In this article I'll 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 part of your stack. This is the kind of situation you'll 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 hands with a slightly positive EV instead of betting all your chips on them, the situation reverses when you are short stacked. To further illustrate this, I've developed an insightful math problem that does a great job of demonstrating this phenomenon. There are other problems that explain this situation more precisely, but they involve very advanced and complex computational knowledge. My little problem does almost as good a job, as well as being a great tool for teaching some probability techniques.
The game is a heads-up poker game, where each player receives 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 you both pay 1 ante chip. Now let's say 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 continually fold, the last hand will have you all-in for the 2 chips, one yours and the other your opponent's, only from before.
If every hand was played blind, you should theoretically call with whichever hand makes you the best bet. With lots of chips, that threshold for calling would be any card like 0.5 or above. But since you know that he's going all-in all the time, and that the game is over as soon as you call, you should obviously wait for a higher number if your stack is large.
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 pay until the end, the chances are 50%-50% of winning the final pot of 2 chips. Your EV will be 1 chip. But suppose that 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 pay (after giving the ante). What number must your card be for the call to be worthwhile?
The answer is that you must pay 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 (a quarter), your EV to call is higher than it would be if you folded the hand (which we've 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 call with a chance of 25% or more. However, this simplicity doesn't hold true as your stack increases. 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 keep 2 chips. So, when we have 2 chips, it's correct to call 75% of the time and fold the other 25%, since we're going to 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'll 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. So 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)
Note that this is more than the 2 tokens we had at the start.
The strategy with 3 chips should now be clearer. Call if your EV is better than 2,125 chips, otherwise wait. Since the pot will have 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 pot odds of 4:2 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 that you won't fold once 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)
The result is approximately 3.31 tokens. This means that with 4 tokens you would have to pay with a real number which would be 3.31 divided by 8, i.e. 0.41 or more.
I'll leave it to the forum readers to continue the table. Until someone does, you might want to try guessing how big a stack needs to be to make folding correct by not being 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"
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