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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? $P(X = k \mid X E {n})$ $E(X) = P(X)E(Y)$

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  • $\begingroup$ Can you write out the probability that $X = 0$? Or $X = 5$? Or arbitrary $X$? $\endgroup$ – TMM Jan 27 '14 at 0:18
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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 $e$.

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  • $\begingroup$ so in the case of a biased coin, e = (1/30)(1 + e) + (1/30)^2(1 + e) + ....+ (1/30)^5(1 + e)? $\endgroup$ – user2635779 Jan 27 '14 at 0:54
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Let $p$ be the chance of a head on a single toss of the coin. Every time you try to throw 5 heads in a row, you either succeed or fail by getting a tail "too soon".

$$\underline{HHT}\ \ \underline{T}\ \ \underline{HT}\cdots\underline{HHT}\ \ \underbrace{\underline{HHHHH}}_{\mbox{success!}} $$

The number of tails $X$ observed is the same as the number of failures before the first success, so $X$ has a geometric distribution with $\mathbb{P}(\mbox{success})=p^5$. Therefore $\mathbb{E}(X)={1\over p^5}-1.$

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