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  #1  
Old 05-20-2007, 08:48 AM
soon2bepro soon2bepro is offline
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Default Adapting the Kelly Criterion to cash NLHE

The main two mathematical problems I see with this adaptation are:

1) If you want to play deep stacked (or any fixed stack-to blinds relation for that matter), you have only a few different options, and the size difference between them is large (it usually doubles or halves with each level)

2) There are some smaller bets than just the ones where you get all in through the course of the hand. I was thinking that this might be overlooked and simply consider those more risky times when you do get all in.

Is my assumption for 2) too far off?

And about 1), how do I imput this restriction into the formula?
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  #2  
Old 05-20-2007, 02:52 PM
curious123 curious123 is offline
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Default Re: Adapting the Kelly Criterion to cash NLHE

See MOP, the Chin article ("What poker players can learn from BJ" or something like that), or the search function.
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  #3  
Old 05-22-2007, 08:10 AM
soon2bepro soon2bepro is offline
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Default Re: Adapting the Kelly Criterion to cash NLHE

MOP?
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  #4  
Old 05-22-2007, 09:23 AM
plexiq plexiq is offline
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Default Re: Adapting the Kelly Criterion to cash NLHE

Just a guess: Mathematics of Poker (Ankenman/Chen)
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  #5  
Old 05-23-2007, 07:54 PM
Jerrod Ankenman Jerrod Ankenman is offline
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Default Re: Adapting the Kelly Criterion to cash NLHE

[ QUOTE ]
The main two mathematical problems I see with this adaptation are:

1) If you want to play deep stacked (or any fixed stack-to blinds relation for that matter), you have only a few different options, and the size difference between them is large (it usually doubles or halves with each level)

2) There are some smaller bets than just the ones where you get all in through the course of the hand. I was thinking that this might be overlooked and simply consider those more risky times when you do get all in.

Is my assumption for 2) too far off?

And about 1), how do I imput this restriction into the formula?

[/ QUOTE ]

[ QUOTE ]
The main two mathematical problems I see with this adaptation are:

1) If you want to play deep stacked (or any fixed stack-to blinds relation for that matter), you have only a few different options, and the size difference between them is large (it usually doubles or halves with each level)

2) There are some smaller bets than just the ones where you get all in through the course of the hand. I was thinking that this might be overlooked and simply consider those more risky times when you do get all in.

Is my assumption for 2) too far off?

And about 1), how do I imput this restriction into the formula?

[/ QUOTE ]

Generally speaking, limiting your bet sizes within a poker game based on risk considerations just gives up too much in expectation. What I advocate if you want to use Kelly or a variant thereof is to just choose limits to play (and/or your stack size) based on win rate and standard deviation at the different limits.

NL cash game results can be a little kurtotic, but I think as long as the stacks aren't like 10,000 BB or something it's not a big problem.

We talk about this in our book, basically you can use the kelly utility function U(x) = ln x. Then you say:

I have bankroll x0. If I play one unit of this game (could be an hour or a hand or whatever), my change in utility is:

<ln(B+x) - ln(B)>

where <> means expected value and x is a random variable taken from the distribution of results of that game.

Now the above simplifies to:

dU = <ln(1+(x/B))>

Expanding this with the Taylor expansion and dropping terms 3+ (because they contain B^3 in the denominator or higher), we get:

dU ~ <x/B - x^2/2B^2>

now <x> obviously equals our win rate w, and <x^2> is w^2 + s^2 where w is our win rate and s our standard deviation.

so we have:
du ~ w^2/B - (w^2+s^2)/2B^2

This is the value of playing a game (in kelly utility terms).

Now suppose we can choose between two games, with win rates and standard deviations specified for each. Somewhere along the line, we would reach a bankroll where we would prefer to play the bigger game (assuming we can win more money at it). Call that c. Then use the above expression for each of the two games and solve. This gets you this:

c ~ (s1^2 - s2^2)/(2(w1 - w2))

Above this bankroll, you should play the higher game (kelly-wise) and below you should play the lower.
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  #6  
Old 05-30-2007, 07:23 PM
Jerrod Ankenman Jerrod Ankenman is offline
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Default Re: Adapting the Kelly Criterion to cash NLHE

[ QUOTE ]
du ~ w^2/B - (w^2+s^2)/2B^2

[/ QUOTE ]

As someone pointed out to me privately, this should be:

du ~ w/B - (w^2+s^2)/2B^2

instead. Later equations are all right, i just transcribed this one wrongly.
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