Another Simulation That Sheds Light on Chips Changing Value
 

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.)

Re: Another Simulation That Sheds Light on Chips Changing Value
 

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.

Re: Another Simulation That Sheds Light on Chips Changing Value
 

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.

Re: Another Simulation That Sheds Light on Chips Changing Value
 

"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?

Re: Another Simulation That Sheds Light on Chips Changing Value
 

Wrong. Think.

Re: Another Simulation That Sheds Light on Chips Changing Value
 

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.

Re: Another Simulation That Sheds Light on Chips Changing Value
 

[ QUOTE ]
Wrong. Think.

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

Re: Another Simulation That Sheds Light on Chips Changing Value
 

http://www.bookshaker.com/images/backpage-happy-e.jpg

Re: Another Simulation That Sheds Light on Chips Changing Value
 

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

Re: Another Simulation That Sheds Light on Chips Changing Value
 

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.

Re: Another Simulation That Sheds Light on Chips Changing Value
 

Ok, I can't stand to see incorrect answers posted, even though I maintain that this is a poor thought experiment since playing-UTG-against-cheating-idiots is so different from poker.

Let value(n) be your equity with a stack of size n. value(1)=1.

The probability of winning with which you should push is value(n-1)/(2n+8), so that you are indifferent to folding to get a stack of n-1 worth value(n-1) and pushing to get a share of the 2n+8 pot. You should push with at least (value(n-1)/(2n+8))^(1/9).

value(n) = value(n-1) ((value(n-1)/(2n+8))^(1/9)) + Integral (2n+8) x^9 dx from (value(n-1)/(2n+8))^(1/9) to 1.

We can use this recursively to compute value(n).

n value(n)
------------
1 1
2 1.882862
3 2.755967
4 3.640068
5 4.542981
6 5.467867
7 6.415821
8 7.386923
9 8.380731
10 9.396536
11 10.433498
12 11.490724
13 12.567311
14 13.662373
15 14.775055
16 15.904540
17 17.050054

100 136.900389
1000 1755.195048

Re: Another Simulation That Sheds Light on Chips Changing Value
 

[ QUOTE ]
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.

[/ QUOTE ]
For that to be the case, you need something to break down with the standard argument that you can ignore the chips beyond your stack. Here, you violated symmetry by giving the player two clear disadvantages, the worst position and that the opponents collude. That is unacceptable.

I don't think you will show that in a reasonable situation, the player needs 30 BB or something similar to break even. (If you try, it will be very easy to discredit your advice as inapplicable to most tournament situations since stacks are often much shorter.) What you can easily show is that some strategies have a superlinear payoff. For example, if I know a player raises UTG to 4 BB with AA only, and will stack off, and I call in late position with 22 to hit a set, I break even with about 50 BB, and gain with more than 50 BB roughly in proportion with the amount my stack exceeds 50 BB. I may expect roughly 1 BB for every 12 BB of the effective stack size over 50 BB, and I may gain 3 times as much with 200 BB as I do with 100 BB. This superlinear payoff doesn't mean I am taking the worst of it when I have a small stack, since I am not forced to call for set value.

Re: Another Simulation That Sheds Light on Chips Changing Value
 

Glad to see the mathmetician in you trumped the crumudgean.

Re: Another Simulation That Sheds Light on Chips Changing Value
 

I'm actually glad you put this up in spite of my horrible wrong answer.

Re: Another Simulation That Sheds Light on Chips Changing Value
 

[ QUOTE ]
Ok, I can't stand to see incorrect answers posted, even though I maintain that this is a poor thought experiment since playing-UTG-against-cheating-idiots is so different from poker.

[/ QUOTE ]

Please show me the error of my ways!?!?

Oh, and why resort to unnecessary recursion when the closed form answer for any n is so easy to compute?

Also, at n=17, your results show that you expect to win chips after the game. This shouldn't be possible. That is there shouldn't be a chip stack beyond which you turn from being a loser on average to a winner on average. Can you explain these counterintuitive results of yours (with a less arrogant tone)?

Another clue your results aren't correct. Notice that you expect to gain 755 chips with a starting stack of n=1000. However, even if you pushed every single hand and won, you could only obtain 2008 chips. How in the world can you rationalize your results to be correct here given that you will never come close to pushing every hand!?!? No chance you on average increase your stack 75% against 9 players all covering you!

Thanks.

[ QUOTE ]

n value(n)
------------
1 1
2 1.882862
3 2.755967
4 3.640068
5 4.542981
6 5.467867
7 6.415821
8 7.386923
9 8.380731
10 9.396536
11 10.433498
12 11.490724
13 12.567311
14 13.662373
15 14.775055
16 15.904540
17 17.050054
100 136.900389
1000 1755.195048

[/ QUOTE ]

Re: Another Simulation That Sheds Light on Chips Changing Value
 

OK, I was gonna go home from work, but I'll tack this on. At x=1, it's true, there is no value in folding. We need to get to a point where there could be value in folding so that the next iteration (x-1) may give us a better chance of doubling up.

I think what we're looking for is the chances you'll have

1) random winning hand
2) with a chip stack you can push
3) where you'll be +EV
4) will get doubled up + 10 chips

There has to be a stack size where your push is profitable. The antes, added to the probability of having the best hand, make it profitable at some stack size. Granted, the stack size is probably pretty high because there's 10 people at the table.

OK everyone, have a good weekend.

Re: Another Simulation That Sheds Light on Chips Changing Value
 

Actually, no, I believe the property of this game that makes it a loser for us with any chip stack is that your opponents can collude in the sense that the best hand of 9 will always play against us. The case for x=1 is trivial. The case for x=2 has the best chance at being a winning game since the antes are largest in propotion to our stack. However, even the OP shows that when x=2 the game is a loser.

Perhaps there's something with the game setup that I'm missing, but these posts I've read make no sense whatsoever. The growth rate in Phzon's EVs seems impossible. At some point it looks as if the expected chip stack after the game will even be larger than the possible pot size! Of course this is hard to tell by looking at a recursive solution. If anyone cares to use his formulae to see if this happens, that would be interesting.

Oh, and while I still believe the steps of my post are correct, the sample numbers I gave might be off. I may have typed (1/10)(1-x^9) in an Excel formula rather than (1/10)(1-x^10) though I may be wrong. I'm not at the computer I used to construct the solution, and I don't care to do it again. Anyone wanting to use these (seemingly useless) figures should check them.

Re: Another Simulation That Sheds Light on Chips Changing Value
 

[ QUOTE ]
[ QUOTE ]
Ok, I can't stand to see incorrect answers posted, even though I maintain that this is a poor thought experiment since playing-UTG-against-cheating-idiots is so different from poker.

[/ QUOTE ]

Please show me the error of my ways!?!?

[/ QUOTE ]
You sound like you don't believe it is possible that I am right and you are wrong. Would you care to wager $1000 on which of us is right? <font color="white">He made multiple minor errors regarding the expected value of a push as well as a major modelling error on the player's decision.</font>
[ QUOTE ]
Oh, and why resort to unnecessary recursion when the closed form answer for any n is so easy to compute?

[/ QUOTE ]
One can compute the integral, but I don't see how to eliminate the recursion.

value(4) =
((7840000 + 9 2^(2/9) 5^(2/3) 7^(8/9) ((9800 + 81 2^(4/9) 35^(8/9) ((8 + 2^(7/9) 3^(8/9)))^(1/9) + 108 2^(2/3) 35^(8/9) ((3 ((8 + 2^(7/9) 3^(8/9)))))^(1/9)))^(10/9))/4900000)

Good luck simplifying that, or the increasingly complicated expressions produced by the recursion.

[ QUOTE ]
Also, at n=17, your results show that you expect to win chips after the game. This shouldn't be possible.

[/ QUOTE ]
You haven't given an argument for that. You have just stated it repeatedly, despite contrary statements by Sklansky and by me.

[ QUOTE ]

Can you explain these counterintuitive results of yours (with a less arrogant tone)?... Another clue ... How in the world can you rationalize ...

[/ QUOTE ]
I've already given an explanation (which agreed with the comments made by Sklansky), but I get the feeling you will call anything I say arrogant until you recognize your own error. Will you owe me $1000 at that point, or just an apology?

I get the feeling that you are wrong infrequently, at least on the essentials, but that you haven't realized that when you disagree with someone who is wrong even less frequently, you will be wrong most of the time. So, you haven't even bothered to check your work, even though a professional mathematician has told you that you are wrong, and come up with a very different answer. Now, which of us is displaying arrogance = unwarranted pride, and which of us is rightly confident? <font color="white">I'm careful enough that I checked my work even though I was confident of being right.</font>

Re: Another Simulation That Sheds Light on Chips Changing Value
 

I think you made the same mistake I did when reading the post. When David highlighted the game ends when you play a hand, he meant make an additional bet besides the ante. Which is clear from his post, but I just glossed over that statement.

Thus, it should be clear a big stack will be able to play this +EV, because he will have time to wait for an excellent hand, and when he gets it, he will always get action.

Edit:
Look at your solution in this light. With a stack of 3, you will be willing to play a hand that isn't justified by your immediate odds, because you know when your stack falls by 1 unit you will be in a -EV situation. So you don't need 1/7 winning chances, you will accept less than that, because falling to 2 will force you to play a -EV game.

Re: Another Simulation That Sheds Light on Chips Changing Value
 

[ QUOTE ]
... When David highlighted the game ends when you play a hand, he meant make an additional bet besides the ante...

Thus, it should be clear a big stack will be able to play this +EV, because he will have time to wait for an excellent hand, and when he gets it, he will always get action.

[/ QUOTE ]
While that is true, it is not an excuse for not seeing that the expected gain is positive on the first hand. The total expected gain is the sum of the probability of blinding down to n chips times the expected gain on the first hand with n chips. The sum can't be positive without the expected value of some of the individual hands being positive.

If n=1,000,000, you could fold except when you are dealt over .999. It's a reasonable approximation to say that you win all of those but ignore the blinds, so you win 1,000,000 1/1000 of the time, and lose 1 chip .999 of the time, for a net gain of about 999 chips. That may not be optimal, but it shows that you can gain chips.

Re: Another Simulation That Sheds Light on Chips Changing Value
 

Agreed. Looking at his notes carefully now, I see that his P(win/push) = (1/10)(1-m^10), which can't be true, because it is maximized at (1/10), obviously, when you push less than every hand that you will be able to win more than 1/10 of the time.

I kept getting that result too when I looked at the answer. I think that P(win/push) should be divided by 1-m. Although, I'm not sure, because I'm not a professional mathematician.

But I was trying to help out, because I don't have enough math skills to Lord over someone how much better at math I am. And because I know what its like when someone tries to make me feel small.

Re: Another Simulation That Sheds Light on Chips Changing Value
 

[ QUOTE ]
I think you made the same mistake I did when reading the post. When David highlighted the game ends when you play a hand, he meant make an additional bet besides the ante. Which is clear from his post, but I just glossed over that statement.

[/ QUOTE ]

You are correct. As this is a poker theory forum, I likened the game to poker and assumed that if we play a hand and win the game is not over. After rereading, I clearly see the sentence in all caps. I also believe this makes the analogy less similar to poker than it even was before.

Re: Another Simulation That Sheds Light on Chips Changing Value
 

[ QUOTE ]
I think that P(win/push) should be divided by 1-m. Although, I'm not sure, because I'm not a professional mathematician.

[/ QUOTE ]
I don't recall using my qualifications instead of an argument. I add it afterwards to illustrate how ridiculous it is for someone to be so sure I am wrong that he is unwilling to rethink his position. See the comments I made in white. My arguments in this thread and elsewhere stand on their own.

[ QUOTE ]
I don't have enough math skills to Lord over someone how much better at math I am. And because I know what its like when someone tries to make me feel small.

[/ QUOTE ]
Reread the thread in order.
[img]/images/graemlins/diamond.gif[/img] Sklansky posted a question.
[img]/images/graemlins/diamond.gif[/img] I gave a correct partial qualitative analysis within 30 minutes.
[img]/images/graemlins/diamond.gif[/img] Djames gave a mass of incorrect calculations leading to a contrary qualitative answer.
[img]/images/graemlins/diamond.gif[/img] I said that incorrect answers had been posted, and I gave a nearly complete and correct quantitative solution. (Asymptotics remain to be determined.)
[img]/images/graemlins/confused.gif[/img] Djames responded by suggesting that I am arrogant, clueless, and irrational, but he offered only the argument that my conclusion disagreed with his intuition.
[img]/images/graemlins/diamond.gif[/img] I said I'm willing to bet that I'm right, and that it is a bad habit to assume you are right when a professional mathematician corrects you.
[img]/images/graemlins/diamond.gif[/img] You tell me I am trying to make djames feel small.

I would object less to the last if you had objected to the abuse he gave me when I posted a correct solution. As is, it looks completely biased and out of place. I had far more reason to attack him than he had to attack me, but I was far more polite, yet you criticized only my tone.

This is a complete waste of my time.

Re: Another Simulation That Sheds Light on Chips Changing Value
 

When I reread the thread in the proper order, I realize my comments were out of line. I apologize.

Re: Another Simulation That Sheds Light on Chips Changing Value
 

pzhon said:

[ QUOTE ]
Ok, I can't stand to see incorrect answers posted, even though I maintain that this is a poor thought experiment since playing-UTG-against-cheating-idiots is so different from poker.

Let value(n) be your equity with a stack of size n. value(1)=1.

The probability of winning with which you should push is value(n-1)/(2n+8), so that you are indifferent to folding to get a stack of n-1 worth value(n-1) and pushing to get a share of the 2n+8 pot. You should push with at least (value(n-1)/(2n+8))^(1/9).

value(n) = value(n-1) ((value(n-1)/(2n+8))^(1/9)) + Integral (2n+8) x^9 dx from (value(n-1)/(2n+8))^(1/9) to 1.

We can use this recursively to compute value(n).

n value(n)
------------
1 1
2 1.882862
3 2.755967
4 3.640068
5 4.542981
6 5.467867
7 6.415821
8 7.386923
9 8.380731
10 9.396536
11 10.433498
12 11.490724
13 12.567311
14 13.662373
15 14.775055
16 15.904540
17 17.050054

100 136.900389
1000 1755.195048

[/ QUOTE ]

I just skimmed through this thread. What pzhon said looks right to me. (Us mathematicians have a habit of agreeing with each other).

Let me just add that it is pretty easy to see that the limit as n goes to infinity of [value(n)]/n is exactly 2. That is, with a big stack you are virtually certain to double up, so that if n is large, then value(n) is `almost' 2n. You can simply wait for a `monster' hand, namely a number 1-f(n) where the function f is chosen so that (a) you are virtually certain to get such a hand after anteing off only a negligible proportion of your stack, and (b) you are virtually certain to win the hand, the crucial point being that one of your colluding/idiot opponents is compelled to call, as the original formulation specified, and their range of hands is independent of n.

I'll let others clarify and quantify.

Re: Another Simulation That Sheds Light on Chips Changing Value
 

[ QUOTE ]
When I reread the thread in the proper order, I realize my comments were out of line. I apologize.

[/ QUOTE ]
Apology accepted. I'll add that I've valued your contributions to this thread, and I just thought the last comment was off target.

Re: Another Simulation That Sheds Light on Chips Changing Value
 

All right, David, here's an answer to your little problem. I'm not an expert mathematician, so this wasn't easy. I get the same results as pzhon, so they must be right! But I want to give a lengthier explanation than he gave. Others feel free to correct any errors or logical gaffes.

First, to restate the problem in brief:

This is a 10-handed no limit [0,1] card game. You begin with x chips. Each player antes 1 chip. Each player is dealt one card. After you see your card, you can move all-in, or fold. If you move all-in, the one other player with the best hand will call. If you fold, you are given the same choice on the next hand. The game is over when you have played one hand.

With optimal play, what is your EV for various values of x? Give a formula, if possible, and the EV for each value up to at least 100.

Because you are only permitted to play one hand, your choice at the beginning of any hand should be the option that has the greater EV--playing that hand, or waiting for the next. Your EV for playing the current hand is calculated as follows.

Let A_n be the event in which player n wins the current hand, and P(A_n) be the probability of player n winning. Let us call you player 1. What we want to determine is P(A_1), if you play. If you play a random card, P(A_1) = 1/10. If you play a hand equal to the value y, your probability of beating one other hand of unknown value, pairwise, is y. Since you must beat 9 opponents pairwise to win the hand, and these are independent events, your probability of winning is the product of the probabilities of beating each opponent pairwise, so P(A_1) = y^9.

Winning the hand yields 2(x-1) + 10 = 2x + 8. Losing the hand yields 0. Therefore, EV(play_x) = P(A_1)*(2x + 8) = (y^9)(2x + 8).

On the other hand, EV(fold_x) = EV(x-1), where EV(x-1) is the expected value of holding x-1 chips, before you have seen your card y on that round.

The breakeven point occurs when EV(play_x) = EV(fold_x) = EV(x-1).

Let z = EV(x-1). Then EV(play_x) = EV(x-1) is equivalent to (y^9)(2x + 8) = z, or y^9 = z/(2x + 8), or y = 9rt(z/(2x + 8)) = (z/(2x + 8))^(1/9). Therefore, optimal play is to play the current hand whenever holding a card y of larger than this value, and otherwise to fold.

The amount one nets from playing a card y whenever the play is breakeven or better is determined as follows. Let us define a function f(x,y) calculating the EV of the yield from playing card y with a stack x. Then EV(play_x) = f(x,y) = (y^9)(2x + 8), as stated earlier. Since we will play whenever y is in the range [(z/(2x + 8))^(1/9),1], and y has an equal probability of holding any value in the range [0,1], the yield from the times we play can be represented by the area under the curve plotting the value of the function against values of y. This is equal to the integral of f(x,y) dy for the range [(z/(2x + 8))^(1/9),1]
= (2x + 8)([1/(9+1)]{1^(9+1) - [z/(2x + 8)]^(1/9)^(9+1)})
= (2x + 8){1 - [z/(2x + 8)]^(10/9)}/10.

Let us define a second function g(x,y) calculating the EV of the yield from folding card y with a stack x. Then EV(fold_x) = g(x,y) = z, as stated earlier. Since we will fold whenever y is in the range [0,(z/(2x + 8))^(1/9)], and y has an equal probability of holding any value in the range [0,1], the yield from the times we fold is simply {[z/(2x + 8)]^(1/9)}z.

The total yield for optimal play of starting stack x is thus the sum of these two partial solutions,
[(z/(2x + 8))^(1/9)]z + (2x + 8){1 - [z/(2x + 8)]^(10/9)}/10.

A starting stack of 1 is a special case. You cannot fold, because you must ante your only chip. You can only play. Therefore, as noted above, your total EV is calculated as 1.

Since the solution is iterative, the easiest way to calculate all of this is with a spreadsheet. I get the following results:

Stack (x) Breakeven y EV for folds EV for plays Total EV Profit (loss)
1 NA 0 1 1 0
2 0.7587 0.7587 1.1241 1.8829 -0.1171
3 0.8002 1.5066 1.2493 2.7560 -0.2440
4 0.8225 2.2667 1.3733 3.6401 -0.3599
5 0.8373 3.0478 1.4952 4.5430 -0.4570
6 0.8482 3.8532 1.6147 5.4679 -0.5321
7 0.8567 4.6842 1.7316 6.4158 -0.5842
8 0.8637 5.5410 1.8459 7.3869 -0.6131
9 0.8695 6.4230 1.9577 8.3807 -0.6193
10 0.8746 7.3295 2.0671 9.3965 -0.6035
11 0.8790 8.2594 2.1741 10.4335 -0.5665
12 0.8829 9.2119 2.2788 11.4907 -0.5093
13 0.8864 10.1859 2.3814 12.5673 -0.4327
14 0.8896 11.1804 2.4820 13.6624 -0.3376
15 0.8926 12.1945 2.5805 14.7751 -0.2249
16 0.8952 13.2273 2.6773 15.9045 -0.0955
17 0.8977 14.2778 2.7722 17.0501 0.0501
18 0.9000 15.3454 2.8655 18.2109 0.2109
19 0.9022 16.4292 2.9571 19.3863 0.3863
20 0.9042 17.5285 3.0471 20.5757 0.5757
21 0.9061 18.6426 3.1357 21.7784 0.7784
22 0.9078 19.7710 3.2229 22.9939 0.9939
23 0.9095 20.9129 3.3087 24.2216 1.2216
24 0.9111 22.0679 3.3932 25.4611 1.4611
25 0.9126 23.2353 3.4765 26.7118 1.7118
26 0.9140 24.4148 3.5585 27.9733 1.9733
27 0.9154 25.6058 3.6394 29.2452 2.2452
28 0.9167 26.8079 3.7192 30.5272 2.5272
29 0.9179 28.0208 3.7979 31.8187 2.8187
30 0.9191 29.2439 3.8756 33.1195 3.1195
31 0.9202 30.4769 3.9523 34.4292 3.4292
32 0.9213 31.7195 4.0280 35.7476 3.7476
33 0.9223 32.9714 4.1029 37.0742 4.0742
34 0.9233 34.2321 4.1768 38.4089 4.4089
35 0.9243 35.5016 4.2498 39.7514 4.7514
36 0.9252 36.7793 4.3221 41.1014 5.1014
37 0.9261 38.0652 4.3935 42.4587 5.4587
38 0.9270 39.3589 4.4641 43.8230 5.8230
39 0.9278 40.6602 4.5340 45.1942 6.1942
40 0.9286 41.9688 4.6031 46.5719 6.5719
41 0.9294 43.2846 4.6715 47.9561 6.9561
42 0.9302 44.6073 4.7393 49.3466 7.3466
43 0.9309 45.9367 4.8063 50.7431 7.7431
44 0.9316 47.2727 4.8727 52.1455 8.1455
45 0.9323 48.6151 4.9385 53.5536 8.5536
46 0.9330 49.9636 5.0036 54.9672 8.9672
47 0.9336 51.3182 5.0682 56.3863 9.3863
48 0.9342 52.6786 5.1321 57.8107 9.8107
49 0.9349 54.0447 5.1955 59.2402 10.2402
50 0.9355 55.4164 5.2584 60.6747 10.6747
51 0.9360 56.7935 5.3207 62.1141 11.1141
52 0.9366 58.1759 5.3824 63.5583 11.5583
53 0.9371 59.5634 5.4437 65.0071 12.0071
54 0.9377 60.9560 5.5044 66.4604 12.4604
55 0.9382 62.3535 5.5646 67.9182 12.9182
56 0.9387 63.7558 5.6244 69.3802 13.3802
57 0.9392 65.1628 5.6837 70.8465 13.8465
58 0.9397 66.5744 5.7426 72.3169 14.3169
59 0.9402 67.9904 5.8010 73.7914 14.7914
60 0.9406 69.4108 5.8589 75.2698 15.2698
61 0.9411 70.8356 5.9164 76.7520 15.7520
62 0.9415 72.2645 5.9736 78.2380 16.2380
63 0.9420 73.6975 6.0303 79.7277 16.7277
64 0.9424 75.1345 6.0865 81.2211 17.2211
65 0.9428 76.5755 6.1425 82.7179 17.7179
66 0.9432 78.0203 6.1980 84.2183 18.2183
67 0.9436 79.4689 6.2531 85.7220 18.7220
68 0.9440 80.9212 6.3079 87.2291 19.2291
69 0.9444 82.3771 6.3623 88.7394 19.7394
70 0.9448 83.8366 6.4163 90.2530 20.2530
71 0.9451 85.2996 6.4700 91.7697 20.7697
72 0.9455 86.7660 6.5234 93.2894 21.2894
73 0.9458 88.2358 6.5764 94.8122 21.8122
74 0.9462 89.7089 6.6291 96.3380 22.3380
75 0.9465 91.1852 6.6815 97.8667 22.8667
76 0.9468 92.6647 6.7335 99.3982 23.3982
77 0.9472 94.1473 6.7853 100.9326 23.9326
78 0.9475 95.6330 6.8367 102.4697 24.4697
79 0.9478 97.1217 6.8878 104.0095 25.0095
80 0.9481 98.6133 6.9387 105.5520 25.5520
81 0.9484 100.1079 6.9892 107.0971 26.0971
82 0.9487 101.6054 7.0395 108.6448 26.6448
83 0.9490 103.1056 7.0894 110.1951 27.1951
84 0.9493 104.6087 7.1391 111.7478 27.7478
85 0.9496 106.1144 7.1886 113.3030 28.3030
86 0.9499 107.6229 7.2377 114.8606 28.8606
87 0.9501 109.1339 7.2866 116.4205 29.4205
88 0.9504 110.6476 7.3352 117.9829 29.9829
89 0.9507 112.1638 7.3836 119.5475 30.5475
90 0.9509 113.6826 7.4317 121.1143 31.1143
91 0.9512 115.2038 7.4796 122.6834 31.6834
92 0.9515 116.7275 7.5273 124.2547 32.2547
93 0.9517 118.2536 7.5746 125.8282 32.8282
94 0.9519 119.7820 7.6218 127.4038 33.4038
95 0.9522 121.3128 7.6687 128.9815 33.9815
96 0.9524 122.8458 7.7154 130.5612 34.5612
97 0.9527 124.3812 7.7619 132.1430 35.1430
98 0.9529 125.9187 7.8081 133.7269 35.7269
99 0.9531 127.4585 7.8542 135.3126 36.3126
100 0.9534 129.0004 7.9000 136.9004 36.9004

The pattern is very interesting. Holding one chip is breakeven. Holding two chips is a loss, and the loss grows up to a stack size of 9. Then the loss starts declining, until you show a profit with a stack size of 17, and the profit continues to grow after that.

Re: Another Simulation That Sheds Light on Chips Changing Value
 

All that math and then you lose to an idiot who limped with K-2s UTG...oh wait, our model doesn't allow that.