Tournament Win Chances Via Pure Logic

In my notorious "cute" problem below, I claimed that the chances of each player winning was exactly proportional to their chips and that you needed no calculations to be sure it was true.


Assume this is a one winner tournament with the prize being all of the chips. If the tournament was played repeatedly, the player with 3 chips would have to win exactly 3 out of six times to break even in the long run. Simlarly for the players with one and two chips. And they must all break even in the long run. Why? Because every individual bet has no edge for anyone. (The poster who brought the point that chips change value as they are accumulated did not realize that the concept does not apply win only situations.)

You are of course right. Here are a more complicated Q !


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Tourney: Sklansky, Andy, Brett and Jack.


You are watching a tourney. When it comes to the final table there are 4 player's left.


The total prize-pool is $ 10.000. The winner gets 50%, number 2 30%, number 3 20%.


Jack has 6.000 in chips - Andy: 2.000 - Brett: 1.600 - Sklansky has only 400. Total= 10.000.


The 4 player's are all eager to go home - so they all offer you to buy them out (before drawing for the button and position).


Assuming you and the 4 player's are all equally good - which player will you buy out AND for how much ?


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This is not an easy Q. Even I don't have the right answer ;-)

I think I must have been led astray by my acquaintances who tried to tell me that there was something in having more chips more than just the fact that you've got more chips (if that makes sense). Well I proved they were wrong, but you are correct, it was much easier than what I did.


Spikey

I think your explanation is incomplete. It is not obvious all players break even. Instead some players might subsidize others. Or some bets might be positive and some negative, cancelling out in the long run.


The logic is to guess the solution (expectation is proportional to chips) and then verify it is correct. If the solution is linear in your stack then each individual bet indeed has no edge. And the proportional solution is correct at the boundaries when you have no chips or all chips. Consequently this solution is correct.

Consider the following game:


Player A has one chip, player B has two chips, player C has three chips. The seating arrangement will be chosen (unbiased) randomly, and then the button will be assigned (unbiased) randomly. On each hand, all of the players will ante one chip, and one player will be awarded the pot. But in order to make the game more realistic, let's give the button an advantage (as would be the case in Hold'em or Omaha). So let's assume that the button has a higher probability of winning the pot than the other players. Then the player with one chip will have less than one chance in six of winning the tournament.

In the game described each player has equal chances of winning the pot and no bet has an edge. In such cases the chip ratio must be the win chances, even by the way if we allowed for the possibilitiy of ties.

The point is that even for very simple situations, real world answers will differ from those of simpler models, sometimes dramatically. For example, one is led to believe that in place paying tournaments, dollar expectation must be concave with respect to chip count, but this is not always the case.