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This is probably a simple one, but I'm trying to figure out how to calculate the probability of victory for player 1 in a static two-player game, where the history of games between the two is (player 1: 6 victories, player 2: 4 victories).

I'm not much good at Maths, so my initial efforts have just been to divide the number of games won by the total number of games played. So, 6/10 = 60%.

What I'd like to avoid is the situation where, after game 1, there is a probability distribution of 100 : 0.

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  • $\begingroup$ Welcome to MSE. Please, show us your effort. $\endgroup$ – Jaroslaw Matlak Sep 5 '17 at 11:47
  • $\begingroup$ I've updated the question with my effort, Jaroslaw. $\endgroup$ – Michael Prestonise Sep 5 '17 at 11:54
  • $\begingroup$ The simplest thing to do is what you did. There are some problems if, say, you stopped after the first game. If player 1 won the first game then after that you would have said that player 1 had a probability of 100% of winning, which is clearly wrong (since B went on to win). There are ways around this sort of problem. Which way is best probably depends on why you're asking this question. $\endgroup$ – Michael Lugo Sep 5 '17 at 14:04
  • $\begingroup$ @michaellugo I was trying to avoid the 100% after 1 game problem, yes. Ideally, the 50% for player 1 / 50% for player 2 would become something more like 52% for player 1 / 48% for player 2. $\endgroup$ – Michael Prestonise Sep 5 '17 at 14:07
  • $\begingroup$ Possible duplicate of Probability of World Series - Using Pascal and Fermat "Problem of Points" $\endgroup$ – mlc Sep 7 '17 at 20:52

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