In this article, I will use a simplistic scenario to give some guidelines for how weak a hand you can play in a tournament if not playing it depletes your stack significantly. This is a situation which if you play a lot of tournaments, you’ll eventually run into.
The sad fact is that even if you are against much weaker opponents, and you would therefore normally fold slightly positive EV hands rather than gamble all your chips on them, the situation reverses when you have a short stack. So to illustrate this, I came up with a cute math problem that does a decent job of demonstrating this phenomenon. There are better ones that more accurately model the tournament situation, but they involve much tougher and more complex computations. My little problem does almost as good a job, and it also is a good way to teach some probability techniques.
The game is heads up “poker” where each player receives a “card” with a random number between zero and one. Higher number wins. You and your opponent both start with X chips and both ante one chip each. We will now say that your opponent moves in every hand. You can only call or fold. If you fold, your opponent wins your dollar and a new hand is dealt. The game ends the first time you call. (You don’t finish the freezeout. If you did the math would be more difficult. And it wouldn’t make a big difference in the results.) If you continually fold, the last hand will have you playing all in for the two chips, one from you and one from your opponent, in antes.
If each hand was played in a vacuum, you should theoretically call anytime you had the best of that bet. With a lot of chips it would be approximately any card of 0.5 or higher. But since you know that he is constantly moving in, and that the game is over the first time you call, you should obviously wait for a much greater number if your stack is large.
But what about with a small stack? I am going to show how to do this problem when you have two or three chips. Then let others take it further on our “Two Plus Two Internet Magazine Forum.”
First, notice that if you don’t call until the end, you will be 50-50 to win that final two chip pot. Your EV will be one chip. But suppose instead of one chip, you start the hand with two chips (before anteing). Your opponent then moves in which in this case will be for just one chip since that is all you have left (after anteing). What does the number on your card need to be to call?
The answer is that you should call the one chip bet as long as your card is above .25. There are four chips in the pot and if you have a one-fourth or better chance to win, then your EV surpasses what it would be if you folded (which we have already shown to be one.)
At this point, you might think this is pretty trivial since you were getting 3-to-1 odds on his bet and you call with a 25 percent chance or more. But that simplicity doesn’t hold as the stack gets larger. (This should be obvious since we know that you shouldn’t take a coin flip with large stacks).
To figure out the proper strategy with three chips, two plus the ante, we have to first calculate our EV if we fold down to two chips. So when we do have two chips, it’s correct to call 75 percent of the time and fold the remaining 25 percent since we are calling whenever our random number is 0.25 or higher. When we fold, our EV is one chip. When we call, we will have a hand ranging from 0.25 to 1.00. This means that when we call, our average hand is 0.6250 since that is the midpoint between 0.25 and 1.00. So when we call, on average, our EV will be 0.6250 times the four chip pot or 2.5 chips.
In other words, if we do in fact play, our EV is now 2.5 chips. Thus our overall EV when starting the hand with two chips is 75 percent of 2.5 chips plus 25 percent of one chip. That equals 2.125 chips.
2.125 chips = (0.75)(2.5 chips) + (0.25)(1 chip)
Notice that this is more than the two chips we started with. As would be expected given his “bad” play.
The strategy with three chips should now be clear. Call if your EV is better than 2.125 chips. Otherwise wait. Since there will be six chips in the pot if you call, you need to have a hand better than 2.125 divided by 6 or 0.3542. Notice that this number is different than the .3333 which corresponds to pot odds of 4-to-2 that the pot would offer you.
To figure out your EV with three chips, first realize that
• The chance you would fold twice is (0.3542)(0.25) = 0.08855;
• The chance you would fold once is (0.3542)(0.75) = 0.26565; and
• The chance you wouldn’t fold at all is 1-0.08855 - 0.26565 = 0.6458
and that your EV if you don’t fold is 6 chips times (1.3542/2) [where 1.3542/2 is the midpoint between 0.3542 and 1.0] or 4.0626 chips.
So your total EV with three chips is
(0.6458)(4.0626 chips) +(0.26565)(2.5 chips) +(.08855)(1 chip)
That's about 3.4 chips. So this means that with four chips you would call with a real number that was 3.4 divided by 8, or about .425 or higher. That's still a significant underdog.
I will leave it to our Two Plus Two readers and posters to extend the chart. But until someone does, you might want to guess how big your stack needs to be to make it right to fold as a small favorite.