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DOTTT
05-11-2004, 05:05 PM
Can someone please explain this concept to me, and how I can use it when putting my opponents on hands? Thanks.

PairTheBoard
05-12-2004, 03:30 AM
Try this explanation. The probability that he has it given that he bets it is the proportion of the time he bets it and has it to the time he bets it for any reason.

Try this example. Consider the game where you and your opponent ante and one card is dealt to your opponent. If he is dealt an Ace he wins. After he is dealt his card he can bet. If he bets you can call and see if he has the Ace, or fold and forfeit the antes.

Also suppose your opponet bets 90% of the time he gets an Ace and bets 30% of the time when he doesn't have an Ace.

Now, the card is dealt and your opponent Bets. What is the chance he has an Ace? Bayes Theorum calculates it as:

[.9(1/13)] / [.9(1/13) + .3(12/13)] = 20%

ie. the proportion of the time he bets and has it to that when he bets it for any reason.

As you can see your opponent is a maniac who bluffs so much you would always call him regardless of the ante structure.

I think poker players apply the Bayes concept by feel most of the time. When an opponent is representing a highly unlikely event it's hard to believe him unless he very rarely bluffs.

PairTheBoard

PairTheBoard
05-12-2004, 01:46 PM
Bayes Theorum works for more than two partitions of the possibilities too. Suppose when your opponent raises you on the turn you figure he has four possible hand Types. T1-A set. T2-Two Pair. T3-A strait or flush draw. T4-One Pair or less. Also, suppose that based on the preflop and flop action (ignoring the turn raise), you figure the chances of his having these hands is 10%,15%,30%,45% respectively. Now suppose the chances of his raising the turn with these hand types is 90%,80%,50%,10% respectively. Based on these assumptions what is the chance he has T1-A Set using the fact he did raise the turn? Bayes Theorum gives the calculation:

.9(.1) / [.9(.1) + .8(.15) + .5(.3) + .1(.45)] = 22%

The chance he has T2-Two Pair is:

.8(.15) / [ same denominator ] = 30%

The chance he is semibluffing with T3-straight or flush draw:

.5(.3) / [same denominator] = 37%

And the chance he is pushing One Pair or less:

.1(.45) / [same denominator] = 11%

As you can see, Bayes Theorum lets you recompute probabilities based on previous probabilities and the current action. In practice, most players do this adjustment of probabilties based on current action by feel and experience. I imagine guys like Sklansky and Malmuth have looked at so many exact calculations of this type that they can virtually do them precisely on the spot if necessary.

PairTheBoard

DOTTT
05-12-2004, 02:52 PM
Great post PairTheBoard!
Thanks for all the help; I'm practicing the theorem right now.

On a separate note I wonder if any of the top tournament players use the theorem when making their decisions, I know that Chris Ferguson did when he decided to call with his A9 in the WSOP.

PairTheBoard
05-12-2004, 06:12 PM
I believe we all apply the theorum DOTTT. Every time we estimate our opponents holdings and then ADJUST our estimate based on his current actions, we are applying Bayes Theorum. Most of us do it by the seat of our pants maybe never having heard of the Theorum. Others do it more precisely.

PairTheBoard

DOTTT
05-20-2004, 04:39 PM
Pairtheboard,
I've been practicing the theorem a lot lately, but I haven’t used it in play yet, not confidant enough. I was wondering if you could help me figure out this problem.

Suppose your opponent goes all in, you hold JJ and suspect the following probabilities.
1. 50% of the time he is holding a smaller pair.
2. 40% of the time he is holding two over cards.
3. and 10% of the time he is holding a bigger pair.

In the first scenario you are an 81% favorite. In the second your a coin flip, an in the third you have a 18% chance of winning.

What's the correct way to work out the math? I've tried doing it on my own, but keep getting some weird results. Thanks.

PairTheBoard
05-20-2004, 07:57 PM
From what you assume this is a straightforward expected value calculation.

(.5)(.81) + (.4)(.5) + (.1)(.18) = 62.3% average winning percent.

PairTheBoard

PairTheBoard
05-27-2004, 01:27 AM
Bumping this Thread to make it easier to find for those I referred here from the "Median Best Hand" thread in Poker Theory.

PairTheBoard