# winning the match (which is first to win n games), if i know the probability of them winning a game?

Say i have two players who each have a certain probability of winning a game, for example for player $1$: $p_1=0.8$ and player $2$: $p_2=0.2$. There are $n$ games in a match, what is the probability of player $1$ being the first to win $n$ games,and therefore win the match? Thanks.

• Hint: Suppose they play out $2n-1$ games (just as the World Series has $7$, though usually they aren't all played). That is, suppose they play all those game regardless of whether or not a winner is settled. Clearly exactly one side will have won $n$ or more games, so the winner is unchanged. – lulu Dec 2 '16 at 18:02
• To be clear: I don't believe there is a simple closed formula for the answer. Just sum up the probability that $1$ wins exactly $n$, exactly $n+1$, and so on. If $n$ is big enough you can use a normal approximation, but if $n$ is small (as it usually is in this context) then you just have to do the sum. – lulu Dec 2 '16 at 18:04
• ok i will try that out, thanks alot :) – whatsupdoc Dec 2 '16 at 18:06
• hi there, when do i stop doing the sum? – whatsupdoc Dec 2 '16 at 18:21
• The posted solution, from @rlartiga , seems complete. The least number of games the winner will win is $n$, the most is $2n-1$. – lulu Dec 2 '16 at 18:34

Calculus for a fixed number of games

If both play a total of $m$ games, then $n$ games must be won for the first player with probability $p$ each one, being them independent: $$p^n$$ The second player must won $m-n$ games. That with probability $1-p$ each one. $$(1-p)^{m-n}$$ Finally to conclude the game the last one must be won by the first player, so we must choose from $m-1$ games which $n-1$ to be won. $${m-1 \choose n-1}$$ So the probability is: $$P(x=m)={m-1 \choose n-1} p^n(1-p)^{m-n}$$

Calculus for the total of games:

The constraint in the number of games is that the second player must win no game or at most $n-1$ so: $$0\leq m-n \leq n-1$$ $$n\leq m \leq 2n-1$$

$$\sum_{k=n}^{2n-1}P(x=k)=\sum_{k=n}^{2n-1} {k-1 \choose n-1} p^n(1-p)^{k-n}=\sum_{k=0}^{n-1} {k+n-1 \choose n-1} p^n(1-p)^{k}$$

If $n=2$:

$$\sum_{k=2}^{3}P(x=k)=\sum_{k=2}^{3} {k-1 \choose 1} p^{2}(1-p)^{k-2}$$ $$=\sum_{k=2}^{3} (k-1) p^{2}(1-p)^{k-2}=(3-2p)p^2$$

As you can see that is not greater than one.

• ok i will try that out :) thanks – whatsupdoc Dec 2 '16 at 18:21
• out of interest what is this method called? – whatsupdoc Dec 2 '16 at 18:23
• @whatsupdoc It looks like a Binomial Distribution. – AlgorithmsX Dec 3 '16 at 3:10
• i'm not sure how this would show which player wins first, could someone explain to me please? :) – whatsupdoc Dec 3 '16 at 20:23
• @whatsupdoc here goes the explanation in the answer. – rlartiga Dec 5 '16 at 16:41