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I was thinking that we seem to be seeing fewer ace matches with the new Bo7 format, I tried to come up with a proof but I don't think I got it exactly right.
n = number of matches before the ace match (4 in a Bo5, 6 in a Bo7)
p = probability of ace match between 2 evenly matched teams
p = 0.5^n * nc(n/2)
where c is the combinatorial. My thoughts are that there are nc(n/2) ways of getting to an ace match, and that the probability of each way happening is the probabilities of each player winning multiplied by each other. (for evenly matched teams, this is 0.5^n no matter who wins each game)
Results: (again, for evenly matched teams)
31.25% chance of an ace match in a Bo7
37.5% chance of an ace match in a Bo5
Given that most teams are not evenly matched, the percentage should not exceed this in reality.
I dunno, something feels wrong here, help?
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United States4053 Posts
The number of ways to get to an ace match with the new format is equal to the number of ways to distribute 3 wins to each team in the first six games, namely 6 choose 3. Therefore the probability of an ace match in a Bo7 is 20/64 = 5/16.
The probability of an ace match in a Bo5 is 3/8.
so what you have looks about right
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The math looks fine. Do we have a reference for what percentage of the time we actually do see ace matches?
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Looks good, that's close to how I thought about it. A cleaner way is to drop the 0.5^n part and count the number of ways you can get to an ace match (6 choose 3) and count the number of ways the first 6 games can play out (2^6).
The only tricky part is that not all 6 games are played in every match. But the way it works out is that win/lose sequences that lead to shorter matches have a higher probability. In particular 5 game sequences occur twice as much and WWWW 4 times as much as any particular 6 game sequence. Which means that for example LWWWW is correctly represented by the sum of 2 fantom matches of LWWWWW and LWWWWL.
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EPT
