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.)

[pzhon]

This is a very poor model of poker, and it does not allow you to conclude what you would like to conclude.

[img]/images/graemlins/diamond.gif[/img] You are UTG. You act first, getting no information from your opponents' actions.

[img]/images/graemlins/diamond.gif[/img] Your opponents collude. I'd be very upset if I pushed in a tournament, and the other players flipped up their hands to figure out which player should call me.

[img]/images/graemlins/diamond.gif[/img] Your opponents act stupidly. If the second highest hand is .5, there is no way this is going to beat a hand worth pushing, but you are assuming the owner (or someone with an even worse hand) will call anyway.

The first two factors dominate when your stack is small, and the third factor dominates when your stack is large. The difference between playing-UTG-against-cheating-idiots and poker, or more symmetric [0,1] games, is too large for you to conclude something about the relative value of chips in large stacks versus small stacks.

[David Sklansky]

The main thing I am trying to show is that you need a minimum number of chips to have the best of it even if you are utilizing skill, in at least some scenarios. That the concept isn't logically impossible.

This model is farfetched to make calculations easy. But I'm betting results would be similar if it wasn't as farfetched.

[wagon30]

"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."

What do you mean by this? To be favored over the field means, you expect to have best hand over 50% of the time. The ninth root of 1/2 is ~.926. Isn't this the limiting probability? If I make the stack arbitrarily large number, I will still call with this hand and above with a slight overlay. What is the meaning of big favorite?

[David Sklansky]

Wrong. Think.

[wagon30]

Oh I just reread the post. I didn't realize we could play another hand if we folded. I thought it was just a one hand scenario.

[Botchman]

[ QUOTE ]
Wrong. Think.

[/ QUOTE ]
David, isn't it a little late for old guys like you???
edit: I am very confused by this post.

[CityFan]

[rachkane]

The answer is x = 223. I can e-mail you the spreadsheet if you PM me.

OOPS, answer is 214
x = 214
total pot = 436 (10 + 213+213)
probability of folding (y) 0.002293
folding frequency (1/y)^(1/9) = 0.5090
calling frequency (1 - folding frequency) = 0.4910
expected return (EV) = (436*0.4910) = 214.07
plus EV? yes

[djames]

For x >= 1, the expected result of the hand is always break even at best. For x = 1, the expected loss is exactly zero. As x increases, so does the expected loss. However, the expected loss as a percentage of your starting stack declines.

I can e-mail a spreadsheet whereever you want.

pot to win = 10 + 2(x-1)

Needed win % for E[push]>=0 = (x-1)/(10 + 2(x-1)), or (stack left)/(pot to win)

Min hand range to achieve win % = [(x-1)/(10 + 2(x-1)]^(1/9) or (needed win%)^(1/9) ... call this m

Prob(Push) = 1 - m

Prob(win|push) = integral from m to 1 of y^9 dy, or simply (1/10)*(1-m^10)

E[chips after game] = (1-Prob(Push))(x-1) + P(push)xP(win|push)x(10+2(x-1))

Profit/Loss = E[chips after game] - x

For numerical verification, here are the values for x=2 and x=100.

x: 2
pot to win: 12
needed win%: 0.0833
m: 0.7587
Push %: 0.2413
P(win|push): 0.0937
E[chips]: 1.03
Loss: 0.97
Relative Loss: 48.5%

x: 100
pot to win: 208
needed win%: 0.4760
m: 0.9208
Push %: 0.0792
P(win|push): 0.0562
E[chips]: 92.02
Loss: 7.91
Relative Loss: 7.9%

From this information it would be trivial to express a closed form for any x. Typing this information in a web form was hard enough for me to do, so I'll leave that last step to you.

Interestingly the relative loss stabilizes very quickly. That is, after about 3905 chips, the relative loss stays at 6.6% up to x = a billion. This is because m converges.