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Jane StreetIn world series (baseball) there are two teams, A and B. You know that each can win 50% of the time (1:1 odds). You also know how the game works, i.e. Whoever wins 4 games first wins. What is the probability of getting to game 7 (i.e. Each team wins 3 games)?
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14 réponse(s)
Assuming no tie. It's the prob of throwing 6 coins and get 3 head and 3 tails, which is (6!/3!3!)/2^6 = 20/2^6 = 5/16
Ben le
to clear up any misconceptions, since there seem to be multiple answers here. 5/16 is the correct result
slin le
Whoever wins the first game does not matter, as it could be either team with a lead after 1 game, so it's the next 5 games that matter. There are 2^5 or 32 possible outcomes for those games. The outcomes for the team that won game 1 that will force a game 7 are WWLLL, WLWLL, WLLWL, WLLLW, LWWLL, LWLWL, LWLLW, LLWWL, LLWLW, LLLWW. This is 10/32 or 5/16.
P le
In order to have 7 games you need a teem that wins exactly 3 games out of 6 Hence, p= 6C3/2^6 = 20/64 =5/16
UCV le
it is 5/16, perhaps an easier way to think about is: P(7 games) = 1  P(6games)  P(5games)  P(4games) [there can be no less than 4 games otherwise the tournament would not be finished] hence the probability = 1  15/64  5/32  1/16 = 20/64 = 5/16. N.B. the probabilities are calculated as 6C2*(0.5^6), 5C1*(0.5^5), 4C0(0.5^4) where C is the choice function [as in nCr] used to determine the number of ways each outcome could happen [i.e. there are 15 ways to get 4 wins and 2 losses in a 6 game tournament]. Hope this helps.
matt le
i'm thinking it's 15/32 for 4 games it's either WWWW or LLLL so it must be 2/16 or 1/8 for 5 games it's LWWWW, WLWWW, WWLWW, WWWLW, and because its symmetrical we have 8/32 for 6 games similar concept we 10/64 that's my idea, what i think the calculation above is wrong is because they added the case WWWWL which is absurd since the game already ended in the 4th game
g le
its absolutely not 5/16... havent figured out yet how to do it mathematically but if you write out all the responses and make sure that the winning team is always last in the sequence, you'll see its about half.
mike le
Now there are 2 scenarios from "my" team's point of view: (1) 4 wins + 3 losses > my team wins (2) 3 wins + 4 losses > my team loses They are symmetrical so simply times 2 in the final result. Consider case 1: the no. of combinations is 7! / (4! * 3!) = 35 You gotta minus the illegal cases where you have 4 continuous wins: there are 3 of them (can you see?) Therefore, the no. of combinations is 35  3 = 32. The probability of having 7 games is 2 * 32 / 2^7 = 64 / 128 = 1/2.
Viet le
http://www.stat.unc.edu/faculty/rs/s11/aip.htm > 31.25% aka 5/16
Utilisateur anonyme le
I got 4/7. Please tell me where my calculation is wrong. # of ways to win in 4 = 1 # of ways to win in 5 = 4 (lose first, second, third or fourth gamecannot lose 5th, as series would be over) # of ways to win in 6 = 10 (5 choose 3, since last game must be W) # of ways to win in 7 = 20 (6 choose 3) That means we have 20/35=4/7 probability of having 7 as opposed to 5/16. Peace out.
Jimmy le
The calculation above is wrong because each the # of ways to win in 4 or 5 is more likely than each case of 6. It doesn't matter that you cannot have WWWWLL because the 1/16 chance of WWWW incorporates that theoretical possibility. Hence the answer is 6C3 * (1/2)^6= 5/16.
Jimmy le
@Ben I dont think theyre equivalent. Coz in this case, the games would stop as soon as one of the teams wins (A wins 4 games or B wins 4 games) and this can happen after 4 or even 5 games are played. I get (6C3) / {(6C3) + (6C4) + (5C4) + 1} = 20/41 Lemme know if you agree with me...
HH le
Each event in the sample space is either a win by A and a loss by B or the other way round, and the probablity of that is 0.5^2 The probablity of this happening until the 6th game is (0.5^2)^6 = 0.5^12
Utilisateur anonyme le
there are only 8 possible outcomes 40 04 41 14 42 24 43 34 only in 2 of 8 they arrive to 7 games, so the probability is 25%
alessandro le