Finishing a Best-of-Seven SeriesDate: 06/19/2009 at 03:05:40 From: Jakob Subject: 2-1 chance of getting to 4? Adam and Britta are playing Backgammon. They are playing to 4 and the current result is 2-1 Adam is in the lead. Sadly they are forced to stop and want to find out who had the biggest chance of winning. Each of them has 50% chance of winning a game. How big is Adam's chance of winning? The combinatorics in this question is a little hard, the tree we use to sort it gets hugely big and in some cases Adam wins after two games, but they might have to play to 3-3 in that case it gets too confusing. The three games they already played have 3 possibilities. AAB, BAA, ABA. Now they play again, then it might look like this: AABA, AABB, BAAA, BAAB, ABAB, ABAA. But adding the next game to the chain makes it very large (might have missed some): AABAA, AABAB, AABBA, AABBB, BAAAAx, BAAAB, BAABA, BAABB, ABABA, ABABB, ABAAAx, ABAAB. Some of those combinations will end the game, Adam winning with 4, those are marked with an x. But they can also play on. From now on it gets too confusing for me. Date: 06/19/2009 at 11:33:14 From: Doctor Ian Subject: Re: 2-1 chance of getting to 4? Hi Jakob, Let's use a simpler representation, like the score. For your problem it doesn't matter how they got to 2:1, we just know they start there. So we start with the score 2:1 in favor of Adam, and the next game is equally likely to go to either of them: +-- 3:1 | 2:1 --+ | +-- 2:2 We can keep building that tree out with two outcomes for the next game in each case. In cases where Adam's score reaches 4, I'll mark it with an A, and where Britta reaches 4 I'll use a B: +-- 4:1 A | +-- 3:1 --+ | | | +-- 3:2 | 2:1 --+ | +-- 3:2 | | +-- 2:2 --+ | +-- 2:3 At this level, we have a win for Adam. What is the probability of that win? It's (1/2)^2, or 1/4. So after two games, the probability is 1/4 that Adam wins, and 3/4 that play continues. Let's go another level, by playing the third game: +-- 4:1 A | +-- 3:1 --+ | | +-- 4:2 A | | | | +-- 3:2 --+ | | | +-- 3:3 | 2:1 --+ | +-- 4:2 A | | | +-- 3:2 --+ | | | | | +-- 3:3 | | +-- 2:2 --+ | | +-- 3:3 | | +-- 2:3 --+ | +-- 2:4 B What are the probabilities associated with these nodes? Let's look at the second 4:2, for instance. How can we get here? Half the time, Britta will win the first game; after that, half the time Adam will win the second game; and half the time, Adam will win the third game. So the probability of ending up here is (1/2)^3 = 1/8. Similar reasoning for the other cases gives us +-- 4:1 A (1/2)^2 | +-- 3:1 --+ | | +-- 4:2 A (1/2)^3 | | | | +-- 3:2 --+ | | | +-- 3:3 (1/2)^3 | 2:1 --+ | +-- 4:2 A (1/2)^3 | | | +-- 3:2 --+ | | | | | +-- 3:3 (1/2)^3 | | +-- 2:2 --+ | | +-- 3:3 (1/2)^3 | | +-- 2:3 --+ | +-- 2:4 B (1/2)^3 As a quick check the combined probability of a win (for either player) at this point is 1/4 + 3/8 = 5/8 And the combined probability of having reached 3:3 without a winner is 1/8 + 1/8 + 1/8 = 3/8 Note that the probabilities of having or not having a winner add to 1, as they should. And if there's a 3:3 tie, that means the next game decides the contest. Let's play that out: +-- 4:1 A (1/2)^2 | +-- 3:1 --+ | | +-- 4:2 A (1/2)^3 | | | | +-- 3:2 --+ | | +-- 4:3 A (1/2)^4 | | | | +-- 3:3 --+ | | | +-- 3:4 B (1/2)^4 | 2:1 --+ | +-- 4:2 A (1/2)^3 | | | | | +-- 3:2 --+ +-- 4:3 A (1/2)^4 | | | | | | +-- 3:3 --+ | | | | | +-- 3:4 B (1/2)^4 | | +-- 2:2 --+ | +-- 4:3 A (1/2)^4 | | | +-- 3:3 --+ | | | | | +-- 3:4 B (1/2)^4 +-- 2:3 --+ | +-- 2:4 B (1/2)^3 And now we can add up the probability of Adam's wins, 1/4 + 1/8 + 1/8 + 1/16 + 1/16 + 1/16 = 4/16 + 2/16 + 2/16 + 3/16 = 11/16 and the probability of Bitta's wins, 1/8 + 1/16 + 1/16 + 1/16 = 2/16 + 3/16 = 5/16 and they add up to 1, which they should--since one of them HAS to win 4 games eventually. Does this make sense? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 06/19/2009 at 12:25:14 From: Jakob Subject: Thank you (2-1 chance of getting to 4?) Thanks for helping, that one bugged me all day. Very simple when you put it like that! |
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