2 player freezeout calculations

[wheeler]

Hi David, Mason, and everyone,

In your excellent books Tournament Poker for Advanced Players and Gambling Theory and Other Topics, you claim that if two players use the same strategy when heads-up in a tournament, the chance of each player winning is directly proportional to the size of his/her stack.

I thought you might be interested to know that this result rests on an implicit assumption that you don't make clear in the books: it only follows if you assume the players use a strategy that does not depend on stack size. Under this assumption, the result follows from the logical argument in TPFAP or otherwise.

But I don't think this assumption is justified. For example, it's not true of the improved "system" in TPFAP and indeed most people employ a strategy that takes into account their stack size. For example, I bully more when I'm the large stack, and move in preflop more when I'm the short stack.

Here's an example to make this clear. To keep things simple, it's rather contrived.

Imagine two players, Alice and Bob, play a heads-up match of Texas Hold'em. Both players employ the following strategy: if you are the short stack, move in with any two cards; if you are the large stack, call with AA, otherwises fold.

There are no blinds. (If you don't like the idea of having no blinds, just assume that blinds are very very small in relation to the stack sizes.) Intially Alice has 40% of the chips; Bob has 60%.

The game starts: Alice moves in every hand, and eventually she is called by Bob. The probability of AA holding up against a random hand is 0.85, so Bob wins the match at least 85% of the time. Therefore Alice's probability of winning is at most 15%, even though she started with 40% of the chips and both players used the same strategy.

Of course, this example is stupid. The modification of the winning probabilities would not be so dramatic with a more reasonable strategy, but the probability is not in general directly proportional to stack size, for strategies that take into account stack size.

- wheeler

[pzhon]

A martingale is a random process with no average drift. Given that the value is V0 at time 0, the average value at time 1 is V0.

If your chips follow a martingale, your probability of winning a tournament equals your fraction of the chip total. The average number of chips you have at the end of the tournament is the number of chips you have now. Since at the end of the tournament you either have all of the chips or none, the probability of having all of the chips must equal the proportion of the chips you have now.

If you disagree with the proportionate chance to win first, you must say that there is some drift, and if the proportionate chance is wrong by a lot, the drift must be significant. Under perfect play heads-up, the only source of that drift is the asymmetry of position, and you can change position at the cost of folding your blind. As long as the blinds are small, the proportion of chips you have should be very close to your probability of winning.

[ QUOTE ]
the probability is not in general directly proportional to stack size, for strategies that take into account stack size.

[/ QUOTE ]

Heads-up, why would it be a good idea to pay attention to the stack size beyond what can be used on the current hand? Undoubtedly, many people play very differently as the chip leader rather than trailer. Many of those differences are clear errors, not proper strategies or responses to opponents' mistakes.

[wheeler]

Hi pzhon,

Thanks for your response!

[ QUOTE ]

Heads-up, why would it be a good idea to pay attention to the stack size beyond what can be used on the current hand? Undoubtedly, many people play very differently as the chip leader rather than trailer. Many of those differences are clear errors, not proper strategies or responses to opponents' mistakes.

[/ QUOTE ]

I completely agree: it is an error to take into account your relative stack size in heads-up play (unless your opponent does).

If you assume that the two combatants play perfectly, then it is true that the probability of winning is proportional to the stack size, as you explain. The point of my post was to show that you cannot conclude this if instead you just assume that both players use the same (possibly imperfect) strategy.

I hope this clarified things...

- wheeler