A local card room has two bonus promotions going on and we have a running argument about which is more likely, and what the odds are. All games are 9 handed if that matters in your calculation. The promotions are as follows...
1. A royal flush where both hole cards play, can be any suit.
2. Four-of-a-kind jacks where the one-eyed jacks are your hole cards.
Our discussion has gone back and forth mainly hinging on whether it is more difficult to have the five cards for a royal...or only four cards for Jacks, but needing two specific Jacks in the hole. Anybody have the math capacity to figure this one out and give me the probability?
Thanks!
No math, but I would bet that its harder to get the 4 jacks.Quote: tmoney941Hello!
A local card room has two bonus promotions going on and we have a running argument about which is more likely, and what the odds are. All games are 9 handed if that matters in your calculation. The promotions are as follows...
1. A royal flush where both hole cards play, can be any suit.
2. Four-of-a-kind jacks where the one-eyed jacks are your hole cards.
Our discussion has gone back and forth mainly hinging on whether it is more difficult to have the five cards for a royal...or only four cards for Jacks, but needing two specific Jacks in the hole. Anybody have the math capacity to figure this one out and give me the probability?
Thanks!
Quote: AxelWolfNo math, but I would bet that its harder to get the 4 jacks.
Yes math, and you would win that bet.
The probability of a Royal that includes both hole cards =
(the probability that the first hole card is part of a Royal)
x (the probability that the second hole card is part of the same Royal)
x (the probability the the five table cards include the other three cards in that Royal)
= 20/52 x 3/51 x 1081 / 2,118,760 = 1 / 86,632
There are (50)C(5) = 2,118,760 possible sets of 5 table cards from the 50 remaining when you remove your 2 hole cards; of these, there are (47)C(2) = 1081 sets that contain the other three cards in the Royal, as the table has 2 cards out of the 47 that are not in the Royal.
The probability of four Jacks where both one-eyes are hole cards =
(the probability that the first hole card is one of the two one-eyed Jacks)
x (the probability that the second hole card is the other one)
x (the probability that the five table cards include the other two Jacks)
= 2/52 x 1/51 x 16,215 / 2,118,760 = 1 / 173,264
The Royal is exactly twice as likely - in no small part because there are so many more possible sets of hole cards.
Quote: ThatDonGuy...The probability of a Royal that includes both hole cards =
(the probability that the first hole card is part of a Royal)
x (the probability that the second hole card is part of the same Royal)
x (the probability the the five table cards include the other three cards in that Royal)
= 20/52 x 3/51 x 1081 / 2,118,760 = 1 / 86,632
...
One small error in the royal calculation: It should be =20/52 x 4/51 x 1081 / 2,118,760 = 1 / 64,974
Quote: ChesterDogOne small error in the royal calculation: It should be =20/52 x 4/51 x 1081 / 2,118,760 = 1 / 64,974
Why, yes. Yes, it should.