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  #1  
Old 11-11-2005, 10:34 PM
NMcNasty NMcNasty is offline
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Default adjusting hand range to flop

Say a player only plays AK, QJ, T9, and 78 preflop.

So when he enters the pot the chance that he has a particular one of those hands is 25%.

Now the flop comes A66. What are the new probabilities adjusted to? Or a better way of putting it is by what percentage should the probability of having AK be reduced by?

AA6?
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  #2  
Old 11-11-2005, 11:49 PM
yellowjack yellowjack is offline
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Default Re: adjusting hand range to flop

Initially there are 16 different ways that a player can have to make AK. When there is an ace on the flop, there are only 12.

Changing the percentages to fractions, the 1/4 * (12/16) = 3/16. All other hands are still 1/4.
P(AK) = (3/16) / ((3/16) + 3*(1/4)) = 1/5
P(QJ) = P(T9) = P(87) = 4/15
The probability of AK is reduced by 1/4 - 1/5 = 0.05 = 5%

For AA6, only 8 ways to make AK.
(1/4)*(1/2) = 1/8
P(AK) = (1/8) / ((1/8) + 3*(1/4)) = 1/7
P(QJ) = P(T9) = P(87) = 2/7
The probability of AK is reduced by 1/4 - 1/7 = 3/28 = 10.7%
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  #3  
Old 11-12-2005, 01:33 PM
AaronBrown AaronBrown is offline
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Default Re: adjusting hand range to flop

Another way to get the same answer, that may be easier to visualize, is to consider the number of ways to form A66. If the other player's hand has no Aces or sixes in it, there are 24. If the other player's hand contains an Ace, there are only 18. So the probability that she holds AK is 18/(18+24+24+24) = 3/15.

I like this method better, because it's more natural to incorporate subsequent information. But Reverend Thomas Bayes proved that you get the same answer either way.

Of course, this is affected by what you hold if you hold a 6 or any card the other player might hold.
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  #4  
Old 11-12-2005, 06:38 PM
yellowjack yellowjack is offline
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Default Re: adjusting hand range to flop

[ QUOTE ]
Another way to get the same answer, that may be easier to visualize, is to consider the number of ways to form A66. If the other player's hand has no Aces or sixes in it, there are 24. If the other player's hand contains an Ace, there are only 18. So the probability that she holds AK is 18/(18+24+24+24) = 3/15.

I like this method better, because it's more natural to incorporate subsequent information. But Reverend Thomas Bayes proved that you get the same answer either way.

Of course, this is affected by what you hold if you hold a 6 or any card the other player might hold.

[/ QUOTE ]

This is much better on the eyes. Thanks.
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