Another Simulation That Sheds Light on Chips Changing Value

[pzhon]

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

[pzhon]

[ 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.

[David Sklansky]

Glad to see the mathmetician in you trumped the crumudgean.

[rachkane]

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

[djames]

[ 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 ]

[rachkane]

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.

[djames]

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.

[pzhon]

[ 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>

[wagon30]

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.

[pzhon]

[ 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.