Crazy Texas hold'em hand at WSOP
June 28th, 2012 at 5:55:32 PM
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I know this isn't a real hand (other than the Royal) but I was at a table where a guy got 8,9,10,J,Q,K,A of spades there was a Flush on the board and he had 8,Q spades in his hand. What is the odds of that hand coming up? has to be over a Billion to 1 right?
June 28th, 2012 at 6:05:55 PM
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For a 7-card Royal Flush holding the Q and 8 and the board coming down AKJ109, four in 133,784,560; in spades, one in 133,784,560.
Beware of all enterprises that require new clothes - Henry David Thoreau. Like Dealers' uniforms - Dan.
June 28th, 2012 at 6:19:48 PM
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Quote: PaigowdanFor a 7-card Royal Flush holding the Q and 8 and the board coming down AKJ109, four in 133,784,560; in spades, one in 133,784,560.
isn't that same odds for just a Royal? what is it for the 7 card Royal Straight flush?
June 28th, 2012 at 6:29:17 PM
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The odds of first being dealt a suited 8Q (any suit), and then the board being AKJT9 of the same suit, are 1 in 702,368,940.
If the suit must be spades, then the odds are 1 in 2,809,475,760.
If the suit must be spades, then the odds are 1 in 2,809,475,760.
June 28th, 2012 at 7:26:23 PM
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Quote: JBThe odds of first being dealt a suited 8Q (any suit), and then the board being AKJT9 of the same suit, are 1 in 702,368,940.
If the suit must be spades, then the odds are 1 in 2,809,475,760.
Amazing! Thank you JB! what is the formula to calculate that?
June 28th, 2012 at 7:57:10 PM
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number of suits * number of ways to get AKJT9 of the same suit = 4 * 1 = 4
divided by
number of 2-card hands * number of 5-card boards = combin(52,2) * combin(50,5) = 1326 * 2118760 = 2809475760
4/2809475760 = 1/702368940 for any suit
if the suit must be spades (or any other specific suit) then change the "number of suits" above from 4 to 1 and you end up with 1/2809475760
Now if you want the odds of getting those 7 cards in any order (that is, the suited hole cards could be 89, 9K, AJ, A8, KQ, etc.) then the math would be:
number of suits * number of 2-card hands from those 7 cards = 4 * combin(7,2) = 4 * 21 = 84
divided by the same figure above yields 84/2809475760 = 1 in 33,446,140
or if it must be in a specific suit, then 21/2809475760 = 1 in 133,784,560
divided by
number of 2-card hands * number of 5-card boards = combin(52,2) * combin(50,5) = 1326 * 2118760 = 2809475760
4/2809475760 = 1/702368940 for any suit
if the suit must be spades (or any other specific suit) then change the "number of suits" above from 4 to 1 and you end up with 1/2809475760
Now if you want the odds of getting those 7 cards in any order (that is, the suited hole cards could be 89, 9K, AJ, A8, KQ, etc.) then the math would be:
number of suits * number of 2-card hands from those 7 cards = 4 * combin(7,2) = 4 * 21 = 84
divided by the same figure above yields 84/2809475760 = 1 in 33,446,140
or if it must be in a specific suit, then 21/2809475760 = 1 in 133,784,560
June 28th, 2012 at 9:27:02 PM
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I came up with 1 in 37. But I may be wrong, am rechecking now.
June 29th, 2012 at 12:03:38 AM
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Quote: buzzpaffI came up with 1 in 37. But I may be wrong, am rechecking now.
I think you are right because we've had the WSOP about 37 times probably [g]
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!” She is, after all, stone deaf. ... Arnold Snyder
June 29th, 2012 at 2:41:01 AM
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there is a video poker machine at Caesars Palace -- only one -- just outside the Shadow bar -- which offers three payoffs including a super bonus for a 7 card royal flush. Pays something like 60-thou or 80-thou coins for 7 coins bet.
My son plays it, and if it weren't 2:40 in the morning I'd call him to get the name. the game pays three different combinations for seven cards, if I recall.
My son plays it, and if it weren't 2:40 in the morning I'd call him to get the name. the game pays three different combinations for seven cards, if I recall.
