I'm doing the probability calculations having some trouble.

Example 7 card flush


6 card flush


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


I think I have 7 card straight flush at


edit 2 ------

the above is wrong I believe this will work


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


I think it would look like this

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

This mimics the form in WIKI.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.