Thank you

Quote:TOPDOGFuther - the payout will happen if any straight flush happens. Could be 2,3,4,5,6 or 3,4,5,6,7 or 4,5,6,7,8 or 5,6,7,8,9. Not just the straight but has to be the straight flush

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Right. The first thing to do is figure out how many straight flushes there are:

2-3-4-5-6 (1)

3-4-5-6-7 (2)

4-5-6-7-8 (3)

5-6-7-8-9 (4)

4 * 4 = 16

The next thing we have to do is figure out how many total cards there are:

2 3 4 5 6 7 8 9 ---> 8 * 4 = 32

The next thing we have to do is figure out how many ways we can pick five of 32 cards:

https://web2.0calc.com/

nCr(32,5) = 201,376

The next thing we have to do is divide the number of combinations that give us a straight flush by the total possible number of combinations:

16/201376 = 0.0000794533608772 or 0.00794533608772%--->1 in 12,586 to be dealt a straight flush.

Significantly more likely than with a full deck.

Quote:TOPDOGThank you so much. That was very helpful. So now I'll throw in a monkey wrench and ask how to calculate the odds if there are 2 Jokers (wild cards) thrown into the mix and off the 34 TOTAL cards in play only 20 cards are drawn to show 4 different 5 card hands

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Perhaps one of the guys good at programming will be along for this one. I could do it, but it would take me longer than I want to spend. Sorry about that.

Quote:TOPDOGThank you so much. That was very helpful. So now I'll throw in a monkey wrench and ask how to calculate the odds if there are 2 Jokers (wild cards) thrown into the mix and off the 34 TOTAL cards in play only 20 cards are drawn to show 4 different 5 card hands

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Do you want to know the probability that a 5-card hand will be a straight flush, or that, if you are dealt four separate hands from the 34-card deck, that at least one of them is a straight flush? The second one is a harder problem.

For the first one, first determine how many ways there are of getting a straight flush:

No jokers: 23456, 34567, 45678, 56789

One joker: for each of the "no jokers" sets, there are 5 ways to replace a card with a joker - but that counts three sets twice (J3456 and 3456J; J4567 and 4567J; J5678 and 5678J), so that's 17 ways

Two jokers is slightly harder - for now, just count them:

From 23456: 234, 235, 236, 245, 246, 256, 345, 346, 356, 456

From 34567: since every set of 3 numbers in 3456 was already counted in 23456, count just the ones with a 7 in them: 347, 357, 367, 457, 467, 567

From 45678: since every set of 3 numbers in 4567 has already been counted, count just the ones with an 8 in them: 458, 468, 478, 568, 578, 678

From 56789: since every set of 3 numbers in 5678 has already been counted, count just the ones with an 9 in them: 569, 579, 589, 679, 689, 789

That is a total of 28

Since each straight flush can be one of four suits, the total is 4 * (4 + 17 + 28) = 196

There are C(34,5) = 278,256 hands, so the probability is 196 / 278,256, or about 1 / 9938

Quote:ThatDonGuyQuote:TOPDOGThank you so much. That was very helpful. So now I'll throw in a monkey wrench and ask how to calculate the odds if there are 2 Jokers (wild cards) thrown into the mix and off the 34 TOTAL cards in play only 20 cards are drawn to show 4 different 5 card hands

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Do you want to know the probability that a 5-card hand will be a straight flush, or that, if you are dealt four separate hands from the 34-card deck, that at least one of them is a straight flush? The second one is a harder problem.

For the first one, first determine how many ways there are of getting a straight flush:

No jokers: 23456, 34567, 45678, 56789

One joker: for each of the "no jokers" sets, there are 5 ways to replace a card with a joker - but that counts three sets twice (J3456 and 3456J; J4567 and 4567J; J5678 and 5678J), so that's 17 ways

Two jokers is slightly harder - for now, just count them:

From 23456: 234, 235, 236, 245, 246, 256, 345, 346, 356, 456

From 34567: since every set of 3 numbers in 3456 was already counted in 23456, count just the ones with a 7 in them: 347, 357, 367, 457, 467, 567

From 45678: since every set of 3 numbers in 4567 has already been counted, count just the ones with an 8 in them: 458, 468, 478, 568, 578, 678

From 56789: since every set of 3 numbers in 5678 has already been counted, count just the ones with an 9 in them: 569, 579, 589, 679, 689, 789

That is a total of 28

Since each straight flush can be one of four suits, the total is 4 * (4 + 17 + 28) = 196

There are C(34,5) = 278,256 hands, so the probability is 196 / 278,256, or about 1 / 9938

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Thank you! I also apologize if this is what he wanted. I thought he wanted to know the probability of at least one, at least two, at least three, all four...no way was I doing that.