Another Simulation That Sheds Light on Chips Changing Value

[David Sklansky]

Expert mathmeticians may not need a computer for this:

You are playing a ten handed no limit game with a stack of x chips. Each player is dealt one of those provebial real numbers from 0 to 1. Each player antes one dollar. All players have you covered.

After you see your card you can move in your (x-1) chips or fold and play the next hand with those x-1 chips. If you do move in you will win the nine dollars in antes plus your ante plus another x-1 chips, presumably from the second best hand, if your hand is the best of the ten players. In other words if you start the hand before the antes with $18, and pick this hand to move in you will rake in a $44 pot if you have the highest of the ten numbers.

ONCE YOU PLAY A HAND THE GAME IS OVER.

The question is what is your profit or loss EV for various values of x? If there is an easily derivable formula for x that would be nice. If not, a chart for various x's up to at least 100 would be good.

To show you that I actually understand some of the math behind these questions let me show you how to start.

Notice that if x is one you can only ante. And that you will have a one tenth chance of having the best hand and winning the ten dollar pot. So you will have an EV of break even. Put differently the EV of your bankroll after this hand is one dollar.

What about if your starting stack size is $2? How good a hand do you need to bet your one dollar? The answer is that you need a one in twelve chance of winning. The pot will be $12 and a 1/12 chance of winning it will give you a bankroll EV of one dollar. With worse hands you are better off taking a chance on the next hand. Notice though that unless your chances are better than one sixth, your bankroll EV will be below two dollars and your overall EV will be negative.

How often will you bet that 2nd dollar? The answer is your hand must have the value, call it y, such that y to the ninth power is larger than 1/12. I think that's about .75. So you will bet about one quarter of the time. (Notice that if you bet it everytime you would win one tenth of the time and wind up with an average of $1.20 or an 80 cent loss. However, since you bet only when your chances are 1/12 or above, your average chances of winning are much better than ten percent.)

If you do bet that 2nd dollar your bankroll EV will range from one dollar to 12. I think you need calculus to average it out exactly. Anyway your overall bankroll average with a $2 starting stack is one dollar times the probability of folding (the ninth root of 1/12) plus, the chances of playing, times your bankroll EV if you do. So that's your TWO DOLLAR BANKROLL EV (TDBE) which I think is a bit under two dollars. Making the final answer negative.

With three dollars to start, you bet your extra two dollars if your chances of winning times the $14 dollar pot is greater than TDBE. I'm guessing you need something like a one in eight chance of having the best hand. Otherwise you give up the dollar.

Hopefully you see how to continue. Notice that with very big stacks there is no reason to play without being favored over the field. With monstrous stacks you should play only as a big favorite (although we now move into an unrealistic scenario since real players will probably simply let you win the ante.)