Poker Odds - Probability of two Royal Flushes in a two handed game.?

Call the two players A and B. It is reasonable to assume that all 10 cards are drawn from the same 52-card deck, without replacement.

There are 52C5 possible hands for player A. Only 4 of these hands are royal flushes, namely 10, J, Q, K, A of all clubs, all hearts, all diamonds, and all spades.

So P(A gets a royal flush) = 4/(52C5).

Once a royal flush is selected for player A, there are 47C5 possible hands for player B. Only 3 of these hands are royal flushes, namely 10, J, Q, K, A of all the same suit, different from the suit for player A.

So P(B gets royal flush, given A gets royal flush) = 3/(47C5).

Thus, P(A and B both get royal flushes)

= P(A gets a royal flush) * P(B gets royal flush, given A gets royal flush)

= 4*3/(52C5 * 47C5)

= 4*3*(1*2*3*4*5)^2 / (52*51*50*49*48*47*46*45*44*43)

= 1/(13*17*49*2*47*46*15*11*43) or about 1 out of 332 billion.

(Note: the answer is not just [4/(52C5)]^2, since all 10 of the cards are drawn without replacement. When the first player's 5 cards are drawn, the 5 cards are not replaced before the second player's 5 cards are drawn.)

Here's another method that gives the same answer. Draw the 10 cards, one at a time without replacement, with the first 5 given to player A and the last 5 given to player B.

Player A's first card can be a 10, J, Q, K, or A of any of 4 suits, which has probability 20/52.

Player A's second card must be one of the 4 remaining values and of the same suit as player A's first card, which has probability 4/51.

Player A's third card must be one of the 3 remaining values and of the same suit as player A's first card, which has probability 3/50.

Player A's fourth card must be one of the 2 remaining values and of the same suit as player A's first card, which has probability 2/49.

Player A's fifth card must be the remaining value and of the same suit as player A's first card, which has probability 1/48.

Player B's first card can be a 10, J, Q, K, or A of any of 3 remaining suits, which has probability 15/47.

Player B's second card must be one of the 4 remaining values and of the same suit as player B's first card,, which has probability 4/46.

Player B's third card must be one of the 3 remaining values and of the same suit as player B's first card, which has probability 3/45.

Player B's fourth card must be one of the 2 remaining values and of the same suit as player B's first card, which has probability 2/44.

Player B's fifth card must be the remaining value and of the same suit as player B's first card, which has probability 1/43.

The overall probability that both hands are royal flushes is

20*15*(1*2*3*4)^2 / (52*51*50*49*48*47*46*45*44*43)

= 4*3*(1*2*3*4*5)^2 / (52*51*50*49*48*47*46*45*44*43), once again.

Lord bless you today!

So we need to select 5 cards from 1 suit and none from any other and we have to do this twice essentially. I guess the way that the deal is done also influences things and how many decks one is playing with. So for simplicity lets say that there is one deck a standard 52 playing cards, and that person A gets dealt all 5 cards first, then person B gets dealt their 5 cards.

So,

P ( A gets Royal Flush ) = (13 C 5) / ( 52 C 5 ) = 33 / 66640

P ( B gets Royal Flush ) = (13 C 5) / ( (52 - 5) C 5 ) = 39 / 46483

Now since they are in different decks the events are independent, and for other reasons too. So we can multiply these two probabilities for the desired answer.

P ( A & B get Royal Flushes ) = ( 33 * 39 ) / (66640 * 46483) = 4.15 * 10^(-7) = 0.000000415

I'm not sure what you mean by 'five a kind suited'...if you mean 5 of the same rank, then that's only possible when wild cards are used (and would not be 'suited' in any case.) If you mean 'straight flush', then a royal flush IS a straight flush...the highest possible one (and the best possible hand when no wild cards are used.)

Probability is defined as the number of successes available divided by the number of possibilities.

Number of successes:

There are 4 suits, so choosing any 2 of those suits there are 4C2 ways, or 4x3 / 1x2 = 6 ways of success. (For those not educated in this type of math, it is a way of counting called combinatorics. 4C2 means the number of combinations of 4 things, if choose 2. Some calculators have keys to perform this. For example, for 7 C 2, the operation is 7! / (2! x 5!). '!' means factorial. 5! means 1x2x3x4x5)

Number of possibilities:

If the royal flush is ace through 10, that is 5 cards per hand. So this is 10 cards dealt. There are 52 C 10 = 15820024220 unique groups of 10 from the deck. For each group of 10 then, how many ways are there to deal two 5-card hands? There are 10 C 5 = 252 unique ways to draw 5 cards from these 10. But since for each selection, its complement is in the other person's hand, we divide by 2, because it doesn't matter who has what hand. So there are only 126 unique pairs of 5-card hands for each of these groups of 10.

So then the number of unique pairs of 5-card hands is 126 x 15820024220 = 1,993,323,051,720.

So then the probability of dealing 10 cards to two people, from a fairly shuffled deck, and getting two royal flushes, is 6 / 1993323051720 = 3.010048971 x 10^-12, or 0.000000000003010048971, basically 3 out of a trillion.

EDITED:

Odds of 1 royal flush = 4 x 52C5 = 649,739:1

Odds of 2 royal flushes:

649,739 x 649,739 :1 or 4.22 x10^11:1 or 422 trillion to one.

Read'em and weep, Jerry <3