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**lllll** A and B play a series of game, the probability of A wining is $\displaystyle p$ and the probability of B wining is $\displaystyle 1-p$. The winner of the series is the player that wins 2 more games.

**A)** Find the probability that A is the final winner.

**B)** Compute the expectation of games played.

**A)** in order for A to be the winner, one of these outcomes would have to occur:

$\displaystyle AA = p^2$

$\displaystyle ABA = (p^2)(1-p)$

$\displaystyle BAA = (p^2)(1-p)$

I think I would have to sum up the values, which would give:

$\displaystyle p^2+2(p^2)(1-p)= p^2(3-2p)$

**B)** I would think that would be around 3, considering that there are a total of 6 possible outcomes, I would think it would look something like this:

$\displaystyle E[N] = E[N=2|Y=1]P(Y=1)+E[N=3|Y=1]P(Y=1)$+$\displaystyle E[N=2|Y=0]P(Y=0)+E[N=3|Y=0]P(Y=0)$

$\displaystyle E[N] = E[N=2|Y=1]p^2+E[N=3|Y=1](p^2(3-p)$+$\displaystyle E[N=2|Y=0](1-p)^2+E[N=3|Y=0](1-p)^2(1+2p)$