Royal Flush over Straight Flush over Quads over Quads ?
What are the odds of a royal flush over a straight flush over quads over quads?
For example, the board is:
QdJdTdQsJs
Player 1: AdKd (royal flush)
Player 2: 9d8d (straight flush)
Player 3: QhQc (quads)
Player 4: JhJc (quads)
1 Reply
We need a connected two-pair board ranging from 5-high to Q-high, and the pairs must each have a suit matching that of the unpaired rank.
The probability of that is 4*8*3*3^2 / C(52,5)
(Numerator: 4 suits, 8 acceptable three-card straights, 3 choices of which rank is unpaired, and 3 choices for each pair's unused suit.)
Then we need the players to have all perfect cards and for the cards to be grouped properly. The chance of that is C(n,4) / [C(47,8)*7!!] where n is the # of players.
All told, the probability is 864*C(n,4) / [C(52,5)*C(47,8)*7!!]
With only 4 players, that's 1 in 9.93 * 10^13
With 9 players, it's 1 in 7.88 * 10^11
