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Background: a friend of mine makes a hobby (as I imagine many do) of trying to predict hockey playoff outcomes. He tries to guess the winning team in each matchup, and the number of games needed to win (for anyone unfamiliar with NHL hockey a series is decided by a best of 7). His record this year after 3 rounds of play (8+4+2=14 best of 7 matchups) is 7 correct/7 incorrect for winning team and 4 correct/10 incorrect for number of games (he only considers the number of games correct if he also picked the winning team).

We got to joking that he's doing no better than blind guessing on the teams question, but that he's substantially beating the odds if one assumes that the probabilities for a 4, 5, 6 or 7 game series are equal (would expect a 12.5% success rate, he's at 28.5%).

This got us wondering what the odds actually are for each possible number of games. I think I've worked it out, but I want to tie up a few loose ends since part of my approach was brute-force scribbling on a big piece of paper. My basic assumption is that the outcome of every game is random with probability $\frac{1}{2}$ for a each team to win.

My conclusion is that:

$$\rm P(4\;games) = \frac{2}{2^4} = 12.5\%\\ P(5\;games) = \frac{8}{2^5} = 25\%\\ P(6\;games) = \frac{20}{2^6} = 31.25\%\\ P(7\;games) = \frac{40}{2^7} = 31.25\%$$

I guided my analysis based on a notion that a 4 game series should have a probability of $\frac{2}{2^4}$, analogous to the odds of flipping 4 coins and getting either 4 heads or 4 tails. The denominators were easy enough to figure out from there. I got the numerators by counting the number of "legal" combinations (WWLWWLL would be illegal since the series would be decided after 5 games, the last 2 games would not be played) of results for a given number of games:

Possible 4 game series (2):
WWWW LLLL

Possible 5 game series (8):
LWWWW WLLLL
WLWWW LWLLL
WWLWW LLWLL
WWWLW LLLWL

Possible 6 game series (20):
LLWWWW WWLLLL
LWLWWW WLWLLL
LWWLWW WLLWLL
LWWWLW WLLLWL
WLLWWW LWWLLL
WLWLWW LWLWLL
WLWWLW LWLLWL
WWLLWW LLWWLL
WWLWLW LLWLWL
WWWLLW LLLWWL

Possible 7 game series (40):

LLLWWWW WWWLLLL
LLWLWWW WWLWLLL
LLWWLWW WWLLWLL
LLWWWLW WWLLLWL
LWLLWWW WLWWLLL
LWLWLWW WLWLWLL
LWLWWLW WLWLLWL
LWWLLWW WLLWWLL
LWWLWLW WLLWLWL
LWWWLLW WLLLWWL
WLLLWWW LWWWLLL
WLLWLWW LWWLWLL
WLLWWLW LWWLLWL
WLWLLWW LWLWWLL
WLWLWLW LWLWLWL
WLWWLLW LWLLWWL
WWLLLWW LLWWWLL
WWLLWLW LLWWLWL
WWLWLLW LLWLWWL
WWWLLLW LLLWWWL

What's a non-brute-force method for deriving the numerators? I'm thinking there may be a recursive definition, so that $\rm P(5\;games)$ can be defined in terms of $\rm P(4\;games)$ and so on, and/or that it may involve combinations like $\rm(probability\;of\;at\;least\;4/7\;W)\times(probability\;of\;legal\;combination\;of\;7\;outcomes)$, but I'm a bit stuck. Initially I thought of some ideas involving $\left(^n_k\right)$ but it seems that only works if the order of outcomes doesn't matter.

Interestingly, another mutual friend pulled out some statistics on 7 game series played (NHL, NBA, MLB 1905-2013, 1220 series) and came up with:

4 Game Series - 202 times - 16.5%
5 Game Series - 320 times - 26.23%
6 Game Series - 384 times - 31.47%
7 Game Series - 314 times - 25.73%

That's actually a pretty good match (at least from my astronomer's point of view!). I'd guess that the discrepancy comes from the outcome of each game having being biased toward a win for one team or the other (indeed, teams are usually seeded in the first round so that the leading qualifying team plays the team that barely qualified, second place plays second last, and so on... and most of the games are in the first round).

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  • $\begingroup$ Am not particularly active on CV.SE, so this may need a bit of re-tagging. $\endgroup$
    – Kyle
    Jun 3, 2014 at 23:31

4 Answers 4

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For a team to win [the series] in game N, they must have won exactly 3 of the first N-1 games. For game seven, there are $\binom{6}{3} = 20$ ways to do that. There are 2 possible outcomes for game seven, and 20 possible combinations of wins for each of the teams that can win, so 40 possible outcomes. For an N-game series a best-of-seven series to end in N games, the number of possibilities is $2 \binom{N-1}{3}$.

Indeed the order doesn't matter, if you're already given the number of games played. Only the last game matters, and the winner must have 3 previous wins, in any order.

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  • $\begingroup$ For an N game series shouldn't it be $2(^{N-1}_{{\rm floor}(N/2)})$, or something like that? Assuming there is an odd number of games, which is only sensible. $\endgroup$
    – Kyle
    Apr 27, 2017 at 15:47
  • $\begingroup$ I was using N as the number of games played in a best-of-seven. Eg. for N=4, $2\binom{3}{3} = 2$ gives you the number of possible ways the series can end in 4 games. ie. for each team, the number of ways to choose 3 wins out of 3 games. $\endgroup$
    – user159099
    Apr 27, 2017 at 19:15
  • $\begingroup$ Yes, the possibilities of an M-game series decided in N games, should be $2\binom{N-1}{\mathrm{floor}(M/2)}$. This will still work if there's an even number of games, if tied series are not considered decided. $\endgroup$
    – user159099
    Apr 27, 2017 at 19:22
  • $\begingroup$ If you are going to be realistic the probability of win should not be 0.5 for each team for each game. There could be a home ice advantage as one example. $\endgroup$ Apr 27, 2017 at 21:20
  • $\begingroup$ @MichaelChernick true, and I touch on this a bit in the last paragraph of the question, but 0.5 as a starting point that can later be adjusted is reasonable. $\endgroup$
    – Kyle
    Apr 28, 2017 at 0:17
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If p is the probability of winning a single game, and N is the number of games one needs to win the series, then the probability "P" of winning the series is given by:

$$ P = \frac{p^N}{\left(N-1\right)!}\cdot\sum_{i=0}^{N-1}\left[\left(\prod_{j=1}^{N-1}\left(i+j\right)\right)(1-p)^i\right] $$

Think of the way a game plays out with points going to each player as a path. Each path that leads to a win has a probability given by the product of the probabilities for the number of wins and losses. The following image tries to illustrate an example. Illustration of the derivation

The pattern for the coefficients is given by (i+N-1) choose (i) where "i" is the number of points the opponent has scored and N is the number of points for the win.

Remember the choose formula is:

$$ nCr = \frac{n!}{r!\left(n-r\right)!} $$

so for us:

$$ nCr = \frac{\left(i+N-1\right)!}{i!\left(N-1\right)!}=\frac{\left(i+1\right)...\left(i+N-1\right)}{\left(N-1\right)!}=\frac{\prod_{j=1}^{N-1}\left(i+j\right)}{\left(N-1\right)!} $$

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    $\begingroup$ Welcome to the site! Can you edit to explain how you arrived at this answer? Why is it correct? Note also that you can use math typesetting via MathJax. More information: math.meta.stackexchange.com/questions/5020/… $\endgroup$
    – Sycorax
    Jan 4 at 20:22
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    $\begingroup$ It's not clear what question this formula answers: it can't apply to the original post, because it gives the wrong values for $N\gt 1.$ $\endgroup$
    – whuber
    Jan 4 at 20:51
  • $\begingroup$ You're right. In my excitement I wrote the form of the equation for N=4. The generic equation involves a product of factors. It was derived from the equation for the choose function. For each N there is a polynomial that gives the proper coefficients from a diagonal of pascal's triangle. I will post the correction when I figure out how to use MathJax $\endgroup$ Jan 6 at 0:52
  • $\begingroup$ Unfortunately, your explanation is vague and your formula produces values that exceed $1$ and therefore cannot be correct. Try it for $N=4$ and $p=1/2,$ for instance. It says the probability is $$(1/2)^4\left(3!(1/2)^0 + 4!/1!(1/2)^1 + 5!/2!(1/2)^2 + 6!/3!(1/2)^3\right) = (6 + 12 + 15 + 15)/16 \gt 1.$$ $\endgroup$
    – whuber
    Jan 6 at 4:18
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    $\begingroup$ Much better, thank you! +1. $\endgroup$
    – whuber
    Jan 6 at 17:41
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An alternate way to look at would be a binomial distribution: You need x=3 (exactly 3 successes) in n = 6 (trails) , so if the probability of winning a game is .5 (both teams equally likely) , binomial would say: P(x=3) = 6C3 * (.5)^3 * (.5)^3 = .3125 This would mean there is 31.25% chance of going to a 7 game series. And the probability you win in the 7th game , would follow negative binomial, how many trails = 7 for 4 success, 7-1 C 4-1 * (.5)^3 * (.5)^4

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Regarding the discrepancy in real-life results, OP noted "I'd guess that the discrepancy comes from the outcome of each game having being biased toward a win for one team or the other (indeed, teams are usually seeded in the first round so that the leading qualifying team plays the team that barely qualified, second place plays second last, and so on." ...

I'll note that even a random ordering would result in such a phenomenon, as many of the games would pair off teams with different levels of skill. In fact, the only system I can think of where the real results would match the theory would be one where #1 is paired off with #2 etc AND where the pairs are coincidentally evenly matched (or the league gives home-arena advantage to the underdog). But even then, the second round would almost certainly feature mis-matched teams even if randomly assigned.

In other words, the theoretical results will always represent a best-case scenario.

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