0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – lulu Dec 2 '16 at 18:02
  • $\begingroup$ 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. $\endgroup$ – lulu Dec 2 '16 at 18:04
  • $\begingroup$ ok i will try that out, thanks alot :) $\endgroup$ – whatsupdoc Dec 2 '16 at 18:06
  • $\begingroup$ hi there, when do i stop doing the sum? $\endgroup$ – whatsupdoc Dec 2 '16 at 18:21
  • $\begingroup$ The posted solution, from @rlartiga , seems complete. The least number of games the winner will win is $n$, the most is $2n-1$. $\endgroup$ – lulu Dec 2 '16 at 18:34
3
$\begingroup$

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: $$0\leq m-n \leq n-1$$ $$n\leq m \leq 2n-1$$

So adding up:

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

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.

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

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.