Hypothetical Heads Up Gambling Situation

[joeg]

Thats wrong jay, for purposes of notation I'm gonna define R as the percentage rank of a hand, i.e. aces has R=100%, and then define W as the win percentage of that hand against a random hand, so in this case aces have W=~80%

The method you show is mathematically correct if for every hand R=W (or if they were linearly related), it could still be correct if this was not the case but it would be a coincidence, I doubt this will be 'correct' for holdem hands, its probably close, but not correct.

I dont really have the skills to prove this, but I'm sure its the case, I'll post a link & question on the probability forum tomorrow.

[SumZero]

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Hey chainsaw , try it yourself .

The first problem has been solved .
Work it out for n=3 , then n=4 and you'll see a pattern emerge .

The final list Nicho provided is accurate .

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No it isn't. See the thread in probability. The list Nicho provides is an approximation, but isn't accurate at the edges. probability thread.

In particular, Nicho says that you want to push on turn two with K3o+. This is because K2o is below his 50% hand list. But this is silly as we know that if you go to the last round where both people are forced to push then you will win 50% of the prize pool as you have equal equity with the opponent. But K2o gives you 50.5087% equity against a random hand so you'd be a fool to fold it if that meant next hand you both pushed blind.

The basic point that joeq is trying to make is that you can't just say what is the probabiliy that I get a better hand next time, but rather what is the probability my equity is better next time. Because equity doesn't scale exactly with hand count the pick the top 1/n-th hands on turn n is not correct, although it is a reasonably close approximation for the early turns. On turn 2 Nicho is missing 22, K2o, Q2s, Q5o, and J6s all of which have greater than 50% equity versus a random hand (50.3340, 50.5087, 50.1690, 50.1201, 50.6059 respectively).

[Nichomacheo]

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On turn 2 Nicho is missing 22, K2o, Q2s, Q5o, and J6s all of which have greater than 50% equity versus a random hand (50.3340, 50.5087, 50.1690, 50.1201, 50.6059 respectively).

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Please explain this to me. If you call with 22 on Hand 2, you are gauranteeing a coin flip. You are better off waiting because the odds that you get a hand that is better than a coinflip on turns 3-10 is very good.

[SumZero]

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On turn 2 Nicho is missing 22, K2o, Q2s, Q5o, and J6s all of which have greater than 50% equity versus a random hand (50.3340, 50.5087, 50.1690, 50.1201, 50.6059 respectively).

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Please explain this to me. If you call with 22 on Hand 2, you are gauranteeing a coin flip. You are better off waiting because the odds that you get a hand that is better than a coinflip on turns 3-10 is very good.

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I mean by turn 2 (using the turn notation from the thread linked where turn 1 is the all-in everybody turn, turn 2 is the hand before that, turn 3 is the hand 2 before that, so on turn n you have n-1 hands left that you can decide to push or not before on the very next hand you have to push), what you called either hand 8 or hand 9

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Hand 8 - Top Half - A2s+, A2o+, 33+, K2s+, K3o+, Q3s+, Q6o+, J7s+, J8o+, T7s+, T8s+, T9, 98s+
Hand 9 - Same

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This list is wrong for the decision right before the push is forced.

[Nichomacheo]

Based on the webpage that Jay provided, where I got my numbers...

Of the 169 starting hands, 89 win more than 50% of the time and 80 win less than 50% of the time. This means that of the top half of hands in terms of equity, four of them will win greater than 50% of the time but you're still correct to fold.

Here's an example:

With 22, there are 86 hands with better equity and 82 hands with worse. If you fold, odds are you will pick up a better hand next time and thats why you should fold 22 on turn 9.

[SumZero]

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Based on the webpage that Jay provided, where I got my numbers...

Of the 169 starting hands, 89 win more than 50% of the time and 80 win less than 50% of the time. This means that of the top half of hands in terms of equity, four of them will win greater than 50% of the time but you're still correct to fold.

Here's an example:

With 22, there are 86 hands with better equity and 82 hands with worse. If you fold, odds are you will pick up a better hand next time and thats why you should fold 22 on turn 9.

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Wrong. Wrong. Still Wrong. That's the very crux of the problem. (And also would indicate a serious flaw in your poker thinking if you were to apply this type of logic to other poker situations. It is as if you are saying if there is a 51% chance that the guy has you beat when he bets $1 into a $100 pot you should fold because odds are he has you beat.). Your goal shouldn't be to ask "is it more likely or not that I get a better hand" but rather it should be to ask "is it more likely or not that I get a better equity situation". Just think about what you are saying. You are saying that you are better off folding a hand (22) that has greater than 50% equity so that you can play all-in blind on the next hand. That is just plain wrong. When you both go all-in blind you are in a 50/50 equity situation. When you go in with 22 your equity is greater than 50%. Therefore going in with 22 is correct if you get 22 on the last hand you are allowed to make a decision.

The reason again is that the quality of hands aren't equally spaced. For instance image there were 6 hand values possible for you to have (this time no opponent, you just get the $ value associated with the hand). Hand 1 is worth $52. Hand 2 is worth $51.50. Hand 3 is worth $51. Hand 4 is worth $50.50. Hand 5 is worth $1. Hand 6 is worth $0. You are told you get to see what hand you have and if you don't like it you can "fold" it and get a random hand (with replacement), but you can only do this once. You want to maximize your $EV. You are looking down at hand 4. Your one buddy, call him nicho, says "dude, there are 3 hands better than this hand and only 2 hands worth so clearly your $EV is maximized by folding this hand and picking a hand at random since it is 3:2 that you pick a more valuable hand". Your other buddy, call him joeq, says "dude, are you insane? Hand 4 has a $EV of $50.50. Picking a hand at random has a $EV of $34.33. Clearly you have better $EV from sticking with Hand 4. Sure 3/6 times you would have ended up with more if you folded and only 2/6 times you would have ended up with less but the miniscule amount more that you end up with doesn't make up for the large amount less that you are risking". Who is right in this game?

Now the hand distribution isn't as extreme as my little 6 handed game above (it is more similar to if hand 5 were $48 and hand 6 were $47 in which case $EV of random would be $50 and sticking with Hand 4 would still be correct), but the distribution isn't completely uniformily spaced which is why your method of calculating what to do gives the wrong answer even though it is reasonably close to the right answer (I.e., your method is wrong, your answer is wrong, but your answer isn't that far from the correct one in terms of what hands to play).

[jay_shark]

I have to think about this some more . I realized after that there was a slight flaw in my reasoning .

The answers are close enough to being correct .

[SumZero]

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Based on the webpage that Jay provided, where I got my numbers...

Of the 169 starting hands, 89 win more than 50% of the time and 80 win less than 50% of the time. This means that of the top half of hands in terms of equity, four of them will win greater than 50% of the time but you're still correct to fold.

Here's an example:

With 22, there are 86 hands with better equity and 82 hands with worse. If you fold, odds are you will pick up a better hand next time and thats why you should fold 22 on turn 9.

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The other reason this is obviously wrong is there are actually 1326 different hands (not counting order of cards) not 169. If you only count 169 hands then each hand is not equally likely as there are 12 kinds of offsuited hands, 4 kinds of suited hands, and 6 kinds of pairs.

[jay_shark]

Sum Zeroes response is excellent and more precise but ours isn't far off from the truth .

Also Nicho , top x % of hands should be out of 1326 which is the total number of ways of selecting 2 cards from 52 .

52c2=1326 Hands like ace king have 16 possible combinations 4*4=16 , while pocket pairs have 6 combinations 4c2=6 .

Suited hands count for 4 additional combos out of 1326 .

[Nichomacheo]

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Based on the webpage that Jay provided, where I got my numbers...

Of the 169 starting hands, 89 win more than 50% of the time and 80 win less than 50% of the time. This means that of the top half of hands in terms of equity, four of them will win greater than 50% of the time but you're still correct to fold.

Here's an example:

With 22, there are 86 hands with better equity and 82 hands with worse. If you fold, odds are you will pick up a better hand next time and thats why you should fold 22 on turn 9.

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The other reason this is obviously wrong is there are actually 1326 different hands (not counting order of cards) not 169. If you only count 169 hands then each hand is not equally likely as there are 12 kinds of offsuited hands, 4 kinds of suited hands, and 6 kinds of pairs.

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Thats actually a very good point. I'll ponder it on my way to Vegas this weekend. Seriously.