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?


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


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


  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.