Let N be an odd, positive number. A best-of-N series of games is played until one team has won a majority of the N maximum possible games in the series. Let p be the probability that team A wins any such game. In a series of up to N games, where each game is won either by A or B (no ties), what is the expectation for how many games will be played to decide the best-of-N series? What is the distribution of number of games? This is a generalization of the Riddler Express problem posted here: https://fivethirtyeight.com/features/how-much-should-you-bid-for-that-painting/

Expansion of \((p+(1-p))^n\) \(\begin{array}{c} & & & & & & & 1 & & & & & & & \\ & & & & & & p & & (1-p) & & & & & & \\ & & & & & p^2 & & 2 p (1-p) & & (1-p)^2 & & & & & \\ & & & & p^3 & & 3 p^2 (1-p) & & 3 p (1-p)^2 & & (1-p)^3 & & & & \\ & & & p^4 & & 4 p^3 (1-p) & & 6 p^2 (1-p)^2 & & 4 p (1-p)^3 & & (1-p)^4 & & & \\ & & p^5 & & 5 p^4 (1-p) && 10 p^3 (1-p)^2 && 10 p^2 (1-p)^3 && 5 p (1-p)^4 && (1-p)^5 & & \\ & p^6 && 6 p^5 (1-p) & & 15 p^4 (1-p)^2 && 20 p^3 (1-p)^3 && 15 p^2 (1-p)^4 && 6 p (1-p)^5 && (1-p)^6 & \\ p^7 & & 7 p^6 (1-p) & & 21 p^5 (1-p)^2 & & 35 p^4 (1-p)^3 & & 35 p^3 (1-p)^4 & & 21 p^2 (1-p)^5 & & 7 p (1-p)^6 & & (1-p)^7 \end{array} \) A wins and B wins after n games. \(\begin{array}{c} & & & & & & & (0,0) & & & & & & & \\ & & & & & & (1,0) & & (0,1) & & & & & & \\ & & & & & (2,0) & & (1,1) & & (0,2) & & & & & \\ & & & & (3,0) & & (2,1) & & (1,2) & & (0,3) & & & & \\ & & & (4,0) & & (3,1) & & (2,2) & & (1,3) & & (0, 4) & & & \\ & & (5,0) & & (4,1) && (3,2) && (2,3) && (1,4) && (0,5) & & \\ & (6,0) && (5,1) & & (4,2) && (3,3) && (2,4) && (1,5) && (0,6) & \\ (7,0) & & (6,1) & & (5,2) & & (4,3) & & (3,4) & & (2,5) & & (1,6) & & (0,7) \end{array} \) Game which decides best-of-N series: \(\begin{array}{c} & & & & & & & N/A & & & & & & & \\ & & & & & & 1 & & 1 & & & & & & \\ & & & & & 3 & & N/A & & 3 & & & & & \\ & & & & 5 & & 3 & & 3 & & 5 & & & & \\ & & & 7 & & 5 & & N/A & & 5 & & 7 & & & \\ & & 9 & & 7 && 5 && 5 && 7 && 9 & & \\ & 11 && 9 & & 7 && N/A && 7 && 9 && 11 & \\ 13 & & 11 & & 9 & & 7 & & 7 & & 9 & & 11 & & 13 \end{array} \)

Thus to decide a best-of-1 series, exactly one game is played taking us from state (0,0) to (1,0) with probability p or to state(0,1) with probability (1-p). To decide a best-of-3 series, after 2 games we have state (2,0) with probability p^2, state (1,1) with probability 2 p (1-p) or state (0,2) with probability (1-p)^2. If the game isn't decided after 2 games, the third will decide it. So the distribution is 2 games with probability p^2 + (1-p)^2 = 1 - 2p + 2p^2 and 3 games with probability 2 p - 2 p^2. The expectation is 2 + 2 p - 2 p^2 games. Generalizing, the expectation is the sum of all expansion terms corresponding to undecided states. \(E = \sum_{i=0}^{2i -1 < N} \sum_{j=0}^{2j -1 < N} { {i+j} \choose i } p^i (1-p)^j \)

Let \( n = \frac{N-1}{2} \geq 0\) Then our expectation value goes like this: \(E = \sum_{i=0}^{2i -1 < N} \sum_{j=0}^{2j -1 < N} { {i+j} \choose i } p^i (1-p)^j = \sum_{i=0}^{n} \sum_{j=0}^{n} { {i+j} \choose i } p^i (1-p)^j = (n+1) \sum_{k=0}^{n} \frac{1}{k+1} { {2k} \choose k } \left( p (1-p) \right)^k \) For p = 1/2 this is approximately \(E \approx 2(n + 1) - \frac{8 n + 3}{4 \sqrt{\pi n} }\) for large n For p = 3/5 this is approximately \(E \approx \frac{5}{3}(n+1) \) for large n For p = 7/10 this is approximately \(E \approx \frac{10}{7}(n+1) \) for large n

...or how many possible games are there? In a game of chess we may number each possible game, and allow the computer to choose which game to play.

Let N be the positive odd number, so the winning player has to win (N+1)/2 games. And the number of games required to be played can be anything between (N+1)/2 to N depending on the strength of players. Unless the relative strength (or some kind of past statistics of games played between these players) is given, finding out the probability as asked in the OP, is a futile exercise.

But it is given in the OP as "p". The expectation takes the largest value for p = 1/2, because as post #6 shows the expected number of games played is a polynomial in \(p(1-p)\) (with maximum value for p=1/2) with all positive coefficients.

For a best of 5 series With a 50-50 probability of winning (like tossing a coin) For the first 4 games LSB is the first game 0000<play 3 games 0001<play 4 games 0010<play 4 games 0011<play 5 games 0100<play 4 games 0101<play 5 games 0110<play 5 games 0111<Play 3 games 1000<Not played 1001<play 5 games 1010<play 5 games 1011<play 4 games 1100<play 5 games 1101<play 4 games 1110<play 4 games 1111<Not played I'm not sure where to go from here, maybe take all the outcomes and divide them by the sum of possible outcomes. Or crash. Crashing... FWIW I reckon... Expect a winner after 3 games on 6 occasions (18) Expect a winner after 4 games on 6 occasions (24) Expect a winner after 5 games on 6 occasions (36)

Continuing the above post but ignoring the last bit I reckon... In a best of 5 series with equally matched teams. 2 end after 3 games with 1/7 chance of victor being declared 6 end after 4 games with 3/7 chance of victor being declared 6 end after 5 games with 3/7 chance of victor being declared So the winner will normally (4/7) be declared without playing all 5 possible matches.