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You and I play the following game: I toss a coin repeatedly. The coin is unfair and $P(H) = p$. The game ends the first time that two consecutive heads $(HH)$ or two consecutive tails $(TT)$ are observed. I win if $(HH)$ is observed and you win if $(TT)$ is observed. Given that I won the game, find the probability that the first coin toss resulted in head.


So far my attempt is:

Let A be the event that I win the game. Using the law of total probability:

$P(A)$=$P(A|H)*P(H)+P(A|T)*P(T)$

Where the probability is conditionally split between the first toss being Heads and Tails.

Now:

$P(H)$= $p$

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

$P(A|H)=P({HH})+P({HTHH})....$=$p+p^2(1-p)$+...= ???

$P(A|T)=P({THH})+P({THTHH})....$=$p^2+p^3(1-p)$+...= ???

I am unsure of how to continue from here. Can someone help?

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I think you're on the right track!

Just one thing, notice that by Bayes:

$$P(H|A) = \frac{P(A|H)P(H)}{P(A)} = \frac{P(A|H)P(H)}{P(A|H)P(H) + P(A|T)P(T)}$$ Then from your calculations, you can notice that $P(A|T) = p \cdot P(A|H)$.

Now plugging things in $$P(H|A) = \frac{P(A|H)P(H)}{P(A|H)P(H) + p \cdot P(A|H)P(T)}$$ Notice you can cancel a lot of things and you should end up at a nice simple answer in terms of $p$

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There is a way to do it without using (sums of) sequences.


Let $\mu$ denote the probability that you win.

Let $\nu$ denote the probability that you win if the first toss results in a head.

Let $\rho$ denote the probability that you win if the first toss results in a tail.

Then we have the following equalities:

  • $\mu=p\nu+\left(1-p\right)\rho$
  • $\nu=p+\left(1-p\right)\rho$
  • $\rho=p\nu$

Solving this we find:

  • $\nu=\frac{p}{1-p\left(1-p\right)}$
  • $\rho=\frac{p^{2}}{1-p\left(1-p\right)}$
  • $\mu=\frac{p^{2}\left(2-p\right)}{1-p\left(1-p\right)}$

If $A$ denotes the event that you win and $H$ denotes the event that the first toss results in a head then under condition $p>0$ we find:$$P\left(H\mid A\right)=\frac{P\left(A\mid H\right)P\left(H\right)}{P\left(A\right)}=\frac{\nu p}{\mu}=\frac{1}{2-p}$$

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