outcomes, there is a tail. So the probability that there is a tail given that there is at least one head is $\dfrac{2}{3}$.

This sort of semi-formal argument can be treacherous. So let's do it more formally. Let $B$ be the event there is at least $1$ head, and let $A$ be the event there is a tail. We want $\Pr(A|B)$ (the probability of $A$ given $B$).

By a standard formula, $$\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}.$$

The probability of $B$ is $\dfrac{3}{4}$.

The probability of $A\cap B$ is $\dfrac{2}{4}$. Divide.

Added: A second part has been added. This one requires interpretation. Answers to similar problems can be quite interpretation-dependent.

Suppose that with probability $\frac{1}{2}$ we get to have a peek at the result of toss $1$, and with probability $\frac{1}{2}$ we get to see the result of toss $2$. Let $S$ be the event we see a head, and let $T$ be the event there is a tail.

We want $\Pr(T|S)$. The computation is much like the one in the answer to the first question. After examination of cases, we find that $\Pr(S)=\frac{1}{2}$ and $\Pr(S\cap T)=\frac{1}{4}$. From that we conclude that $\Pr(T|S)=\frac{1}{2}$.

  • $\begingroup$ There's been an update to the question. $\endgroup$ – Zev Chonoles Feb 10 '13 at 18:17

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