Poker Odds - Probability of two Royal Flushes in a two handed game.?
There was a story on strip poker on the literotica website.......
"I picked up my cards and stared at them. Only once before in my life had I ever seen a royal flush.. but that was what I held in my hands. I had the ten through the ace of clubs but, before I could lay the cards down on the table and watch the sexy lady strip off her panties, she beat me to it.
"Read 'em and weep, George," she announced, and placed her cards on the table facing me.
That was the third time I had seen a royal flush, and hers was in hearts. The odds against one such hand are immense; the chances of two at the same time in a two-handed game have probably never been calculated......"
I guess it could happen but the odds must be astronomical just how astronomical I would love to know.
Thanks
Story here: http://www.literotica.com/s/more-strip-poker-with-kelly-monaco
?2012-07-19T10:12:47Z
Favorite Answer
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.
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.