# Best of $2n+1$ game

Two teams are playing a best of $2n+1$ game: the team who has won $n+1$ games (as first) is the winner. Both teams have a probability to win of 50%. However, Team 1 has made an agreement with their sponsor that they only have to win $n$ games, to be the winner (team 2 still has to win $n+1$ games). What is the probability that Team 1 will be the winner?

• Did your textbook specify if the order matters or not? In my answer, I assumed that for any case, all orders (of teams winning) are equivalent since they all result in winning of a match for team A, which is desired. If order does matter (specified in book?), we need to specify additional parameters for ordering of teams. – Gaurang Tandon Mar 9 '17 at 9:50
• @GaurangTandon The words "as first" indicate that the order matters. – drhab Mar 9 '17 at 10:05
• @drhab Could be. It seems like I misunderstood the question. Thanks! – Gaurang Tandon Mar 9 '17 at 11:59
• In many cases the teams stop playing if one team wins $n+1$ games. Does that happen here? In that case the only advantage to team 1 is if the series gets tied at $n-n$ and they lose the last game. – Ross Millikan May 16 '18 at 23:52

Let the teams play all $2n+1$ games (also if there is a winner earlyer).

Then team 1 will win the match iff team 1 will win at least $n$ of the first $2n$ games.

Probability on that:$$p:=2^{-2n}\sum_{k=n}^{2n}\binom{2n}{k}$$

Also observe that: $$2^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}=2\sum_{k=n}^{2n}\binom{2n}{k}-\binom{2n}{n}$$

This leads to:$$p=\frac12\left(1+2^{-2n}\binom{2n}{n}\right)=0.5+2^{-2n-1}\binom{2n}{n}$$

Other approach:

Team1 will win according to the modified rules iff one of the following disjoint events occurs:

• team 1 wins if the original rules are applied.
• team 2 wins if the original rules are applied, but looses if the modified rules are applied.

The probability of the first event is $0.5$ by symmetry. The second event occurs iff after $2n$ games the score is $n-n$ and team 2 wins the last match. The probability on that is $$q:=2^{-2n-1}\binom{2n}{n}$$

Then (again) we find $$p=0.5+q=0.5+2^{-2n-1}\binom{2n}{n}$$

Hint: Team A will be the winner if it wins:

1. first $n$ consecutive games
2. $n-1$ games and B wins one game, then A wins the last one (why?)
3. $n-1$ games and B wins two games, then A wins the last one (why?)

last case: $n-1$ games and B wins $n$ games, then A wins the last one

All these are mutually exclusive events. Can you figure out the probability now?

Hint 2: $$Probability = \frac{\text{favourable ways (listed above)}}{\text{total ways}}$$

• If I'm getting this right, I will get something like $(1/2)^n+(1/2)^{n-1}(1/2)(1/2)+(1/2)^{n-2}(1/2)^2(1/2)+... right? – jbuser430 Mar 9 '17 at 9:46 • @JBIBB$(1/2)^n+(1/2)^{n-1}(1/2)(1/2)+(1/2)^{n-2}(1/2)^2(1/2)+...$yes that's right. Also notice the GP summation there. – Gaurang Tandon Mar 9 '17 at 9:48 • @JBIBB You might need to put combinatorics for ordering of people, as stated in other answers. – Gaurang Tandon Mar 9 '17 at 11:59 P(A win$n$games and B win$0$games)$=\dbinom{2n+1}{n}\left(\dfrac{1}{2}\right)^n\left(\dfrac{1}{2}\right)^0$P(A win$n$games and B win$1$game)$=\dbinom{2n+1}{n}\left(\dfrac{1}{2}\right)^n\left(\dfrac{1}{2}\right)^1\cdots$P(A win$n$games and B win$n$game)$=\dbinom{2n+1}{n}\left(\dfrac{1}{2}\right)^n\left(\dfrac{1}{2}\right)^n$Required probability$\dbinom{2n+1}{n}\left(\dfrac{1}{2}\right)^n\left[\left(\dfrac{1}{2}\right)^0+\left(\dfrac{1}{2}\right)^1+\cdots+\left(\dfrac{1}{2}\right)^n\right]$• the${2n+1 \choose n}$is because you pick$n$games out of the$2n+1$? – jbuser430 Mar 9 '17 at 9:50 • yes, you are right – Kiran Mar 9 '17 at 9:50 • okay but why doesn't this change when A wins$n\$ games and B 1? – jbuser430 Mar 9 '17 at 9:53