Game theory exercise: Bet size with two rounds to bet
 

Suppose you are playing NL, and there are two rounds of betting. There is already $1 in the pot. You each have $x. There is no drawing. You can assume that you are playing high card with at least 5 cards, and your opponent has the second highest card face up.

[img]/images/graemlins/diamond.gif[/img] How much should you bet on the first round, as a function of the stack size x?
[img]/images/graemlins/diamond.gif[/img] What is the optimal strategy?
[img]/images/graemlins/diamond.gif[/img] Do you gain from having multiple rounds to bet?

Re: Game theory exercise: Bet size with two rounds to bet
 

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You can assume that you are playing high card with at least 5 cards, and your opponent has the second highest card face up.

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I don't think I understand that; could you please elaborate on the rules of the game.

Re: Game theory exercise: Bet size with two rounds to bet
 

For a similar question and partial spoilers, see Jazza's question.

At showdown, whoever has the higher card wins.

It is common knowledge that your opponent has the second-highest card.

If you have the highest card, you can value bet. If you have anything else, you might bluff.

Re: Game theory exercise: Bet size with two rounds to bet
 

i've read the spoiler, so i'll hold back for a while, but one question: what increments are you allowed to bet in?

Re: Game theory exercise: Bet size with two rounds to bet
 

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i've read the spoiler, so i'll hold back for a while, but one question: what increments are you allowed to bet in?

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You are welcome to comment. The same ideas apply, but the answer is a bit more complicated.

You can assume the bets are continuous.

Re: Game theory exercise: Bet size with two rounds to bet (Answer)
 

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Suppose you are playing NL, and there are two rounds of betting. There is already $1 in the pot. You each have $x. There is no drawing. You can assume that you are playing high card with at least 5 cards, and your opponent has the second highest card face up.


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In the optimal strategy, you bet some amount y some of the time, including every time you have the nuts and some bluffs. When you are called, you either push for the remaining x-y on the second round or give up.

You should bluff so that your opponent is indifferent to calling or folding.

On the first round, the pot odds are (y+1):y, so you should bet and give up y times for every y+1 times you plan to push on the second round.

On the second round, the pot odds are (x+y+1):(x-y), so you should bluff x-y times for every x+y+1 times you push with the nuts.

You get to bet (1+y/(y+1))(1+(x-y)/(x+y+1)) times as frequently as you have the nuts. This is proportionate to your equity. To maximize your equity, set y=(sqrt(1+2x)-1)/2. Qualitatively, you should bet so that your opponent gets the same pot odds on each round.

Here is a concrete example drawn from this thread: Suppose the pot is $25, and you have $40 left to bet. If you push, you can only bluff 40/65 = .615 as often as you value bet. If you bet $20 at a time, you can bluff .784 times as often as you value bet. The optimum is to make your bets on the first round $13.11, which allows you to bluff .807 times as often as you value bet. (You bet about half of the pot on each round.) Of course, the model is not perfectly accurate, but it shows that it can be good to use the extra round of betting.

Re: Game theory exercise: Bet size with two rounds to bet (Answer)
 

did you just give the answer!? i didn't get a chance to try it [img]/images/graemlins/mad.gif[/img]

but thanks for the post, i enjoy game theory posts like these

Re: Game theory exercise: Bet size with two rounds to bet (Answer)
 

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did you just give the answer!? i didn't get a chance to try it [img]/images/graemlins/mad.gif[/img]

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I posted the answer about 36 hours after I posted the problem. This thread had received plenty of views.