This is driving me insane because it does not seem too difficult of a problem and solving it would ease my life a little.
Some background: I make estimates in esports and try to find edge over the betting markets. Most of the games are 3-game series so visualising team power ratings is easiest that way. I end up having a BO3 probability and then run into the problem of changing it to BO1 probability. As of now I figure it out by "testing" it in my excel spreadsheet, since I know how to make a BO1 probability into BO3.
In this case we also assume that all games in the BO3 series are identical probabilities.
Right now I know that:
$X=Y²+(2*((1-Y)*Y²))$
and
$1-X=(1-Y)*(1-Y)+(2(Y*(1-Y)*(1-Y)))$
Where X is BO3 probability and Y is BO1 probability. I don't think Y is solvable here though.
Just to be clear: BO3 means best of 3 game series and BO1 means a single match.
Thanks a lot in advance to anyone that bothers to help me on this one.
I'm going to assume winning a 3 game series refers to winning 2 out of 3 and they will continue playing even if they win the first 2. Let $X$ be the event that team A wins and Y be the event they lose, there are 4 scenarios where they win $XXX, XXY, XYX, YXX$. Now let $P(X) = x, P(Y) = 1-x$ be the probability they win a game and lose a game respectively.
We know that $x^3 + 3x^2(1-x) = 0.6$ which gives us the solution $x \approx 0.57$
I think this should be correct!