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I became interested in analyzing and predicting my performance in put putt golf. The course has 18 holes and I have collected data of enough rounds to be able to make a statement about the probability to play an ace at each specific hole. That probability p(i) is vastly different for each hole i (i = 1..18) because the obstacles are very different. Let's for simplicity assume that the probabilities are mutually independent (scoring an ace in one hole has no effect on subsequent holes).

Is there a formula, given the ace probability p(1), p(2), ... p(18) for each hole, for computing the probability to play 0,1,2, ... 18 aces over the entire course? I was unable to come up on my own with an algorithm; using "ad hoc" combinatorics seems already too cumbersome for that many holes (although I admit it's probably immediately programmable). Also, my Google-Fu failed: I didn't even know what to enter as search terms.

The general structure of the problem seems ubiquitous enough: for example, Presidential election outcomes in the U.S. have similar probabilities per state (but with the additional complication of different weights for each state).

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    Can we assume that scoring an ace in one hole is independent from scoring an ace in another hole? As in, scoring an ace in one hole gives us no more information about the probability of scoring an ace in another hole. – mhdadk Aug 03 '21 at 22:39
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    What you are describing follows a [Poisson binomial distribution](https://en.wikipedia.org/wiki/Poisson_binomial_distribution). For theoretical details, see https://stats.stackexchange.com/questions/9510/probability-distribution-for-different-probabilities and https://stats.stackexchange.com/questions/93852/sum-of-bernoulli-variables-with-different-success-probabilities. For how to compute the probabilities using Python or R, see https://en.wikipedia.org/wiki/Poisson_binomial_distribution#Computational_methods. – mhdadk Aug 03 '21 at 23:06
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    Note that 0 holes-in-one is also possible, not just 1-18 (and if my experience is typical, is by far the most likely outcome) – Glen_b Aug 04 '21 at 00:33
  • @mhdadk Thank you. If you make that an answer, I'd like to upvote it. – Bernhard Aug 04 '21 at 06:48
  • @Bernhard Actually my question is basically a duplicate of the mentioned ones, one of which is also marked as a dup of https://stats.stackexchange.com/questions/9510/probability-distribution-for-different-probabilities (also my thanks to @mhdadk). – Peter - Reinstate Monica Aug 04 '21 at 08:14
  • So, then we do not need this thread anymore. – Bernhard Aug 04 '21 at 10:01

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