Consider a best-of-31 game where team A won 16 times and team B won 14 times. Assuming a rematch in which all conditions are the same, is there any way to infer (or guesstimate) what the probability of team A winning is?

My attempt: I used team A's score and the total number of rounds to find the probability of A winning a single given round, i.e. A score / (A score + B score). In this case it was 0.533. Then, I wrote a Python script which gives the probability of a team winning a best-of series, given their probability of winning a given round. The results for this example using 0.533 was 63.2%.

This seems rather high for a close match, and indeed using a score of 16:12, team A has a massive probability of winning of 80%.

Is inferring a probability from the score complete nonsense, and is there a better way? Can my attempt be considered useful?

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    $\begingroup$ When you say recurring are you saying that A probability of winning any game is statistical independent of winning any other game with probability 0.53 for A winning? $\endgroup$ Dec 12 '17 at 23:45
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    $\begingroup$ Apologies Michael, I simply meant 0.53333333 etc. $\endgroup$ Dec 12 '17 at 23:54
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    $\begingroup$ My point wasn't about the exact value of 16/30 but rather the independence assumption using 16/30 as the win probability for A. $\endgroup$ Dec 12 '17 at 23:59
  • $\begingroup$ I'm not sure what you're asking but I was inferring the probability of team A winning a round against team B from the score of the match alone. $\endgroup$ Dec 13 '17 at 0:05
  • $\begingroup$ By independent I just mean that a win or lose on one game will not affect the probability of a win on the next game. Statistical independence P(C|D)=P(C) where C and D are two events such as C is {A wins game 2} and D is {A wins game 1}. $\endgroup$ Dec 13 '17 at 0:34

$16/30$ is the maximum likelihood estimate for the single game win probability of team A given your data, so it is very reasonable to use this value. In fact, without additional data or assumptions, it is the only reasonable estimate.

Team A's probability of winning the match is equal to the probability that the binomial random variable $X \sim B(n, p)$ is greater than or equal to $16$, for $n=31$ and $p=16/30$:

$$ \Pr(X \geq 16) = \sum_{k=16}^{31} {31 \choose k}(16/30)^k(1-16/30)^{31-k} \approx 64.6\% $$

This explicitly calculates the probability team A will win at least 16 games if all 31 games are played. In practice the teams will stop playing as soon as one team reaches 16 wins, but this does not change the resulting probability, because even if the match continues beyond that point team A's status of having either less than 16 wins or 16 or more wins cannot change.

The reason this seems high is because the teams are playing a lot of games. The more games the teams play, the closer team A's probability of winning the series will get to $100\%$. This is due to The Law of Large Numbers.


If teams A and B have played games in the same player pool, then you can use the Elo rating system to compute their ratings and estimate the probability of team A beating team B in a single game. This calculation will include the 30 head-to-head games the teams played as well as additional games against other opponents, and it will give you a more accurate and robust estimate of team A's win probability than the 16/30 you are using.

According to the Elo system, the probability of team A beating team B is:

$$ P(A\ beats\ B) = \frac{1}{1 + 10^{(R_B-R_A)/400}} $$

Where $R_A$ is team A's rating, and $R_B$ is team B's rating.

  • $\begingroup$ Why is it 31 choose k ? If I recall correctly in a best of 31 as soon as a team wins 16 game then no other games need to be play. If team A wins a straight 16 game then the match is over. $\endgroup$
    – Fina Cg
    Dec 13 '17 at 5:24
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    $\begingroup$ The probability of being the first to $16$ in a best-of-$31$-games match is the same as the probability of winning $16$ or more out of a full $31$-game match $\endgroup$
    – Henry
    Dec 13 '17 at 8:14
  • $\begingroup$ Good question @FinaCg and thanks for the clarification, Henry. I updated my answer to make this explicit. $\endgroup$
    – Imran
    Dec 13 '17 at 11:45

If you want to take into account your intuition that a $16-14$ victory in a best-of-${31}$ match could easily have happened by chance from a probability of winning of about $\frac12$, possibly even less than $\frac12$, you could take a Bayesian approach:

  • start with a sensible prior $\pi_0(p)$ density for the probability $p$ of Player A winning an individual game;
  • update with the $16$ wins from $30$ observations with likelihood proportionate to $p^{16}(1-p)^{14}$ to get a posterior distribution; and
  • use the calculation probability for the next best-of-$31$-games match $\mathbb P (\text{win match}\mid p)= \sum_{i=0}^{15}{15+i \choose 15}p^{16}(1-p)^i = \sum_{j=16}^{31}{31 \choose j}p^{j}(1-p)^{31-j}$
  • combine these three steps, which (dividing through by constants) will give $$\mathbb P (\text{win match}\mid \text{seen }\tfrac{16}{30}; \pi_0) = \dfrac{\int_0^1 \pi_0(p)\, p^{16} (1-p)^{14} \mathbb P (\text{win match}\mid p) \, dp}{\int_0^1 \pi_0(p)\, p^{16} (1-p)^{14} \, dp}$$

Typical $\pi_0$ priors often used for $p$ are (a) uniform, which would give an estimate of about $0.59567$; (b) Jeffreys, with a density proportional to $\sqrt{\tfrac{1}{p(1-p)}}$, which would give an estimate of about $0.60258$; or (c) improper, with a density notionally proportional to ${\tfrac{1}{p(1-p)}}$, which would give an estimate of about $0.60512$. All of these are close to each other and rather lower than $\mathbb P (\text{win match}\mid p=\frac{16}{30}) \approx 0.646$ and may better fit your intuition

But in any case $0.646$ is not a particularly high number: it clearly has to be more than $0.533$ and playing more games increases the change of the better player winning overall. This table shows how with $p=\frac{16}{30}$ the probability changes with the length of the match

    bestof probwinoverall
       1      0.5333333
       3      0.5499259
       5      0.5623151
       7      0.5725935
       9      0.5815471
      11      0.5895695
      13      0.5968908
      15      0.6036588
      17      0.6099757
      19      0.6159151
      21      0.6215324
      23      0.6268706
      25      0.6319637
      27      0.6368391
      29      0.6415194
      31      0.6460237

Considering a higher probability:

  • To have a $95\%$ probability of the better player winning the match where winning each game has an independent probability of $\frac{16}{30}$ would require a best-of-$607$-games match
  • To have a $95\%$ probability of the better player winning a best-of-$31$-games match would require winning each game to have an independent probability of over $0.643427863$

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