What is the expected number of coin flips, if you stop when the first coin flip is the same as the last?

Question

In order to calculate the $\text{E}[X]$ where $X$ is the number of total coin flips, this is the approach I took:

The probabilities are:

$Pr(H) = p$

$Pr(T) = (1-p)$

Define indicator random variables:
$H_i$ = 1 if $i$'th flip is heads = 0 if $i$'th flip is tails

$T_i$ = 1 if $i$'th flip is tails = 0 if $i$'th flip is heads

Now, there are two possible events that could occur:

$A =\{\ H ...TTT...H \}\ ~~~~or~~~~~B=\{\ H ...TTT...H \}$

so $\text{E}[X] = Pr(A)\cdot \text{E}[A] + Pr(B)\cdot \text{E}[B]$.;

$\text{E}[A]=\sum_{k=1}^\infty Pr(H_i)\cdot \text{E}[H_{i}]=\sum_{k=1}^\infty pkp^k = p\sum_{k=1}^\infty kp^k$ $\text{E}[B]=\sum_{k=1}^\infty Pr(T_i)\cdot \text{E}[H_{i}]=\sum_{k=1}^\infty (1-p)k(1-p)^k =(1-p) \sum_{k=1}^\infty kp^k$

Is this right so far? Because I'm hinted that I should get and use the formula for $\sum_{k=1}^\infty kx^{k-1}$, but I don't understand why there should is a $k-1$ as an exponent. Where did the $-1$ come from?

Answer 1

A few days have passed, so it's probably okay to give bigger hints (in spite of the OP not really fixing up the question as requested).

1. You can never stop on the first flip. In fact it merely serves to identify which face counts as a "success" in later flips.

2. After the first toss, we're looking for the number of tosses to the first success. This is a standard problem which is solved in any number of texts (but which I imagine you've also covered in class or as an exercise).

So the success-defining toss in "1." adds 1 to the standard problem "what's the total number of tosses to the first success"; which itself can be thought of as 1 more than "what's the total number of failures to the first success?".