Another Simulation That Sheds Light on Chips Changing Value

[wagon30]

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.

[djames]

[ 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.

[pzhon]

[ 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.

[wagon30]

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

[thylacine]

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.

[pzhon]

[ 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.

[WRX]

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.

[Shandrax]

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