Trying to work through an exercise, hoping you guys can help me verify what i've done and keep moving forward:

What is the probability of quad fours being beaten by a royal flush? Given the following constraints:

5 players are dealt in and see the hand to the end
Both hands must use both hole cards
The quads must have a pocket pair

I figure I need to find two things:
-probability of one player getting 44 and another getting two royal flush cards
-probability of the bad beat condition triggering given that the right hole cards are present
And then I can multiply my way to the answer

I think I'm alright on the second one. Assuming one player has 44, another has two royal flush cards, the board must contain 5 exact cards from the deck - the other 2 4s, and the other 3 royal flush cards. There are 48 cards not yet spoken for, so I do 48-choose-5 and I get 1712304, so the probability is 1 in 1712304.

My concern there is that I'm bringing in the wrong data... should I be accounting for the times that I deal 44 and more than one royal flush possibility (i.e. 44, JsTs, KhJh, xx, xx)?

Which brings me back to the first half of the problem. I'm not really sure how to approach the hole card problem... I was thinking of finding the probability of 44 and a royal flush combo in just 2 hands, and then multiplying by 5 choose 2. But I don't know if that gets me what I'm looking for, or that I'm looking at the right thing.

Thoughts?

2nd

Your approach is correct since only 2 players can hold these 2 hands.

5*4*6*4*C(5,2)/C(52,2)/C(50,2)/C(48,5)

= 1 in 579,454,375.5

There are 5 ways to choose the player with the quads, times 4 ways to choose the player with the royal, times 6 ways to choose the pair of 4's in the hole, which determines the pair of 4s on the board, times 4 suits for the royal, times C(5,2) ways to choose the hole cards for the royal, which determines the 3 royal cards on the board, all divided by C(52,2)*C(50,2)*C(48,5) total ways to choose the hole cards and the board.

Note that you could also just count flops, 6*4*C(5,3) for the 6 pairs, 4 suits for the royal, C(5,3) royal cards, and then the hole cards are determined. This gives the same result since C(5,3) = C(5,2).


Yes, since only 2 players can have these 2 hands, we can multiply the probability for 2 players by C(5,2). I multiplied by 5*4 above instead of C(5,2) in order to allow each player of every pair to have both the royal and the quads.

Bruce, you're the man, thanks.

I'm with you until the very final sentence, where I'm not entirely up to speed:


I notice that 5*4 is equivilant to P(5,2)... is that why you're using those numbers instead of C(5,2)? We're dealing with a permutation instead of a combination?

Thanks,
2nd

Right, because we want to distinguish the case where player 1 has quads and player 2 has the royal from the case where these hands are reversed. If the rest of your calculation already counts both of these cases, then you just have to multiply by C(5,2). Mine assumed that a particular player has the quads while the other has the royal, so I needed an extra factor of 2.