Head Up Theory Question

[xpsyuvz]

chapstick,

I'm thinking that the trend would be similar to the ideas that refer to the "golden mean of poker/ 41.42%". I don't really know anything about it, but Underlord's post on page 24 seems to explain it a little.

[marv]

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[I have a program for these sorts of games.]

With 100 cards (and at most 25 raises), an optimal strategy with a 99 is to open the betting 95% of the time and then call after the 4th bet/raise:


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Marv, can your code do more than a 100-card deck? Could you do the series of decks: 10, 20, 40, 80, 160, 320 and see if there is a trend, to extrapolate to a million, as to when you should stop raising and call?

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These optimal strategies do involve checking first and even folding, but the number of raises in any scenario appears to be roughly the same, and in line with the [0,1] game analysis.

[max 25 raises]
10 rrc
20 crrrc or crrrf (20%)
40 crrrc or rrrrf (13%)
80 rrrrc
160 crrrrrc

[max 20 raises]
320 rrrrrrc or rrrrrrf (11%)


Marv

[Jerrod Ankenman]

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I think David's question is basically where Chen and Jerod left off, that is the infinite bet finite pot game, which is hard to solve.

The halving method is incorrect because you should only be reraising about the upper 41.4% of the your hands inthe range that your opponent could be raising.

I found a quick and dirty approximation I think that would give the optimum number of reraises by Hero, in an ex-showdown value sense.

0.414^x = 0.000001 (x is the number of bets, 0.000001 represents the percentile of hand that 999,999 is in.)
x = ln(.000001)/ln(.414) ~= 15.6658103

I round down to 15 because the nth bet the first player (hero) must put in must be an odd integer, and you can't round up in this circumstance.

The problem with this approach is that it doesn't take into account check-raising, bluffing, bluff-raising, or folding. I assumed that folding wouldn't enter the equation because the pot odds will grow increasingly greater (diminishing the likelihood of folding). Furthermore, it would only impact the answer on the final bet, (earlier bets don't apply, since it would still be correct to put in 15 bets regardless of when the opponent calls/folds) so I don't think it will have much effect on the answer.

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This is the right answer (15 bets). I haven't had the patience to read all the responses to see if anyone actually got it right, but anyway. We have solved the infinite bets allowed finite pot games, btw. They are sorta hard, but not impossible. The game with check-raise allowed can be easily solved for Y, but getting X into closed form is extremely hard so we just leave X in terms of Y and solve for the Y values for a specific game before trying to get an X value.

Anyway, you can read about it in our book "The Mathematics of Poker", which is close to complete and should be published by a different publishing house sometime this fall.

Jerrod

[raisins]

If you have solved it, we would appreciate the answer to the pot limit game as well.

regards,

raisins

[PairTheBoard]

So 15 bets means 8 raises for yourself and 7 for your opponent? And this is the $1 at a time version?

PairTheBoard

[mes]

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4. So the question is when to call? It seems to me that the optimal strategy is to raise until the pot size equals or exceeds the number of my card and then call. In this case, I would raise until the pot is at least $999,999 and would then call.

If I follow this strategy, it seems to me that villian cannot exploit it. Since the odds of him having a higher card than I is 50/50, the strategy should assure me in the long run of winning at least half the pots for any given card number.

Let's say that instead of 999,999, I'm dealt 314,159 or maybe 271,828. I keep raising til the pot hits my card and then call. After playing a few million hands, the villian's going to figure out my strategy pretty quickly. In fact, I'll save him the trouble and tell him upfront.

What's he gonna do about it? If he holds a higher card, he'll win the pot when I just call. But if he holds a lower card, is he really going to keep raising beyond the value of his card when he knows that I'll call at some point and he's throwing money away? I suggest the answer is no. If he does that consistently, over time I'll come out ahread, but since he's an expert, I can assume he won't do that.

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Yes, but your strategy is not optimal. Look at how much you are giving him when you have a poor hand. My counterstrategy is as follows: I will fold to the first bet when my number is less than 333,334. If my card is between 333,334 and 666,667, I will call the first bet. If my card is 666,667 or greater, I will raise until the pot is 100,000 and then call if you haven't already.

So 1/3 of the time, I will fold and lose 1 bet (ante). 1/3 of the time I will call and my range of hands is such that I should win exactly 50% against a random hand on average. Since I win 2 bets when I win and lose 2 bets when I lose, I will break even on these calls. The last third is more difficult. 10% of the time, your card will be less than 100,000 and I will win your card value for an average win of 50,000. 90% of the time, your card will be >100,000. 566,667 cards you could hold will be lower than my minimum threshold, so I will win 100K when you hold those (62.963% of the 90%) for an average EV of 56,667. When you hold a hand equal to or higher than my minimum threshold, we break even (we hold the same range of hands in that scenario).

So when I choose to gambool we have these EVs:

.1 * 50,000 == 5,000
.9 * .62963 * 100,000 == +56,667
.9 * .37037 * 0 == 0

For a total EV of 61,667 when I have 666,667 or better.

Partial EVs on the whole bet:

1/3 * -1 + 1/3 * 0 + 1/3 * 61,667 == 20555.333...

I'm going to win over $20,000 per trial on average using this simple counterstrategy. Your strategy is clearly not optimal.


Michael

[Jerrod Ankenman]

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So 15 bets means 8 raises for yourself and 7 for your opponent? And this is the $1 at a time version?

PairTheBoard

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Oh, it's actually 15 bets (although when I actually solved for all the R_p terms instead of just assuming they converged superfast it actually is pretty close on whether the 15th bet goes in.) So like, you bet, your opponent raises, now you should be willing to put in the 15th bet and then play some sort of mixed strategy in terms of calling the 16th one. Also, if the game is sampled without replacement, which I assume, you should put in more raises with hand #2 as a bluff to protect your hand #1 value bets; these are demibluffs since when you get called you'll win half the pot and their frequency goes to zero as you put in the nth bet as n -> infinity.

I saw that someone asked for the pot limit case. I don't have time to work on it right now, even though it's quite a bit easier, because R is constant. I'll try to get to it tonight, or maybe I can talk Bill into doing it for me. He's better at these things than I.

Jerrod