Poker Games Betting Strategy

Suppose you are playing in a holdem or lowball tournament and find yourself down to a very short stack. Say you have just enough to call the bigs blind. And there is just one or two hands before you will be forced to blind it.

I have frequently been asked in this situation if it is correct to loosen up your calling requirements. Many players have suggested to me that since you have to blind it anyway on the next hand, you should play any hand that is better than the average random hand you will get on your blind.

Others say that the upcoming hand in any poker games makes no difference. It should not influence you to take the worst of it on the present hand.

While mulling over this problem, I realized that a simply stated mathematical question can be to solved to give some insight into this dilemma.

Suppose you have $100 in your pocket with no way of getting more for the time being. Meanwhile, in five minutes, you will be forced to bet this $100 with someone on a sports event or some such thing.

You can only get out of this bet, if you play with proper betting strategy like if you lose the $100 before he shows up. (However, if you have increased your bankroll, your bet still be exactly $100.)

The question is – what should you do if someone else offers to bet $100 on something right now?

If you take him up on it, you will either wind up with $200 ($100 of which you will have to bet back in five minutes, so that your final bankroll will be either $100 or $300) or you will go broke, which eliminates your obligation to make the second bet.

Before deciding whether to take the first bet or whether to pass it and just wait to make the second bet, we have to know the chances of winning each bet.

Call your chances of poker winning the first bet “X” and your chances of winning the second bet (if you get the chance to make it) “Y.”

To solve this problem we must calculate the expected value of your bankroll and pot odds if you take the first bet, and also if you pass it. The second option is easy to figure.

If you pass the first bet, and only make the second bet, the expected value of your bankroll is simply $200 Y. What about would be your betting strategy if you do take the first bet?

Now you will wind up with either $0 (chances of this are [1-X], $300 (chances of this are XY), or $100 (chances of this are X [1-Y]. So the total expected value of your bankroll if you take the first bet is:

0(1-X) + 300XY + 100X [1-Y]

This reduces to:

200XY + 100X

the expected value of your bankroll if you accept the first $100 bet and then make the second bet if you win it. Notice that both options give you the exact same expected value if.

200Y = 200XY +100X

Solving for X:

200Y = X(200Y +100)

X =      200Y
        200Y + 100
=      2Y
     2Y + 1

So if the probability of winning the first bet is greater than:

2Y
2Y + 1

where Y is the probability of winning the second bet, then you should accept it.

Tournament Play

Razz Problems

FAQ