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Originally Posted by ImSavy
So the combinations are 12W, 12W1L, 12W2L, 11W3L, 10W3L, ..., 1W3L 0W3L
In each of these combos the final result has to be a W when 12W and a L when 3L so can be thought of as a combination of all results except the last.
Each of these can be worked out using (total outcomes)!/ [(W outcomes)!(L outcomes)!]
Example - 8W3L can be thought of as a combo of 8Ws and 2Ls so 10!/(8!2!)
Each outcome has a probability of (x)^W * (1-x)L where x is your win%
By multiplying the # combos with the probability of each happening we get how likely that result is.
edit
Then I assume we can work out our avg # of wins just by doing # wins * how likely for every result.
I didn't see this when I posted my initial response to your question.
This reasoning is spot on.
Your result of 10!/(8!2!) is equivalent to my notation of C(10,8). This is the binomial coefficient.
By symmetry, C(10,8) = C(10,2); just like 10!/(8!2!) = 10!/(2!8!)
To restate my solution here, using your notation of W for number of wins and L for number of losses:
For cases where you lose 3 matches
EV(W,L=3,x) = C(W+3-1,2)*X^W*(1-X)^3
For cases where you win 12 matches
EV(W=12,L,x) = C(12+L-1,L)*X^12*(1-X)^L
For all the "inner" cases, your EV of being in any state of 0 <= W <= 11 and 0 <= L <= 2
EV(W,L,x) = C(W+L,L)*X^W*(1-X)^L
or, by symmetry of the binomial coefficient
EV(W,L,x) = C(W+L,W)*X^W*(1-X)^L
The only notable difference is the "-1" in the binomial coefficient for the "end states" that is not present in the "intermediate states."
I still find that you need x > 81.3% to get a 50% shot at winning 12 matches.
Of course, this all assumes that all matches are of equal difficulty and your winrate taken as an aggregate of all matches is acceptable as an estimator as the Arena progresses to different matches. I think this is not too bad a model, since the arena matches seem to be always against players of roughly your rank +/- 1, therefore roughly equal skill, no matter how many W/L you have.
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