3 cards to a royal
October 7th, 2013 at 7:36:35 AM
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One of the Wizard of Odds questions computed the odds of getting a 3 card royal at VP:
"What is the probability of being dealt three to a royal flush in video poker?
— Anonymous
There are 4 suits to choose from for the 3 to a royal. There are combin(5,3)=10 ways to choose 3 out of the 5 ranks. There are combin(47,2)=1081 ways to choose the other 2 cards. There are combin(52,5)=2,598,960 ways to choose 5 cards out of 52. So the probability of getting 3 to a royal is 4×10×1081/2,598,960 = 1.66%."
My question is in the 2 card combinations there will be 4 that complete a royal flush and a lot that will make it a 4 card royal. How are these being excluded in the computation?
"What is the probability of being dealt three to a royal flush in video poker?
— Anonymous
There are 4 suits to choose from for the 3 to a royal. There are combin(5,3)=10 ways to choose 3 out of the 5 ranks. There are combin(47,2)=1081 ways to choose the other 2 cards. There are combin(52,5)=2,598,960 ways to choose 5 cards out of 52. So the probability of getting 3 to a royal is 4×10×1081/2,598,960 = 1.66%."
My question is in the 2 card combinations there will be 4 that complete a royal flush and a lot that will make it a 4 card royal. How are these being excluded in the computation?
October 7th, 2013 at 7:41:28 AM
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larwiz,
As I read it, they're coming from the combin(5,3) being used to only take 3 out of the 5, and then the combin(47,2) - not (49,2) - being used to determine the other 2 cards. Two of the 5 initially subtracted from 52 are not considered as "other 2 cards", regardless of which two are missing from the royal.
As I read it, they're coming from the combin(5,3) being used to only take 3 out of the 5, and then the combin(47,2) - not (49,2) - being used to determine the other 2 cards. Two of the 5 initially subtracted from 52 are not considered as "other 2 cards", regardless of which two are missing from the royal.
If the House lost every hand, they wouldn't deal the game.
October 7th, 2013 at 7:47:07 AM
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Thank for the quick answer. In the time it took me to post the question and then go sit in another chair then come back over here and log on this exact thought crossed my mind. I suddenly remember the 47 vrs 49.
October 19th, 2013 at 7:52:16 PM
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After a review of recent posts, I stumbled upon this thread...
Hmmm, but the answer given allows 2-Pair, A Straight, or a Flush to be the initial deal... all of which would be keepers, negating the chance of a royal. I would therefore presume a JoB VP game, and would like to ask what are the odds that I am dealt 3-RSF, and keep them in JoB?
Hmmm, but the answer given allows 2-Pair, A Straight, or a Flush to be the initial deal... all of which would be keepers, negating the chance of a royal. I would therefore presume a JoB VP game, and would like to ask what are the odds that I am dealt 3-RSF, and keep them in JoB?
Some people need to reimagine their thinking.
