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I'm doing the probability calculations having some trouble.

Example 7 card flush

4*(C(13,7)-217)

6 card flush

4*(C(13,6)-71)*39

I am pretty sure these are correct I am having trouble with the 7 and 6 card straight without flush.

What i have so far is 217 * 15540 but I believe this is 7 distinct cards five of which make up straight I need to adjust this to get the odds 7 distinct cards that are all ranked in order of straight without flush happening.

Any help would be appreciated

Thanks Mark

----------------------edit-------

I think I have 7 card straight flush at

4*13C(46,7)

edit 2 ------

the above is wrong I believe this will work

C(4,1)[C(1,1)C(47,2)+C(9,1)C(47,2)]

The last C(47,2) should allow for the card that should allow for last two cards to be the correct rank for the 7 card straight flush

---------------------edit 3---------------------

Thanks grung and paparazzo

Let's see -- 4*(C(13,7) - c(8,1))

I'm pretty new to this but this looks like it might work for a non flushed 5 card straight the idea being 13 available ranks minus the 8 ranks that aren't continuous leaving the five ranks that will work to complete a straight times the four suits.

Anyways I think I need to further explain what I'm trying to figure out. Imagine 7 card draw where you can get seven more cards and keep the first seven also. So now you will have 14 cards and I want the probabilities of making all of the usual 7 card hands (best 5 of 7 cards) plus I'm trying to figure the probabilities of some 7 card hands like a (7 card straight) and a (4 of a kind 3 of a kind) using this C(52, 14) divisor.

I am pretty sure I have the straight flush right.

Straight Flush

-- C(4,1)[C(1,1)C(47, 9) + C(9, 1)C(46, 9)] ---

Thanks all

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I think it would look like this

4*(C(13,7) - c(8,1))

This mimics the form in WIKI.

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  • $\begingroup$ This is being automatically flagged as low quality, probably because it is so short. At present it is more of a comment than an answer by our standards. Can you expand on it? You can also turn it into a comment. $\endgroup$ – gung - Reinstate Monica Oct 6 '18 at 0:20
  • $\begingroup$ @gung Is this enough. If you want to move it to a comment OK by me. I might be able to expand more tomorrow. $\endgroup$ – paparazzo Oct 6 '18 at 0:55
  • $\begingroup$ Could you just explain the reasoning behind how you came up w/ 4*(C(13,7)-c(8,1))? $\endgroup$ – gung - Reinstate Monica Oct 6 '18 at 0:59

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