Poker Theory Question

[inlemur]

I apologize if this is an old question, because it seems like the kind of thing that has probably been posed several times, but it can't really be searched for.

Suppose I design a robot that plays poker, and you are going to play a heads-up NL hold-em tournament with it. The robot's strategy is to push all in with any two cards without even looking at them.

What strategy should you adopt to maximize your equity? That is, what hands should you call with? How does this change as a function of the blind structure of the tournament? Finally, what is your equity against the robot assuming perfect play?

Calling with any favored hand does not seem correct unless each player has a very small number of BBs. For example, many hands are slightly +EV against a random hand but if it costs very little to wait for a hand with a larger +EV vs random, then it seems correct to do so. But how can you relatively accurately calculate your equity as a function of the structure?

This is a summarization of a question a friend asked while I was teaching them how to play poker. I was trying to explain that unlike say, basketball, where I would lose to LeBron 100% of the time, in poker, even a two line program could win a heads up game against Phil Ivey some percent of the time. I could not, however, come up with that percent chance of winning figure, though I suspect it's around 20% for most reasonable blind structures.

Thanks for any replies.

[AaronBrown]

Well, I can give you a start to this question.

Begin with a simpler question. You have chips for n hands, paying the antes and blinds. Your goal is to win one hand, but you lose with game over if you lose one hand. The solution is to wait for a hand such that your chance of losing is less than 3^(k-1)/2^(2*k-1), where k is the number of hands have left if you fold this one.

For example, if you can afford one more hand after this, k=1 and you would call the bet if your chance of losing was less than 3^0/2^1 = 1/2. If you could afford 4 more hands, you would call the bet if your chance of losing was less that 3^3/2^7 = 27/128. If this is the last hand you can afford, the formula does not work, you have to call on anything.

Going back to your game, you have a choice on the first hand: either call or put yourself in a position that you'll have to call and win two hand to win the tournament. There is some probability of winning such that you would call the first bet.

If you don't call the first bet, you know you have to win twice. If you lose another half of your initial stake, you'll have to win three times.

I think you can get a pretty good approximation by treating the number of folds before you increase the number of wins you need as the limit in the formula above. For example, suppose you start with enough chips for 10 folds. You fold the first hand and have chips for 9 folds. If you get down to 4 folds, you'll have to win three times instead of twice. That's such a disadvantage, that it makes sense to play similar to what you would do if you needed a win in the next five hands. That's the 27/128 chance of losing from above. It's not exact, but my intuition says it's pretty close.

I'm sure there's a way to figure the optimal calling win percentage in each of these cases. You're right that it depends on the number of rounds you can afford to fold.

[Louie Landale]

Can't do the math but suspect your formula does match YOUR situation. But the situation you gave is NOT the one of the original post.

Your objective is to maximize your chances of winning the one hand you intend to play. The original objective is to win the tournament. [1] The sooner you play a hand the more chips you have and therefore the more you can win. If you have enough for 10 hands you would MUCH prefer to win the first than win the 8th. [2] If you are short on chips you are going to have to win at least two hands.

So if you have enough for only 2 antes, you are much more likely to play the first hand since if you don't, you only have half as much money left. If you don't you have to win the next two hands. Not sure how to calculate it, however.

- Louie

[AaronBrown]

I assumed both players started with the same size stack. So you can call on the first hand and win (or lose) the tournament immediately; or else resign yourself to having to win at least two hands to win the tournament. In the second case, it doesn't matter when your two hands occur, unless they happen so late that you start needing three winning hands to win.

For example, if you had a 60% shot on the first hand, but thought you could wait and get two 80% chances, you would fold because that gives you a 64% chance of winning.

[AaronBrown]

Well, I simulated a lot of hands, and got a surprising (to me) answer. It's darn hard to beat the strategy of calling on any pair or any hand with an Ace. If your chips get down so you can only afford to fold ten or fewer hands then you should loosen up a bit (in the extreme, with your last bet in the pot, you call with anything of course), but doing that doesn't change your odds much assuming you start with enough chips to fold at least 30 hands.

You win about 90% of the time with this strategy, as long as you start with a reasonable stake relative to the antes/blinds. Even with an infinite stake (no antes or blinds) you can't be 100% sure of winning. You'd call only with A A, but sometimes that loses to a random hand.

There are smarter strategies, but they don't seem to do much better.

[piggity]

[ QUOTE ]
It's darn hard to beat the strategy of calling on any pair or any hand with an Ace.

[/ QUOTE ]

One obvious improvement to this strategy would be to call with the first n best starting hands in a heads-up game, as opposed to arbitrarily picking pairs and Ax...

[AaronBrown]

Actually, those are the 25 best starting hands in heads up. It does very slightly better to add KQ and KJ, the 26 and 27th best, but I wanted to keep it simple.

[piggity]

[ QUOTE ]
Actually, those are the 25 best starting hands in heads up. It does very slightly better to add KQ and KJ, the 26 and 27th best, but I wanted to keep it simple.

[/ QUOTE ]

Actually, I'm fairly certain they're not. I don't have my results handy, but I'll post the first 25 when I get home.

Note that the "best" in this case needs to be measured purely as preflop pot equity vs. a random hand, since there is no betting strategy.

[bigjohnn]

Hi,

Just wondering how you derived this formula?

Cheers, John

[piggity]

[ QUOTE ]
[ QUOTE ]
Actually, those are the 25 best starting hands in heads up. It does very slightly better to add KQ and KJ, the 26 and 27th best, but I wanted to keep it simple.

[/ QUOTE ]

Actually, I'm fairly certain they're not. I don't have my results handy, but I'll post the first 25 when I get home.

Note that the "best" in this case needs to be measured purely as preflop pot equity vs. a random hand, since there is no betting strategy.

[/ QUOTE ]

Alright, the first ten can be found here (option #4).

The next ten include both KQs (16th) and KJs (20th), and the venerable 22 is all the way down at 87th place.