Does Playing Best-of-N Games Help the Better Player?
Kevin Carter writes:
So, given a clear superiority of one player over another it is by no means certain that a Best-of-N match will be won by the better player.
On the other hand, more challenging conditions will help to nullify the 'luck' element, notably one good hit-in in easy conditions resulting in an all-round break, a missed lift and subsequently a triple peel win. So, if Player X has a 60% chance in easy conditions the probability of his winning each game might rise to, say, 70% in challenging conditions.
Or, to put it another way, a single game win at Surbiton [difficult conditions] could be worth more than a Best-of-5 at Hurlingham ...
Sam Murray supplied a graph showing the probability of match wins as a function of game win %, and number of games in match:
Joel Taylor gives the background to the calculation
Let us assume player A has a probability of p of winning any given match. Thus, A’s opponent, B has a probability of q = (1 − p) of winning each game.
The probability of player A winning a Best-of-3 match can be determined by considering the total number of games as different cases. A can win (W) with the following score lines:
The probability of player A winning the Best-of-3 match is therefore: p2 [ 1 + 2q ]
The 2 comes from the binomial distribution of the first two games. This idea can be extended to Best-of-5 games.
The same analysis for the Best-of-5 games gives:
So the probability of player A wining the Best-of-5 match is the sum of the individual probabilities.
We use the shorthand nCk for Binomial Coefficients, which corresponds to n!/k!(n-k)!, and is the number of ways of picking k unordered outcomes from n possibilities. E.g. 3C2 is the number of ways you can pick 2 objects out of three.
In a Best-of-n games match (where n is odd), to win, a player must win ½(n + 1) games.
So the probability of winning in straight games is p½ (n+1). Following the pattern of before, the probability of winning in any other number of games and thus the total probability of player A winning is:
p½(n + 1) [ 1 + ½(n + 1)C½(n - 1) q + ½(n + 1) + 1C½(n - 1) q2+ ½(n + 2) + 1C½(n - 1) q3 + ... + (n - 1) C½(n - 1) q½(n - 1) ]
This mess looks slightly simpler if we let k =½(n + 1), so k is the number of games the player must win to win the match is:
pk [ 1 + kC(k - 1) q +(k + 1) C(k - 1)q2+ (k + 2) C(k - 1)q3 + ... + 2(k - 1) C(k - 1)q(k - 1) ]
Samir Patel adds:
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