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A biased coin has probability p of showing heads. We toss the coin repeatedly until two heads or two tails occur in row. What is the probability that two heads occur before two tails?

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Consider the four states "HH", "HT", "TH", "TT", representing always the results of the last two tosses. You want the probability of the event $A$ that "HH" occurs before "TT".

After the first two tosses you are in each of these with probabilities $p^2$, $pq$, $pq$ and $q^2$ respectively, where $q=1-p$. Of course if it's "HH" your event does occur, and if it's "TT" it doesn't. So

$$\mathbb P(A) = p^2 + pq \; \mathbb P(A \mid HT) + pq \; \mathbb P(A \mid TH) $$

Given you're in the state "HT", with probability $p$ the next toss is heads, leaving you in the state "TH", while with probability $q$ it's tails, leaving you in "TT", where the event A does not occur. Thus

$$ \mathbb P(A \mid HT) = p \; \mathbb P(A \mid TH) $$

Similarly, $$\mathbb P(A \mid TH) = p + q\; \mathbb P(A \mid HT) $$

Solve this pair of equations for $\mathbb P(A \mid HT)$ and $\mathbb P(A \mid TH)$ ...

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Suppose you do it right off the bat. (in 2 filps)

$P(hh) = p^2$

in 3 flips? $P(thh) = p^2(1-p)$ (this is the only way to win in exactly 3 flips)

4 flips? $P(hthh) = p^3(1-p)$ (this is the only way to win in exactly 4 flips)

In order to win in $n+2$ flips, the first $n$ flips must alternate $hththththth$

$2k$ flips $= p^{k+1}(1-p)^{k-1}$

$2k+1$ flips $p^{k+1}(1-p)^{k}$

$\sum_\limits{k=1}^{\infty} p^{k+1}[(1-p)^{k-1}+(1-p)^k]$

$\frac {p^2(2-p)}{1-p(1-p)}$

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