June 26th, 2013 at 3:40:33 PM
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Hello Wizard,

I like to know what the chance is that somebody has pocket aces, given that there is an ace on the flop or turn. I think this problem must be solved by making use of conditional probability. Something like this, P(Player has Aces|Ace on flop) / P(Playes has Aces|Ace on turn). P(Player has Aces)= There are 52*51/2 = 1326 ways to arrange 2 cards out of 52. There are 4*3/2=6 ways to arrange 2 aces out of 4. So the answer is 6/1326 = 1/221. But what is the chance that there is an ace on the flop or turn? And do you think that this is the correct way to analyze this problem?

Thank you in advance for your help!

I like to know what the chance is that somebody has pocket aces, given that there is an ace on the flop or turn. I think this problem must be solved by making use of conditional probability. Something like this, P(Player has Aces|Ace on flop) / P(Playes has Aces|Ace on turn). P(Player has Aces)= There are 52*51/2 = 1326 ways to arrange 2 cards out of 52. There are 4*3/2=6 ways to arrange 2 aces out of 4. So the answer is 6/1326 = 1/221. But what is the chance that there is an ace on the flop or turn? And do you think that this is the correct way to analyze this problem?

Thank you in advance for your help!

June 27th, 2013 at 5:05:46 AM
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Quote:scf2Hello Wizard,

I like to know what the chance is that somebody has pocket aces, given that there is an ace on the flop or turn. I think this problem must be solved by making use of conditional probability. Something like this, P(Player has Aces|Ace on flop) / P(Playes has Aces|Ace on turn). P(Player has Aces)= There are 52*51/2 = 1326 ways to arrange 2 cards out of 52. There are 4*3/2=6 ways to arrange 2 aces out of 4. So the answer is 6/1326 = 1/221. But what is the chance that there is an ace on the flop or turn? And do you think that this is the correct way to analyze this problem?

Thank you in advance for your help!

Disclaimer: I am not The Wizard.

Aces-No known cards

Imagine dealing your opponent two cards face down and you haven't looked at your cards yet, then you are right that the probability of your opponent holding pocket Aces is 1/221.

4/52 * 3/51 = 0.004524886877828055 = 1/0.004524886877828055 = 1/220.99999999999997

Aces-Two known cards

Okay, if you know your two cards and neither of your cards are Aces, then the effect of removal is such that your opponent has a greater probability of holding Pocket Rockets:

4/50 * 3/49 = 0.004897959183673469 = 1/0.004897959183673469 = 1/204.16666666666668

Aces-Five known cards

Now, if neither of your cards are Aces, but an Ace comes on the flop:

3/47 * 2/46 = 0.0027752081406105457 = 1/0.0027752081406105457 = 1/360.33333333333337

Aces-Six known cards

It is more likely, however, for your opponent to hold Pocket Rockets if there are no Aces on the flop, but you do see an Ace on the turn:

3/46 * 2/45 = 0.002898550724637681 = 1/0.002898550724637681 = 1/345

NOTE

I'm not sure why you are interested, but from a practical standpoint for playing poker, this information is nearly meaningless. How your opponent behaves (i.e. how he bets) will generally either increase or decrease, intuitively, the likelihood that he has Pocket Rockets. For example, if he pushes out a huge bet pre-flop, then a hand such as 7/2, Off, I'm going to take for granted that he doesn't have. If he must not have that hand, then that is a hand eliminated from the set of hands he could have, which increases the probability that he has Pocket Rockets or any other good hand.

Vultures can't be choosers.