Dr Ian Plummer
Technical
Does Playing BestofN Games Help the Better Player?
Kevin Carter writes:
I think that the effect of playing Bestof3s and Bestof5s is often overestimated. If you crunch then numbers you will find that:
When player X has a 60% chance of beating player Y in each game then (ignoring stamina and psychological issues) the probability of X beating Y in a Bestof3 rises to only 65% and a Bestof5 pushes the figure up a bit more to 68%.
If instead of 60% you put in 70%, then Bestof3 gives 78% and Bestof5 84%.
So, given a clear superiority of one player over another it is by no means certain that a BestofN match will usually 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:
Diagram created by Teodora Baeva
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 Games
The 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:
Score 
Probability 
= 
WW 
p * p 
p^{2} 
WLW 
p * q * p 
p^{2}q 
LWW 
q * p * p 
p^{2}q 
Probability of A winning = 
p^{2} + p^{2}q + p^{2}q 
= 
p^{2} [ 1 + 2q ] 
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 Games
The same analysis for the Bestof5 games gives:
Score 

Probability 
= 
WWW 

p * p * p 
p^{3} 
LWWW 

q * p * p * p 
p^{3}q 
WLWW 

p * q * p * p 
p^{3}q 
WWLW 

p * p * p * q 
p^{3}q 
LLWWW 

q * q * p * p * p 
p^{3}q^{2} 
LWLWW 

q * p * q * p * p 
p^{3}q^{2} 
LWWLW 

q * p * p * q * p 
p^{3}q^{2} 
WLLWW 

p * q * q * p * p 
p^{3}q^{2} 
WLWLW 

p * q * p * q * p 
p^{3}q^{2} 
WWLLW 

p * p * q * q * p 
p^{3}q^{2} 
Probability of A winning = 
p^{3} + 3p^{3}q + 6p^{3}q^{2} 
= 
p^{3} [ 1 + 3q +6q^{2 }] 
= 
p^{3} [ 1 + (^{3}C_{2}) q + (^{4}C_{2}) q^{2 }] 
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.
^{n}C_{k} 
n!/k!(nk)! 
= 
^{2}C_{2} 
2 / 2*1 
1 
^{2}C_{1} 
2 / 1 
2 
^{3}C_{2} 
6 / 2*1 
3 
^{4}C_{2} 
24 / 2*2 
6 
Bestofn Games
In 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)} ]
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