Someone has AA when I have an ace or two

Well, given that you have aces, there are 50 remaining cards in the deck. The first opponent to your left either haha aces (probability 1/50C2 = 1/1225) or not (1/1224). Player two, assuming player one doesnÂ’t have aces would have aces with probability 1/48C2 or 1/1128. He does not with probability 1127/1128. Continuing the pattern, we see that the probability of seven opponents not having aces would be
(50C2-1)/50C2 x (48C2-1)/48C2 x (46C2-1)/46C2Â…(38C2-1)/38C2, which equals 99.243%. Thus if you have aces the probability that someone else does is 0.757%.

For the other question, itÂ’s a more complex calculation that IÂ’m not going to tackle exactly. To apporoximate it, realize that there are now three combos of aces an opponent could have rather than the one combo when we have aces. This means that approximately the chance that opponent 1 has aces is 3/50C2 so the chance he does not is (50C2-3)/50C2. For player 2 the chance he does not have aces is approximately (48C2-3)/48C2. This is only approximate because player 1 could have one ace, which would lower this probability. But using this approximation we get a 97.744% chance that nobody has aces so a 2.256% chance that someone does.

If there are two events with probabilities A and B, the probability at least one occurs is A+B-AB. AB is the probability both occur but that is included in A and B so it is to be subtracted once. If A and B are small, AB can be very small so the estimate of events A or B = A+B is likely to be very close to exact.

Thus, with small probabilities a very good approximation for estimating the probability of at least 1 of n successes (aces) is to calculate the occurrence probability for 1 opponent and multiply that by n, the relevant number of opponents. Since the events here are mutually exclusive, the joint probabilities = 0 so we can simply add the individual player probabilities, which are equal.

Let’s do the case of hero having Ax and we are interested in the probability of at least one of n opponents having AA (yes, I know there can be only one at most)

If hero has Ax, there are 3 aces left. The number of ways of selecting AA out of 3 aces is C(3,2)=3. For one opponent, there are C(50,2) =1225 possible hands he can have. Thus, the probability one opponent has AA given hero has Ax is 3/1225 = 0.245%, a very small number.

For 7 opponents, the probability is estimated to be 7 * 0.245% = 1.71%.

For hero having AA, the probability one opponent has AA is 1/1225. Then 1 out of 7 having aces is estimated to be 7*(1/1225)= 0.576%.

This result differs from Stremba's. I think it’s because he didn’t account for the possibility that a player not having AA can have Ax, thus reducing the hit AA probability for other players. But Stremba is pretty damn good, so we need some expert decider.

[QUOTE=statmanhal]Since the events here are mutually exclusive, the joint probabilities = 0 so we can simply add the individual player probabilities, which are equal.[/QUOTE]This is key and it makes this problem easy. However, you and stremba missed that the question is about Omaha.

If AAA doesn't count as AA, then what OmahaDonk did is right except he forgot it's 8-handed meaning only 7 villains.
7*3(45C2) / (48C4) ≈ 10.68%

Another approach: there are 48C3 ways to place the remaining aces, of which 7(4C2)44 result in AA being within someone's 4 cards.

If AAA does count as AA, then: 7[3(45C2)+45]/(48C4) or 7[(4C2)44+4]/(48C3) ≈ 10.85%

Next problem: if Hero has exactly AA, then: 7(4C2)/(48C2) or 7(46C2)/(48C4) = 7/188 ≈ 3.72%

Yep. I missed that it’s Omaha. I calculated based on 2 card hands. My calculations are indeed approximate - I did not account for opponents possibly having Ax in either case as that would make a fairly complex probability tree for the exact calculation.