When Antes are “eating” your Stack

In this article, I will use a simple scenario to show how you can play weak hands in a tournament in situations where choosing not to play the hand means losing a significant portion of your stack. This is the type of situation you will encounter if you decide to play a lot of tournaments.

The sad fact is that even if you are playing against very weak players, and therefore you would normally fold slightly positive EV hands rather than going all in on them, the situation is reversed when you are short stacked. To illustrate this further I have developed a clever mathematical problem that does a great job of demonstrating this phenomenon. There are other problems that explain this situation more accurately, but they involve very advanced and complex computational knowledge. My little problem does almost as good a job, and is also a great tool for teaching some probability techniques.

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The game is heads-up poker, where each player is dealt a “card” with a random number between 0 and 1. The highest number wins. You and your opponent start with X amount of chips and both pay 1 ante chip. Let’s say now that your opponent goes all-in every hand, your only options are to 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. If you fold continuously, the last hand will have you all-in for the 2 chips, one yours and one your opponent’s, from the antes only.

If every hand were played blind, you should theoretically call with whatever hand makes you better off betting. With a lot of chips, that threshold for calling would be any card like 0.5 or higher. But since you know he's always going all-in, and the game is over as soon as you call, you should obviously wait for a higher number if your stack is big.

But what if your stack is small? I'll show you how to solve this problem when your stack is 2 or 3 chips. Other values can be calculated if you wish!

What to do with a small stack

First, note that if you don't call all the way, the odds are 50%-50% of winning the final 2-chip pot. Your EV will be 1 chip. But suppose instead of 1 you start the hand with 2 chips (before putting the antes in the pot). Your opponent then bets all in, in which case you only have 1 chip left to call (after putting the ante in). What number do you have to have on your hole card for the call to be worthwhile?

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The answer is that you should call your 1 chip when your card is higher than 0.25. There are 4 chips in the pot, so if you have a chance equal to or greater than ¼ (one quarter), your EV to call is higher than it would be if you folded the hand (which we have already shown to be 1).

You might think this conclusion is obvious, since you had 3:1 odds on your opponent's bet and should have called with a chance of 25% or better. However, this simplicity doesn't hold true as the stack gets bigger. This should be obvious, since we know that you shouldn't risk a coin flip against a big stack.

To develop the best strategy with 3 chips, 2 plus 1 ante, we first need to calculate our EV if we decide to fold the hand to get 2 chips. So, when we have 2 chips, it is correct to call 75% of the times and fold the other 25%, since we will call whenever our random number is equal to or greater than 0.25. When we fold, our EV is 1 chip. When we call, we will have a hand with a range between 0.25 and 1.00. This means that when you call, your average hand will be 0.6250, since this is the midpoint of 0.25 and 1.00. Therefore, when we call, our EV will be, on average, 0.6250 times the 4 chips in the pot, or 2.5 chips.

In other words, if we decide to play, our EV is now 2.5 chips. So overall, our EV at the start of the hand with 2 chips is 75% of 2.5 chips plus 25% of 1 chip. This equals 2,125 chips.

2,125 tokens = (0.75) * (2.5 tokens) + (0.25) * (1 token)

Realize that this is more than the 2 tokens we had at the beginning.

The 3-chip strategy should now be clearer. Call if your EV is better than 2.125 chips, otherwise hold. Since the pot will be 6 chips if you call, you need to have a hand better than 2.125 divided by 6, or 0.3542. Note that this number is different from 0.3333, which corresponds to the 4:2 pot odds that the pot would offer you.

To calculate your EV with 3 tokens, note that:

• The chance of you folding twice is: (0.3542) * (0.25) = 0.08855;
• The chance of you folding once is: (0.3542) * (0.75) = 0.26565; and
• The chance of you not folding at all is: (1 – 0.08855 – 0.26565) = 0.6458

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 3 tokens is:

(0.6458) * (4.0626 tokens) + (0.26565) * (2.125 tokens) + (0.08855) * (1 token)

Calculating, we get approximately 3.31 chips. This means that with 4 chips you would have to pay with a real number that would be 3.31 divided by 8, that is, 0.41 or greater.

I'll leave it to the forum readers to continue the table. Until someone does, you might want to try to guess how big a stack needs to be to make the correct fold if you're not the favorite in the hand.

Written by: David Sklansky

Editor's note: This article was originally published in 2008.

Translated and adapted from: Sklansky: When the Antes are “Eating you Up”