Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Shootin' pool

  1. Jun 5, 2010 #1
    Many pool games, like 8-ball and 9-ball, are played with the "winner breaks" rule. That is, winning a game gives you the right to break (make the opening shot) in the next game. Players usually flip a coin or lag to decide who breaks on the first game. For decent players, getting the break can be a real advantage, because they can sometimes "break and run", winning the game in one turn. And if not, they still may be able to gain control of the game by leaving their opponent a difficult shot. This amounts to saying that the probability r of winning the game on your break is greater than the probability s of winning the game on your opponent's break.

    The winner breaks rule means that a series of games is not very well modeled as a series of independent bernoulli trials. The outcomes of games i and j are not independent, although their dependence gets smaller as |j-i| increases. It would seem that in a very long series of games, the fraction of games you win would very likely approach some value f. This sounds like a law of large numbers claim, but it isn't immediately obvious to me how you would prove it. The fraction of games that you win is also the fraction of games where you are breaking, so

    f = f*r + (1-f)*s

    f = s/(s+1-r)

    I think this is the probability of interest when deciding what the odds are that player A wins a long match against player B, say a race to 9 or 11 games. Of course, that assumes you have some way of knowing r and s.

    Is there a name for this kind of process, where there are two different success probabilities, r and s, one applying when the previous trial was successful and the other when it failed?
     
  2. jcsd
  3. Jun 5, 2010 #2

    Borg

    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    If this statement were true, whoever broke first would win all of the games.
     
  4. Jun 5, 2010 #3
    Okay, my statement was a little ambiguous. I didn't mean by that that you win every game where you break, just that the fractions of games won and games where you broke are the same. For every game where you break, there is a game that you won, namely the previous game. They are in a one to one correspondence, except for the very first game. But the first game doesn't affect the fraction in an infinite series of games.
     
  5. Jun 7, 2010 #4
    If it's a rule that the winner of a game gets to break the next game in some sequence, then it seems this is a tautology after the first game.
     
    Last edited: Jun 7, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook