Card Sharks

[BeerMoney]

I would bet nothing, and wait for a better spot.. I would bet pretty small. I would hammer away on the 2's and the aces. Particulary once the last 2 and the last ace appear. The more cards we've seen, the better our chances.

[bobman0330]

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Bet everything and let it ride.

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I was going to disagree, but then I did the following math:

The EV of your fourth bet:
EV (4) = edge * bet + winning percentage * (EV (5) w/ winning bankroll) + losing percentage * (EV (5) w/ losing bankroll).
= (W-50%) * X + W((avg. edge) * (BR + X)) + (1 -W)((avg. edge) * (BR - X))
W = winning percentage; X = bet size; BR = bankroll

This equation is maximized by betting it all, even when it is a 50-50 proposition. (Most 8's are not even money because of prior cards).

I find this result surprising, as it this situation is exactly analogous to what Sklansky describes in TP4AP. It seems that the risk of not being able to take advantage of future +EV gambles is more than offset by the ability to take fuller advantage of those gambles by having a bigger roll. I guess in poker contexts, (assuming you're one of the few best players in the tournament), you'll usually have one of the biggest stacks, so having more money won't allow you to win more from future gambles. Not to hijack the thread, but does anyone else have comments on this situation v. TP4AP?

[pzhon]

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Not to hijack the thread, but does anyone else have comments on this situation v. TP4AP?

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I don't recall any contradiction with what TPFAP says.

TPFAP: The very best players should be willing to give up small edges when they come with a lot of variance.

Standard misinterpretation: I can become a great player by passing up large advantages.

TPFAP: Be prepared for your opponents to value survival more than chips, whether or not this is correct for them.

[Wacken]

My feeling says you should always go all-in, but i cannot think of mathematical proof right now.

[bobman0330]

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[ QUOTE ]
Not to hijack the thread, but does anyone else have comments on this situation v. TP4AP?

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I don't recall any contradiction with what TPFAP says.

TPFAP: The very best players should be willing to give up small edges when they come with a lot of variance.


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This game suggests that you should be willing to gamble it up even when your edge is tiny (or nonexistent). This is indeed a contradiction, at least on the surface.

[AaronBrown]

I think the answer is more complicated. phzon is correct that you should always bet all or nothing. If you're trying to maximize expected value, each dollar you have is equivalent. If it's right to put your first dollar in, it's right to put your last. Kelly maximizes the expected value of the logarithm of final result, not the expected value of final result.

Looking at things another way, it's true that losing everything on the first card costs you the positive expected value you get from additional cards. But that's true of every dollar you lose on the first card individually, whether you bet $1 or $1,000. Either it's right to bet, or it's not.

If the cards were independent, phzon would be correct that the maximum expected value comes from betting everything you have every time you have the edge (that is, every time you aren't looking at an eight). But they are not independent and that makes a difference.

To see this, suppose you played with three cards, two betting rounds. The first card is a seven. There are five possibilities among the 51*50 = 2,550 cards:

(1) The next card is an eight, meaning you win the first bet and make no further bet (200/2,550).

(2) The next card is higher than a eight, and the third card is higher still, so you win two bets (780/2,550).

(3) The next card is higher than a eight, and the third card is the same or lower than the second, so you win the first but lose the second bet (220/2,550).

(4) The second card is seven or lower and the third card is higher than the second, so you lose the first but win the second bet (960/2,550).

(5) The second card is seven or lower and the third card is the same or lower than the second, so you lose both bets (390/2,550).

You always bet everything on the last card if you have the advantage, so (3) and (5) are the same, you lose everything. Suppose you bet X on the first card, the seven. Your possible outcomes are:

(1) 1,000 + X (p = 200/2,550)
(2) 2,000 + 2*X (p = 780/2,550)
(4) 2,000 - 2*X (p = 960/2,550)

A quick inspection shows that this gives an expected value of (3,680,000 - 160*X)/2,550, so you want X = 0.

Although you have the advantage when the first card is a seven, you don't want to bet anything because your average opportunity if you lose the first bet is better than your average opportunity if you win. You want to preserve your dollars for the situation in which you have better opportunities. The same is true of a first card of nine. Eights you don't bet because you have a negative expected value. Two to six and ten to Ace you bet the maximum.

[bobman0330]

brilliant!

My TP4AP question remains though.

[fnord_too]

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brilliant!

My TP4AP question remains though.

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If you are talking about where he mentions passing up one bet today that is +EV for a better bet tomorrow that is more +EV, I don't think you had the option of parlaying your winnings from one to the other. That is, say you can bet $100 today at 52% and can bet $100 tmw at 80% (even money payout to you). Then playing both gives you EV of:
0.52*(100 + 0.8*100 - 0.2*100) - 0.48*100 = +35.20
vice EV of +60 if you just wait.

If you can parlay though, your EV is:
.52*.8*400 - 100 = +66.40

[randomstumbl]

I'm just spitballing here, but if the marginal utility of a dollar is less than a dollar then you have to discount chips won.

In other words, if losing 100 chips hurts your expected value more than winning a 100 chips, then marginal situations are actually money losers.

[pzhon]

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If the cards were independent, pzhon would be correct that the maximum expected value comes from betting everything you have every time you have the edge (that is, every time you aren't looking at an eight). But they are not independent and that makes a difference.

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That is a good point: In theory, there could be a sequence of dependent wagers so that losing would mean you get a better subsequent wager than if you win, and in that case you might not want to bet on a +EV wager. However, in your example, you should still bet everything and let it ride when you don't get an 8. It looks like your calculations are wrong.

[ QUOTE ]

To see this, suppose you played with three cards, two betting rounds. The first card is a seven. There are five possibilities among the 51*50 = 2,550 cards:

(1) The next card is an eight, meaning you win the first bet and make no further bet (200/2,550).


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Ok.

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(2) The next card is higher than a eight, and the third card is higher still, so you win two bets (780/2,550).

(3) The next card is higher than a eight, and the third card is the same or lower than the second, so you win the first but lose the second bet (220/2,550).


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First, I think you switched the labels on these cases.

Second, I get 4*(27+31+35+39+43+47)=888 ways to win twice and 4*(23+19+15+11+7+3)=312 ways to win and then lose. As should be expected, these total 888+312 = 1200 = 6*4*50, since there are 6 ranks that win for you, 4 cards in each of those ranks, and 50 possible third cards given the first two.

Did you assume aces are low?

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(4) The second card is seven or lower and the third card is higher than the second, so you lose the first but win the second bet (960/2,550).

(5) The second card is seven or lower and the third card is the same or lower than the second, so you lose both bets (390/2,550).


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I get 864=3*28+4*(31+35+39+43+47) and 286=3*22+4*(19+15+11+7+3). These add up to 864+286 = 1150 = 23*50, which should be expected since there are 23 losing second cards, 6 ranks of 4 cards minus the initial 7.

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You always bet everything on the last card if you have the advantage, so (3) and (5) are the same, you lose everything. Suppose you bet X on the first card, the seven. Your possible outcomes are:

(1) 1,000 + X (p = 200/2,550)
(2) 2,000 + 2*X (p = 780/2,550) <font color="blue">888/2550</font>
(4) 2,000 - 2*X (p = 960/2,550) <font color="blue">864/2550</font>

A quick inspection shows that this gives an expected value of (3,680,000 - 160*X)/2,550 <font color="blue">(3,704,000 + 248*X)/2,550</font> , so you want X = 0 <font color="blue">X=1000</font> .

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To simplify the calculation, you can pair the ranks (ace,2), (king,3), (queen,4), (jack,5), and (ten,6). In the first bet, you get the former versus the latter in each pair with equal probability, winning with one, and losing with one. The probability to win the second bet doesn't depend on whether you won or lost the first bet, so you break even on these cases, having a total of 1000+X + 1000-X = 2000 to bet if you get one of each. So, in these cases, X doesn't matter. All of the benefit comes from getting 7, 8, or 9. You do have a slightly better wager if you lost with a 7 than if you won with a 9, but there are 4 9s and 3 7s, so if you know a 7 or nine will come up, you want to bet as much as possible. Finally, if you know an 8 will come up, you want to bet as much as possible. So, the set of possible second cards can be divided into subsets in which you are indifferent to the amount of your first bet and subsets in which you prefer to bet as much as possible.