Technical
Does Playing BestofN 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 BestofN 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 hitin in easy conditions resulting in an allround 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 Bestof5 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. Bestof3 GamesThe probability of player A winning a Bestof3 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 Bestof3 match is therefore: p^{2 }[ 1 + 2q ] The 2 comes from the binomial distribution of the first two games. This idea can be extended to Bestof5 games. Bestof5 GamesThe same analysis for the Bestof5 games gives:
So the probability of player A wining the Bestof5 match is the sum of the individual probabilities. We use the shorthand ^{n}C_{k} for Binomial Coefficients, which corresponds to n!/k!(nk)!, and is the number of ways of picking k unordered outcomes from n possibilities. E.g. ^{3}C_{2} is the number of ways you can pick 2 objects out of three.
Bestofn GamesIn a Bestofn 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 p^{½ (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: p^{½(n + 1) }[ 1 + ^{½(n + 1)}C_{½(n  1) }q + ^{½(n + 1) + 1}C_{½(n  1) }q^{2}+ ^{½(n + 2) + 1}C_{½(n  1) }q^{3} + ... + ^{(n  1) }C_{½(n  1) }q^{½(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: p^{k } [ 1 + ^{k}C_{(k  1) }q +^{(k + 1) }C_{(k  1)}q^{2}+ ^{(k + 2) }C_{(k  1)}q^{3} + ... + ^{2(k  1) }C_{(k  1)}q^{(k  1)} ] CommentsSamir Patel adds:
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