What are the odds for a Straight Flush when...

Two players are head to head. (two players are being dealt 2 cards each and of course 5 table cards)

You have a single deck with:

Example 1
All 13 Diamonds
Just 4 Hearts
Just 4 Clubs
Just 5 Spades

What are the odds of ONE OR BOTH OF THE TWO PLAYERS being dealt a Diamond Straight Flush in Texas Holdem?


And then Example 2
All 13 Diamonds
Just 5 Hearts
Just 6 Clubs
Just 7 Spades

What are the odds of ONE OR BOTH OF THE TWO PLAYERS being dealt a Diamond Straight Flush in Texas Holdem in this situation?

First to answer correctly gets a cookie from Google!

Even money if I am allowed to deal.

Quote: Buzzard

Even money if I am allowed to deal.

Only if you're a Mech.

Anyone with the maths?

This is much harder than I'd originally thought. Cant figure out how to handle the short suits with length >4.

Quote: rdw4potus

This is much harder than I'd originally thought. Cant figure out how to handle the short suits with length >4.

Those would have to be just dead cards that couldn't form a SF. it would have to be only 11 out of all combinations possible with 26 cards in groups of 7 (10 diamond SF, 1 spade), and that assumes the spades are sequential, which he didn't state. He didn't exclude the board from having the SF, so you shouldn't have to subtract those hands. The second group would form up similarly, but again, he didn't say the partial suits were sequential, so it may be a trick question.

Allow me to clarify.

The only Straight Flush I am looking for are ones via Diamond flush.

EDIT- I will edit the OP to include this. I should also include that this would be a "heads up" sort of situation. So it would actually be a single flush possible out of 9 cards (two players hands and the table's 5)

Under those conditions;

There are 5 cards a person could hold with a SF missing 1 card on the board.
There are 10 combinations a person could hold with 2 in hand and 3 cards on the board.
There are 10 SF's in a single suit (diamonds).

so, (5+10)*10 = 150 configurations where at least one of two players could have the SF.

This excludes the board having a SF, which would fall under "other hands" like any other junk, but does not exclude the other player from having a combination which also includes a diamond SF. For example, board holds A 3 4 5 x; one player has the 2, other player has the 6 7. Both have SF's, so that hand would have to be taken into consideration and subtracted from total hands somehow, because the condition is that only one player has a SF. Which means that 150 configurations is not the same as total hands where a straight flush is possible.

Gonna have to think more about how to structure this.

Getting there!

So, saying hypothetically that the number of possible configurations for a straight flush is 150. You would then need to find out how many possible hands do NOT create a straight flush. That... is the harder number to figure. I'm sure there is a set of equations that can bring up these results quite quickly, but my mathematics ended at College Statistics.

Summoning the math geniuses!

Quote: Rorry

Getting there!

So, saying hypothetically that the number of possible configurations for a straight flush is 150. You would then need to find out how many possible hands do NOT create a straight flush. That... is the harder number to figure. I'm sure there is a set of equations that can bring up these results quite quickly, but my mathematics ended at College Statistics.

Summoning the math geniuses!

well, as to the first group,

combin(26,2) + combin(24,2) + combin(22,5) = 325 + 276 + 26334 = 26935 total hands possible from the stripped deck.

26935 - 150 would give you something, but as I said above, that would include configurations where BOTH players would have a SF. So those two numbers are, at the moment, apples and oranges. (One is total hands, one is a configuration of cards). That's the part I'm still mulling over.

I'm going to step back now and let the math geniuses shine the light for you. I'm just learning, myself, and that's who you're asking for, not me. Thanks for your patience.

Both of them getting it is fine. Any time one of them get's it is counted into the likely hood. So if both of them get it, then one of them get's it.

The original questions asked for the odds of one and only one player having a Diamond Straight Flush. Here are my results for that:

With the 26-card (13-4-4-5) deck: 12782304/2362159800 ≈ 0.005411 ≈ 1 in 185
With the 31-card (13-5-6-7) deck: 32707664/15241016700 ≈ 0.002146 ≈ 1 in 466

The questions have been revised to ask for the odds of one or both players having a Diamond Straight Flush. Here are my results for that:

With the 26-card (13-4-4-5) deck: 13285872/2362159800 ≈ 0.005624 ≈ 1 in 178
With the 31-card (13-5-6-7) deck: 33886732/15241016700 ≈ 0.002223 ≈ 1 in 450

Well, thanks, JB. I was waayyyy off, looking at those numbers. Back to the books.

Quote: beachbumbabs

Well, thanks, JB. I was waayyyy off, looking at those numbers. Back to the books.

No you were strangely close. 26935 - 150 = 26785. Take that and divide 150 by it and you get 0.0056.

Compare that to JB's calculation: With the 26-card (13-4-4-5) deck: 13285872/2362159800 ≈ 0.005624 ≈ 1 in 178

I'm not sure if this was coincidence or you were actually on the right track?

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At any rate thank you JB! Here is your cookie! --> http://images1.wikia.nocookie.net/__cb20130324185051/creepypasta/images/4/42/Cookie.gif