basic probability question
 

Sorry if this is a basic question. If I have AJ (suits don't matter)and there are 2 callers pre flop and the flop is A84. How do I figure the probability that one of the 2 remaining players are holding a better ace (AQ, AK or AA)?

Thank you in advance,

Yoda

Re: basic probability question
 

[ QUOTE ]
Sorry if this is a basic question. If I have AJ (suits don't matter)and there are 2 callers pre flop and the flop is A84. How do I figure the probability that one of the 2 remaining players are holding a better ace (AQ, AK or AA)?

[/ QUOTE ]

The answer is affected by the caller's ranges, so this is definitely not a basic probability question.

The chance that there are two aces in the 4 cards is 47*46/12 -- roughly 1 in 200.
The chance of one ace is about 1 in 10, and the chance that it's with an AKQ is 9 in 46 which works out to about 1 in 46.
Adding the two together works out to about 6 in 200 or about 1 in 33.

Another way to look at it is that there are 1081 possible holdings, and you're afraid of 17 of them. So 17 holdings*2 people /1081 possible = 34/1081 also roughly 1 in 33.

So, based on just the cards, the odds are 1 in 33. In practice, your opponents are very unlikely to have called with many of those 1064 other holdings, and the effect of their range is going to dwarf the cards-based odds calculation.

Re: basic probability question
 

thanks rufus this helps me put it in better perspective.