Write probability in terms of shape parameters of beta distribution

[hjam24]

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:

$$ P(A_{wins} \cap y|\alpha, \beta) =\binom{10+10}{10}\frac{B(12+\alpha, 10+\beta)}{B(\alpha, \beta)} \cdot \space _{...} $$

 

[bpet]

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:

$$ P(A_{wins} \cap y|\alpha, \beta) =\binom{10+10}{10}\frac{B(12+\alpha, 10+\beta)}{B(\alpha, \beta)} \cdot \space _{...} $$

It's not quite a beta-binomial distribution. However you can do a similar integral $$\int_0^1 P(A_{wins}\cap y|\alpha,\beta)Beta(p|\alpha,\beta)$$. Note also that the expression for P further simplifies before you do the integral.