Specific hand odds
November 5th, 2015 at 9:42:05 AM
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Can you tell me how often one can expect to get a specific hand (lets say in DDBP) like four 8's with an Ace?
November 5th, 2015 at 10:05:12 AM
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If you have VPW, you can figure it out yourself. I don't have a computer with VPW on it right now.
But basically you create your own custom paytable, including hands like 8888 w/ A kicker, paying 50-for-1 (or 250-for-5). It'll generate the strategy, the same strategy for regular DDB. But when you go to the screen where it shows hand frequencies, you'll have a list of those other hands you're looking at.
Out of curiosity, why?
But basically you create your own custom paytable, including hands like 8888 w/ A kicker, paying 50-for-1 (or 250-for-5). It'll generate the strategy, the same strategy for regular DDB. But when you go to the screen where it shows hand frequencies, you'll have a list of those other hands you're looking at.
Out of curiosity, why?
November 5th, 2015 at 11:21:23 AM
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Do you mean, on the deal, or after the draw?
After the draw, it depends on what strategy you are using.
On the deal, of the 2,598,960 different possible deals, there are 4 that are four aces and an 8 (four aces & 8 of spades; four aces & 8 of hearts; four aces & 8 of clubs; four aces & 8 of diamonds).
It would also help if you had a list of the hands you were interested in. There are 134,459 possible results.
After the draw, it depends on what strategy you are using.
On the deal, of the 2,598,960 different possible deals, there are 4 that are four aces and an 8 (four aces & 8 of spades; four aces & 8 of hearts; four aces & 8 of clubs; four aces & 8 of diamonds).
It would also help if you had a list of the hands you were interested in. There are 134,459 possible results.
November 5th, 2015 at 12:19:08 PM
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On the deal it's quite easy to find any 5 card combination you want...Quote: JDLVCan you tell me how often one can expect to get a specific hand (lets say in DDBP) like four 8's with an Ace?
Odds of Card #1
Odds of Card #2
Odds of Card #3
Odds of Card #4
Odds of Card #5
Thus, if you want 8-8-8-8-A... P(8888A) = (4/52)*(3/51)*(2/50)*(1/49)*(4/48) = .0769 * .0588 * .04 * .0204 * .0833 = .00000031, or about 1 in 3.2 million.
Do note: I didn't specify any particular suit of the 8's... so if you had to have them in a specific order (which I highly doubt) that would compound the probability greatly. Just a disclaimer, I don't think that's the actual case.
After the deal it's more of the same... given you hold X-X and you need Y-Y-Y. For example, you get dealt 8-8-x-x-x... you want to know the odds of re-drawing 8-8-8-8-A...
Well you hold 8-8, and assuming the 3 throw aways were NOT 8's or A's, then the probability is just that of drawing 8-8-A from the REMAINING deck:
P(88A) = (2/47) + (1/46) + (4/45) = ...you can take it from here =).
Playing it correctly means you've already won.
