tmoney941
tmoney941
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April 8th, 2015 at 7:35:40 AM permalink
Hello!

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!
AxelWolf
AxelWolf
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April 8th, 2015 at 8:09:15 AM permalink
Quote: tmoney941

Hello!

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.
♪♪Now you swear and kick and beg us That you're not a gamblin' man Then you find you're back in Vegas With a handle in your hand♪♪ Your black cards can make you money So you hide them when you're able In the land of casinos and money You must put them on the table♪♪ You go back Jack do it again roulette wheels turinin' 'round and 'round♪♪ You go back Jack do it again♪♪
ThatDonGuy
ThatDonGuy
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April 8th, 2015 at 8:45:58 AM permalink
Quote: AxelWolf

No 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.
ChesterDog
ChesterDog
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April 8th, 2015 at 8:57:10 AM permalink
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
ThatDonGuy
ThatDonGuy
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April 8th, 2015 at 9:48:04 AM permalink
Quote: ChesterDog

One 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.
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