How do I find the PMF of $X$ when $X$ equals number of flips of a fair coin that are required to observe the same face on consecutive flips?
The hint was to draw some sort of a tree diagram, but I'm not quite sure as to how it helps. I know that $P(X=x)=\sum(f(x))$, but I don't even know where to start for this particular question. Any help would be greatly appreciated!
Thank you!
Let $X$ be the number of tosses required. It is clear that $X$ must be $\ge 2$.
In order for $n$ to be the first place where we observe two consecutive identical flips, (i) heads and tails must alternate in the first $n-1$ tosses, and (ii) the $n$-th toss must match the $(n-1)$-th toss.
The probability of alternation in the first $n-1$ tosses is $2\cdot \frac{1}{2^{n-1}}$. For we can get alternation in $2$ ways: HTHTHT$\dots$ or THTHTH$\dots$. So our required probability is $2\cdot \frac{1}{2^{n-1}}\cdot\frac{1}{2}$, which is $\frac{1}{2^{n-1}}$.
Another way: After the first toss, the probability of "success" each time is $\frac{1}{2}$. Thus the number of tosses required is $1$ more than the number of trials until the first success, where the probability of success is $\frac{1}{2}$.
It follows that $X=1+Y$, where $Y$ has geometric distribution with parameter $\frac{1}{2}$. Thus $\Pr(X=n)=\Pr(Y=n-1)=\frac{1}{2^{n-1}}$.