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Two players $A$ and $B$ play a series of games that ends when one of them has won $n$ games.

Suppose that each game played is, independently, won by player $A$ with probability $p$. Let $X$ be the number of games that are played. Compute the distribution of $X $ when $n = 2$ and $n = 3$.

How would one solve this?

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I give an answer for $n=2$. (The case when $n=3$ goes the same way.)

First of all $$P(X=0)=P(X=1)=P(X>3)=0.$$ What is the probability that $X=2$? That event takes place if we have $AA$ or $BB$. (Meening that $A$ wins twice or $B$ wins twice.) Then

$$P(X=2)=p^2+(1-p)^2.$$

If $X=3$ then there are the following possibilities $BAA, ABA, ABB, BAB$. That is,

$$P(X=3)=2p^2(1-p)+2p(1-p)^2.$$

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  • $\begingroup$ Note that the sum of these is $1$, so you could avoid the computation of $P(X=3)$ and subtract $P(X=2)$ from $1$ instead $\endgroup$ – Ross Millikan May 16 '19 at 16:14

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