Quote:
Originally Posted by plz  If the question is phrased as "I have KK in a 7-handed game. What are the odds that at least one opponent has AA?" then the answer is 2.9322%, or about 33:1. This is typically the way the question is asked, since you want to know the probability that someone has aces against your kings. The calculation of these odds is a bit intricate but understandable - I can post an explanation if anyone is interested.
The odds would be different if the question were "What is the probability that, on any specific deal in a 7-handed game, exactly one player has AA and exactly one other player has KK?" since you're starting without the assumption that someone already has a pair of kings. I haven't calculated these odds but this situation doesn't seem very interesting except in the abstract...
If you ask the question "Tonight I will have pocket kings twice in a 7-handed game. What are the odds that both times, the same opponent will have pocket aces?" I would calculate it as follows: From my odds above, the first time is about 33:1 = 1/34. The odds of it happening twice is 1/34 * 1/34 *1/6, since there is only a 1/6 probability that the second time it happens, it's the same opponent as the first time. This is about 6978:1. Pretty long odds. |
correct. should have phrased it as "I have KK in a 7-handed game. What are the odds that at least one opponent has AA?"
not that you answered that question as 33:1, have another question -
how does the probablility change when 2 player after me fold, leaving only 5 possible players (5 includes me) in the hand? does it drop to 1.9% at this point?
AND - im interested in your explanation of the first situation - and the second if you care to expand on one for that also.
thx
