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A fair coin is tossed repeatedly and independently until two consecutive heads or two consecutive tails appear. What is the PMF of the number of tosses?

edit: in italic

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Let $X$ be the random variable that counts the number of tosses until we get two successive tosses of the same type. Although you did not say so explicitly, I assume that you want the probability mass function of $X$.

What is the probability that $X>n-1$? We need to have a sequence of tosses of length $n-1$ and of type HTHTHT and so on (alternating heads and tails, starting with head), or THTHTH and so on. The probability that the first type of sequence is followed up to and including the $(n-1)$-th toss is $\frac{1}{2^{n-1}}$. The second type of sequence has the same probability, for a total of $\frac{2}{2^{n-1}}$.

The number $X$ of tosses is equal to $n$ if $X>n-1$ and the last toss matches the next to last toss. The probability of that is $\frac{1}{2}$, so $$P(X=n) =\frac{2}{2^{n-1}}\cdot\frac{1}{2}.$$ This simplifies to $\dfrac{1}{2^{n-1}}$. Note that $n$ ranges over the integers that are $\ge 2$.

Another way: The first toss doesn't matter. After that, it is just a matter of matching the previous toss. So it is very much like tossing a fair coin and winning if you get (say) heads. The only difference is the "wasted" toss at the beginning. If we are tossing a fair coin, and $Y$ is the number of tosses until the first head, then $P(Y=n)=\frac{1}{2^n}$ ($n=1, 2, \dots$). But $X$ has the same distribution as $Y+1$. So $P(X=n+1)=P(Y=n)=\frac{1}{2^n}$, or equivalently $$P(X=n)=P(Y=n-1)=\frac{1}{2^{n-1}}\quad (n=2,3,\dots).$$

Generalization: Suppose that the probability of a head is $p$, and the probability of a tail is $q=1-p$. Either analysis of the problem can be extended to the more general situation. We win on the $n$-th toss if we did not win on the first $n-1$, and then the $n$-th toss matches the previous one.

It is convenient to split the analysis into two cases, $n$ odd and $n$ even. We do a full analysis of the odd case. So let $n=2m$. As is the first analysis, there are two ways to win on the $(2m+1)$-th toss. Either (i) we went HTHTHT and so on, with a tail on the $(2m)$-th toss, and then a tail again on the $(2m+1)$-th toss, or (ii) we went THTHTH and so on, with a head on the $(2m)$-th toss, and a head again on the $(2m+1)$-th toss.

The probability of (i) is $p^mq^m q$ and the probability of (ii) is $q^mp^mp$. Add them together. We get $p^mq^m(p+q)$, which is $p^mq^m$ ($m=1,2,\dots)$.

Next we deal with the case $n$ is even, say $n=2m$. The calculation is very similar to the previous one, so it will be omitted. The probability is the slightly more complicated-looking $p^mq^{m-1}p+q^mp^{m-1}q=p^{m-1}q^{m-1}(p^2+q^2)$, where $m$ ranges over the positive integers.

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Edited the question. Thanks for the detailed solution. Any hints on how to generalize if $P(H) = p$? –  anon19 Feb 24 '12 at 3:27
    
brian: you could try to adapt the steps of André's detailed solution to cover this case as well. The beginning is identical, since $[X\geqslant n]$ still corresponds to exactly two H/T sequences of length $n-1$. Now, you must compute the probability that each of these occurs and sum the two results... –  Did Feb 24 '12 at 6:45
    
@brian: I added a couple of things, including a generalization to the case $P(H)=p$. Same idea works, but it is a bit more complicated, and things are much smoother if we treat $n$ odd and $n$ even separately. –  André Nicolas Feb 24 '12 at 7:16
    
@André: Separating the even and odd case was the trick here, I guess. I was completely stuck thinking how to write a common expression for the PMF. Thanks! –  anon19 Feb 24 '12 at 8:08
    
@André: In your shorter solution, are you using a property like: if $X = f(Y)$, then $P(X = f(y)) = P(Y = y)$? If yes, why is it true? Is it true only for one-to-one functions? –  anon19 Feb 24 '12 at 8:18

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