probability of higher ace-x
 

Came across this blurb in wiki's poker probability page :


If the player is holding Ax against 9 opponents, there is a probability of approximately 0.0218 that one opponent has AA.

Where x is the rank 2–K of the second card (assigning values from 2–10 and J–K = 11–13) the probability that a single opponent has a better ace is calculated by the formula

P = (3/50 * 2/49) + (3/50) * (13 - x) * 4) /49 * 2)

= 3/1225 + (12 * (13 - x))/1225

= (159 - 12x)/1225

The probability 3/50 x (13-x) * 4 / 49 of a player having Ay, where y is a rank such that x < y <= K, is multiplied by the two ways to order the cards A and y in the hand.


I suck - I'm confused as to whether this problem also refers to a 9 handed table or heads up, and what portion(s) of it I would use.

For example, if I'm dealt A8, how would I figure out the likelyhood of someone at a 9 handed table holding a higher kicker? I tried plugging the 8 into x, but the answer of .05 doesn't seem right unless it is referring to heads up, or more likely, I don't understand how to utilize the problem.

Re: probability of higher ace-x
 

Here is a simple explanation .

If you're at a 10-handed table and you hold A-x , then the approximate probability that someone has pocket aces is :

1-[(1225-3)/1225]^9 ~ 0.021826135

The probability player i doesn't have pocket aces is (1225-3)/1225 . We may approximate the solution by assuming independence . So the probability neither of the 9 players holds pocket aces is [(1225-3)/1225]^9 . Now the last step is to take the complement which is 1 - [(1225-3)/1225]^9 .

Re: probability of higher ace-x
 

[ QUOTE ]
Here is a simple explanation .

If you're at a 10-handed table and you hold A-x , then the approximate probability that someone has pocket aces is :

1-[(1225-3)/1225]^9 ~ 0.021826135

The probability player i doesn't have pocket aces is (1225-3)/1225 . We may approximate the solution by assuming independence . So the probability neither of the 9 players holds pocket aces is [(1225-3)/1225]^9 . Now the last step is to take the complement which is 1 - [(1225-3)/1225]^9 .

[/ QUOTE ]

You don't neeed to approximate in this case since at most 1 player can have AA, so the exact probability is 9*3/1225 =~ 0.0220. However, your approximation method works in the more general case where more than 1 player can hold the hand in question.

EDIT: Fixed typo 0.220 -> 0.0220

Re: probability of higher ace-x
 

That's right Bruce .

By the way , you have a typo .

9*3/1225 ~ 0.02204