- #1
hjam24
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- TL;DR Summary
- Initially we have a probability of someone winning a game with certain scoring rules. The probability is of winning depends on the probability of winning a point, p, (which is assumed to be constant). The goal is to draw p from a beta distribution and change the formula accordingly
Assume that players A and B play a match where the probability that A will win each point is p, for B its 1-p and a player wins when he reach 11 points by a margin of >= 2The outcome of the match is specified by $$P(y|p, A_{wins})$$
If we know that A wins, his score is specified by B's score; he has necessarily scored max(11, y + 2) points
In the case of y >= 10 we have
$$ P(A_{wins} \cap y|p) = \binom{10 + 10}{10}p^{10}(1-p)^{10}
\cdot[2p(1-p)]^{y-10}\cdot p^ 2$$
The elements represents respectively:
- probability of reaching (10, 10)
- probability of reaching y after (10, 10)
- probability of A winning two times in a row
I would like to change the constant p assumption and draw p from a beta distribution.
The first part can be rewritten as as [beta-binomial](https://en.wikipedia.org/wiki/Beta-binomial_distribution) function:
$$ P(A_{wins} \cap y|\alpha, \beta) =\binom{10+10}{10}\frac{B(10+\alpha, 10+\beta)}{B(\alpha, \beta)} \cdot \space _{...} \cdot \space _{...}$$
The original formula can be simplified to
$$P(A_{wins} \cap y|p) = \binom{10 + 10}{10}p^{12}(1-p)^{10}
\cdot[2p(1-p)]^{y-10}$$
Is it correct to combine the first and third element as follows:
$$ P(A_{wins} \cap y|\alpha, \beta) =\binom{10+10}{10}\frac{B(12+\alpha, 10+\beta)}{B(\alpha, \beta)} \cdot \space _{...} $$
If we know that A wins, his score is specified by B's score; he has necessarily scored max(11, y + 2) points
In the case of y >= 10 we have
$$ P(A_{wins} \cap y|p) = \binom{10 + 10}{10}p^{10}(1-p)^{10}
\cdot[2p(1-p)]^{y-10}\cdot p^ 2$$
The elements represents respectively:
- probability of reaching (10, 10)
- probability of reaching y after (10, 10)
- probability of A winning two times in a row
I would like to change the constant p assumption and draw p from a beta distribution.
The first part can be rewritten as as [beta-binomial](https://en.wikipedia.org/wiki/Beta-binomial_distribution) function:
$$ P(A_{wins} \cap y|\alpha, \beta) =\binom{10+10}{10}\frac{B(10+\alpha, 10+\beta)}{B(\alpha, \beta)} \cdot \space _{...} \cdot \space _{...}$$
The original formula can be simplified to
$$P(A_{wins} \cap y|p) = \binom{10 + 10}{10}p^{12}(1-p)^{10}
\cdot[2p(1-p)]^{y-10}$$
Is it correct to combine the first and third element as follows:
$$ P(A_{wins} \cap y|\alpha, \beta) =\binom{10+10}{10}\frac{B(12+\alpha, 10+\beta)}{B(\alpha, \beta)} \cdot \space _{...} $$