Case 1 - PLO

Say I am holding AQK9 and flop comes out Q Q 9...
Ignoring preflop action and player-specific ranges, what are the odds of my opponent holding the case Q? What about the odds multiway? One of 2 opponents holding the case Q? One of 3 opponents holding the case Q? 4 opponents?

Case 2 - PLO6
Say I am now holding AKQJ97. Flop comes out Q Q 9 again. Same question as case 1... HU? Vs 2 opponents? 3 and 4 opponents?

I know I am asking a lot, but can you show some procedure? I would really like to learn to do this myself so I don't have to ask again and so I can help others with similar questions.

Prob of Event E = Pr(E) = Number of ways E can occur / Number of total ways
Odds = Pr(E)/Pr(not E)

For Case 1, E is an opponent holding the 1 remaining queen. There are 7 known cards dealt, the 4 cards hero holds and the 3 flop cards. Therefore, the remaining deck has 45 cards. For one opponent, he is dealt 4 cards. The easiest way to calculate the probability he has the queen is to first calculate the probability that he doesn’t have it and subtract that result from 1.0. There are 44 cards other than the case queen.

To avoid using combinatorics, which is the easiest way, I’ll use direct probabilities:

Pr(no Q) = 44/45 * 43/44 * 42/43 * 41/42 = 43/45 * 41/43 = 41/45 [= C(44,4)/C(45,4)]

Then Pr(Q) = 1- 41/45 = 4/45

Odds against a queen = Pr(no Q)/Pr(Q) =(41/45) / (4/45) = 41/4 = 10.25 to 1

For multiway the same approach can be used; e.g., for two players, 8 rather than 4 dealt cards are to be evaluated.

Note: this problem is relatively simple because there is only 1 “success” card, the one queen. For more than one possibility it gets more complicated but the same general approach applies.