Tourbament deal - a split descision.

[EjnarPik]

I once followed the final stage of a large tournament on pokerstars. I think it was $200, with more than 100

When they got down to the final 3, they started to negotiate a deal. Somebody suggested each taking money for third place, and split the rest according to chip count.

The player with the lowest chip count refused, saying that split according to chip count, was unfair to the low stack.

They played on, and later that player became chip-leader, and now he was eager to make the deal. So it looks like he means it.

Anybody know the arguments, abd do they hold true?

Best Regards

Ole Berg

[Louie Landale]

If there were only two players left then each taking 2nd and splitting the rest of the 1st place prize money based on relative chip count would be perfectly fair.

With 3 folks left is may or may not be fair depending on the exact payoff amounts. I cannot do the calculations but suspect that the closer the 1st and 2nd prizes the worse it would be for the low chip count player.

I believe in general he's telling the truth because his chips are "worth" more vis-a-vis face value than the chip leader's chips.

- Louie

[Bozeman]

This chip count split (McEvoy method), decidedly favors the big stacks. In many cases, a big stack can get more than 1st place money, which is obvously incorrect. Say the big stack has 80% and payouts are 50%, 30%, 20%: then it would give .2+.8*.3+.8*.1=52%. Even in cases not this extreme one can show that its result is incorrect. It gives the correct fraction of first place (which also implies Louie's statement that it gives correct split headsup), but for a big stack his percentage of 2nd's will be lower than his 1st's so this gives him too much. Any time that P1, the fraction of total chips held by the big stack is greater than .5 it gives him the $ he would get if he had NO probability of finishing 3rd.

For a review of fair splits and better approximations, see TFP

Craig