I'm trying to solve an expected value problem where a biased coin is flipped until a run of five heads is achieved. I need to compute the $E(X)$ where $X$ is the number of tails expected before the run of five heads.
Would this require conditional expectation, since $E(X)$ is dependent on $P(Y)$ which is the probability of a run of five heads?
I know how to calculate the expected value of flips, but I'm pretty lost on counting the number of tails.
If $E(Y)$ is value $n$, then would I solve like so? $P(X = k \mid X E {n})$ $E(X) = P(X)E(Y)$
The following is a conditional expectation argument. We first deal with an unbiased coin, and then a biased coin. Let $e$ be the required expectation.
Unbiased Coin: If the first toss is a tail (probability $\frac{1}{2}$) then the expected number of tails is $1+e$.
If first toss is a head and the second is a tail (probability $\frac{1}{4}$, then the expected number of tails is $1+e$.
If first two tosses are head and the third is a tail, then the expected number of tails is $1+e$.
Same for first three heads, and fourth a tail.
Same for first four heads, and fifth a tail.
If first five tosses are heads, then expected number of tails is $0$.
Thus $$e=\frac{1}{2}(1+e)+\frac{1}{4}(1+e)+\cdots +\frac{1}{32}(1+e).$$ Solve for $e$.
Biased Coin: The same idea works for a biased coin. Let the probability of head be $p\ne 0$. Then the probability of tail is $1-p$, the probability of head followed by tail is $p(1-p)$, the probability of two heads followed by tail is $p^2(1-p)$, and so on. Thus $$e=(1-p)(1+e)+p(1-p)(1+e)+\cdots +p^4(1-p)(1+e).$$ Solve for $e$.