Help with Probability and expected value…

[imcastleman]

It’s down to 3 players in a tournament and there are 10,000 chips in play. Player A has 4000, player B has 3500, player C has 2500 chips. Assuming all players are equal in skill level, what are the probabilities for each of them that they get 1st, 2nd, and 3rd? The numbers I used are wrong but I don’t know why. Here are the numbers I came up with:

Player a:
P(1st) = 40%, P(2nd) = 35%, P(3rd) = 25%

Player b:
P(1st) = 35%, P(2nd) = 40%, P(3rd) = 25%

Player c:
P(1st) = 25%, P(2nd) = 35%, P(3rd) = 40%.

It may be obvious how I came up with these numbers, but I do realize there is a flaw but don’t know how to come up with the correct numbers. I'm interested in how you come to the solution as opposed to the actual solution so please don't just give me the answer.

Thanks in advance for any help!!

[pzhon]

The independent chip model can be described in a few ways. One is that the chips are randomly removed one by one. When your last chip is eliminated, you bust out. Another way to describe it is that the chips are shuffled. The player with the first chip wins. This happens to each player with probability equal to their fraction of the chip pool. Then, among the survivors, the player with the highest chip wins, etc.

By the independent chip model, the player with 4000 chips finishes 1st 2/5 = .400, 2nd 68/195 = .349, and 3rd 49/195 = .251. The player with 3500 finishes 1st 7/20 = .350, 2nd 7/20 = .350, and 3rd 3/10 = .300. The player with 2500 finishes 1st 1/4 = .250, 2nd 47/156 = .301, and 3rd 35/78 = .449.

One plausibility check is that the probabilities of placing 2nd add up to 1. 0.349 + 0.350 + 0.301 = 1.

The independent chip model does not necessarily fit the dynamics of a real poker table. If the game is limit and the blinds are small, a diffusion model would be better.