Head Up Theory Question

[M.B.E.]

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Ok, I have read the thread and understand why something around 20 is much more likely to be right than the obvious 499,998 or so.

However, what I don't understand is whether the "correct" answer assumes that our opponent never overvalues his hand, or whether that doesn't matter.
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Are we to amend the original question to assume an opponent who won't overvalue his hand and plays correctly?

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Original question said we are playing against an expert. I take that to mean there are no exploitable weaknesses in his/her strategy so we are trying to calculate the game-theoretically optimal strategy.

[M.B.E.]

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What if the ante, instead of $1, were $10,000,000 (but bets and raises were still in increments of $1).

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I purposefully made no refernce to any numbers.

It seems that there would be a lot more value betting, value calling and raising for value.

So Xo would be really high. And so, the reraising would go on for a really long time, each hand.

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No, I don't think so. Suppose you are first to act. The pot to start out is $20,000,000. The range of hands you can bet for value is not any larger than if the ante was $1. For example, if you have card 490,000 and you are first to act, you certainly can't bet it for value since chances are your opponent's hand is better than yours.

The question I was thinking about is, if antes are $10-million each, would you ever fold? Let's say our opponent is first to act, he bets $1, and the card we were dealt is card 3. So there's only two cards in the deck we can beat. But our pot odds are 20,000,001:1. It would be strange to fold, in a $20,000,001 pot, for only $1 more. Yet if our opponent's opening bet means he has a card greater than 600,000 (or whatever), doesn't that mean we can fold everything smaller than that, despite the huge pot odds?

The answer is that even when the pot is so extremely high because of the antes, it would still be correct to bluff sometimes, but very rarely. Therefore, if the player first-to-act bets, and the other player has card 3, the other player should still call $1 because of the tiny chance that the first player was dealt card 1 or 2 and decided to bluff-bet it.

[FishAndChips]

For eveyone that thinks it's a large number of raises, 900,000 or more (as opposed to ~20), it seems to me that you are looking at your odds of having the best hand against a random hand by your opponent, and then figuring it is correct to bet and raise a number of times proportional to the number of hands you beat initially. However, everytime your opponet raises it's for what he perceives as value(as discussed previously he's an expert and not making exploitable mistakes like bluffing with zero fold equity etc.) Therefore you should only raise the number of times that you are the favorite when called. If your opponent raises on the 20th bet for value look out! He can really only have one hand. He would have just called at some point sooner if his hand was any weaker than yours, because it would not be a raise for value against what he would narrow down as your range of possible hands.

Take an example from Hold'em. You have KK heads-up against a single opponent. The odds of your opponet having a better hand, AA, is 203.17-1. While this is not as large as the odds in the Sklansky example, 999,998-1, the number of raises is somewhat proportional. Now using the logic of the large number of raises, you would raise with KK something like 200 times. Now it's interesting to note that in actual practice, most of you would probably put your opponnet on AA after only a few bets and raises as opposed to 200+ and would subsequently just call much sooner.

[SparkyDog]

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.

[donkey69]

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This is getting almost cruel watching peope trying to develop game theory from scratch. Will someone who knows how the well developed theory of games works please do this and put us out of our misery.

PairTheBoard

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I believe the optimal strategy is to bet the fair value of your cards. Playing fair value against all other strategies I can think of the EV=0. Note that EV=0 is a perfectly acceptable result against expert opponents.

The ante is $1 and since you have 999,999 the fair value of your card is $999,999. If you get 1 you cap the betting at $1. If you get 500,000 you cap the betting at $500,000. If both players use fair value strategy, at the end of the day for both players EV=0. Half the time one player wins, the other half the other player wins.

Fair value v Bluffer
What if one of the players bets the fair value and the other bluffs all the time, that is, plays as if he has the nuts 1,000,000 everytime? Since the fair value player is able to cap the betting by calling it will make no difference if the other player is bluffing. 1/2 the time the fair value player will win, the other 1/2 the time the bluffer will win. EV=0 for both players.

Fair Value V Super Tight
What if the opponent only bets the nuts? The Super Tight player will win when he gets the nuts. The other 999,999 times he loses the ante. EV=0 again.

Fair Value v Randomised Strategy
What if the opponent plays a mixture of Super Tight, Fair Value and Bluffing? Sometimes he bluffs, sometimes he bets fair value and sometimes he Bluffs. xEV(0)+yEV(0)+zEV(0)=0. Where x,y,z are the probabilities he plays that strategy.

Any comments appreciated.

[felson]

What you wrote is almost exactly what I was thinking, which is why my first post said 15 bets. Except that I think you should probably use 0.000002 rather than 0.000001, so so I changed my answer to 13 bets. I haven't thought about it enough to be completely sure though.

David's problem is exactly like the [0,1] game with infinite bet finite pot, except that there are a finite number of hand values rather than a continuum.

[felson]

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What if the ante, instead of $1, were $10,000,000 (but bets and raises were still in increments of $1).

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I purposefully made no refernce to any numbers.

It seems that there would be a lot more value betting, value calling and raising for value.

So Xo would be really high. And so, the reraising would go on for a really long time, each hand.

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No, I don't think so. Suppose you are first to act. The pot to start out is $20,000,000. The range of hands you can bet for value is not any larger than if the ante was $1. For example, if you have card 490,000 and you are first to act, you certainly can't bet it for value since chances are your opponent's hand is better than yours.

The question I was thinking about is, if antes are $10-million each, would you ever fold? Let's say our opponent is first to act, he bets $1, and the card we were dealt is card 3. So there's only two cards in the deck we can beat. But our pot odds are 20,000,001:1. It would be strange to fold, in a $20,000,001 pot, for only $1 more. Yet if our opponent's opening bet means he has a card greater than 600,000 (or whatever), doesn't that mean we can fold everything smaller than that, despite the huge pot odds?

The answer is that even when the pot is so extremely high because of the antes, it would still be correct to bluff sometimes, but very rarely. Therefore, if the player first-to-act bets, and the other player has card 3, the other player should still call $1 because of the tiny chance that the first player was dealt card 1 or 2 and decided to bluff-bet it.

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I am pretty sure that the game-theoretic strategy never bluffs when the pot is this big, since your opponent will call every time.

This question is pretty much the infinite pot, infinite raise case treated in Part 3 of the [0,1] game. Answer is that you never fold, and that you bet/raise with top 41.4% of hands, reducing hand values with each additional raise.

[SparkyDog]

If one solved this, then they would be very close to solving the 0,1 game. The only two differences is that 0,1 is lowball and has an infinite number of hand values, where this is played for high and has a finite number of hand values. The principles remain the same though.

But 15 bets is the correct number of bets to put in disregarding folding and bluffing. I got that answer from basically working backwards from the examples given in the 0,1 game. So setting r^x = .000001 (the fraction of hands that is equal to or greater than 999,999).

[reubenf]

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Any comments appreciated.

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Would you actually use this strategy in a real game (at different stakes obviously, unless your'e rich) without thinking a bit harder about it? I don't understand why you wouldn't at least calculate the EV of a couple of hands against a few strategies (including your own). Say, calculate your EV against your own strategy if you are dealt 500,000.

[donkey69]

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Any comments appreciated.

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Would you actually use this strategy in a real game (at different stakes obviously, unless your'e rich) without thinking a bit harder about it? I don't understand why you wouldn't at least calculate the EV of a couple of hands against a few strategies (including your own). Say, calculate your EV against your own strategy if you are dealt 500,000.

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Actually I was wrong about Super Tight. Super Tight is EV+ against Fair Value Betting. Conclusion comes from exhaustive listing of a game with cards numbered 1 to 9. Super Tight in this game would mean folding immediately if you get a 1, calling/capping the next bet if you have a 2 to 8. Bet to the end if you have 9 (the nuts). Playing Super Tight against a Fair Value better results in net profit of $20.

To understand what I mean just do a 9x9 matrix of scenarios and the related payoffs. In this illustrated example you play Super Tight but your opponent bets for value.

Your Card Your Opponent Payoff
1 2 -1
1 3 -1
1 4 -1
1 5 -1
1 6 -1
1 7 -1
1 8 -1
1 9 -1
2 1 1
2 3 -2
2 4 -2
2 5 -2
2 6 -2
2 7 -2
2 8 -2
2 9 -2
3 1 1
3 2 2
3 4 -2
....
9 1 1
9 2 2
9 3 3
9 4 4
9 5 5
9 6 6
9 7 7
9 8 8

In conclusion it appears that the Super Tight strategy dominates both the Fair Value and Bluffing strategies. Obviously Super Tight is EV=0 against Super Tight.