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The probability that A wins when he/she has $\alpha$ points and player B having $\beta$ points is equal to

$$ p_\text{win}(\alpha,\beta)=\binom{\alpha+\beta+1}{\alpha-1}p^\alpha(1-p)^\beta $$

where $p$ is the probability that A wins a point.

Next I implement this into the probability mass function of the binomial distribution

$$\operatorname{Binomial}(n\mid N, p_\text{win}) $$

As I result I can infer from $n$ and $N$ the probability that A wins a point, $p.$ However, during the sampling there are a significant amount of zeros and ones for $p.$ resulting in the following error:

Probability parameter is 1/inf/nan, but must be in the interval [0, 1]

To overcome this I would like to transform $p$ to its unconstrained equivalent, $x.$ I use logit parameterization for this

$$\operatorname{BinomialLogit}(n\mid N, x) = \operatorname{Binomial}(n\mid N, \operatorname{logit}^{-1}(x))$$

I am unsure how to rewrite $p_\text{win}$ in terms of $x.$ to the above. Please advise

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  • $\begingroup$ psst... you didn't hear this from me... but you could always $p\gets\textrm{clip}[p,0.0001,0.999]$ $\endgroup$ Apr 27, 2023 at 13:58
  • $\begingroup$ Perhaps more work, but you might consider building upon the Continuous Bernoulli distribution $\endgroup$
    – Firebug
    Apr 27, 2023 at 14:37
  • $\begingroup$ Will zero/one-inflated beta (ZOIB) solve your problem? $\endgroup$
    – wzbillings
    Apr 27, 2023 at 14:39
  • $\begingroup$ Can you work in log space (sampling $\log(p)$)? Sampling $\log(p)$ directly is fairly straightforward if $p$ is Beta-distributed. $\endgroup$
    – jblood94
    Apr 27, 2023 at 21:10
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    $\begingroup$ This question doesn't make sense to me --- you say you are inferring $p$ (which is presumably an unknown parameter?) but then you refer to having values of 0 and 1 for $p$. Please clarify what parts of your problem are data and what parts are unknown parameters. (If you already have known values of zero or one for $p$ then the resulting win probabilities are trivial, so I don't see the problem.) $\endgroup$
    – Ben
    May 18, 2023 at 1:32

1 Answer 1

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For some combinations of $\alpha, \beta$, the beta distribution concentrates its mass near 0 or 1. One option is to impose a prior on $p$. For example, you could consider

$$ p \sim \text{Beta}(a + \alpha, b + \beta) $$ where $a > 0$ and $b >0$, which will bound $p$ away from the extremes. Choosing $a=b$ might make sense, if you believe the teams are evenly matched prior to playing the game. The larger $a$ and $b$ are, the lower the variance of $p$. The same is true for $a + \alpha$ and $b + \beta$.

Alternatively, you could implement your idea & re-parameterize the model: draw $x$ from some distribution, and then transforming it to the $[0,1]$ interval. For instance, $x \sim \text{Normal}(\mu, \sigma^2)$ is symmetric about 0 for $\mu = 0$ and therefore $p=\text{logit}^{-1}(x)$ is symmetric about 0.5 on the probability scale.

The re-parameterization solution is not a panacea, though, because you may not want a symmetric distribution (if the teams are not evenly matched). However, setting $\mu$ very far from 0 will risk $p$ getting so close to 0 or 1 that the you have the same problem that you do now.


From the perspective of coding, you could write a series if if/then statements to detect p close to 0 or 1. In the case of p=0, you know a binomial distribution will have 0 successes; for p=1 a binomial distribution will have $N$ successes.

Or set p = median([p, eps, 1 - eps]) where eps is a small value, such as 1e-6. There's nothing statistically principled about this, it's just a kludge.

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