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.
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