Playing 7-card rummy with 4 players, I was dealt a 7 card Spade flush (not straight) and on the first card picked from the deck pulled an 8th card spade and immediately won (it was 23456 8910).

What are the odds of being dealt all 7 cards in the same suit?

3 Answers 3


What are the odds of being dealt all 7 cards in the same suit?

The odds of pulling 7 spades in 7 cards the number of ways to pull 7 spades (from among the 13 spades) divided by the number of ways to pull 7 cards (from among the 52 cards).

13c7 = 1716

52c7 = 133784560

1716 / 133784560 = 0.0000128 = 0.00128 %

For general suit (not just spades), multiply that by four:

= 0.0000513 = 0.00513 %

If you want to continue to the eighth card:

The odds of pulling 8 spades in 8 cards the number of ways to pull 8 spades (from 13) divided by the number of ways to pull 8 cards (from 52).

13c8 = 1287

52c8 = 752538150

1287 / 752538150 = 0.00000171 = 0.000171 %

And again multiply by four for general suit:

= 0.00000684 = 0.000684 %

(subtract the number of ways to get a straight if you want to exclude straights from "all cards in the same suit".)

Notation explanation:

13c8 is read as "13 choose 8" and means "the number of ways to choose 8 things from a set of 13 things". It is computed as 13!/((13-8)! * 8!).

In general, NcM = N! / ( (N-M)! * M! )

  • 1
    Adjusted for the 7-card version ( I got excited about the 8th card matching when I read the description and somehow focuesed on that ). Commented Apr 20, 2020 at 11:41
  • +1 argument is correct, but to someone who doesn't know maths, it must surely not look very intuitive. Your first sentence is good. Then you just state three things. It's not at all clear what these mean without knowledge of how to do the problem in the first place. For non-mathematicians, I think it would be really helpful if you add a few words -- eg, "13 of the cards are spades, so we wish to know 13c7, namely the number of distinct (unordered) ways of choosing 7 objects from an original collection of 13", and then similarly for 52c7. I think this would be illuminating for many people :)
    – Sam OT
    Commented Apr 20, 2020 at 15:02
  • 2
    Also, a little bugbear of mine: one should not think of NcM as N! ( (N-M)! * M! ). (I wanted to write an exclamation mark there for emphasis, but thought it would be confusing with the factorial! Also, you have typos in yours.) Instead, it is N (N - 1) ... (N - M + 1) / M!. Indeed, how many ways are there to choose M from N when you care about ordering? Precisely the numerators. Now divide by the number of reorderings, which is precisely the denominator. Simple! :) -- your formula (after being corrected) is just one way of writing this. Not that helpful for working out by hand either!
    – Sam OT
    Commented Apr 20, 2020 at 15:05
  • @SamT thanks for the catch. I've corrected the typos in the formulation. Commented Apr 20, 2020 at 16:53
  • @BenjaminCosman Exactly so, as I said. "multiply by four if you don't care that it's spades in particular" Commented Apr 20, 2020 at 16:56

Some slightly watered down math (that still gives the right answer).

You have a 100% chance of your first card being a "good" card.
There are then 51 cards left, of which 12 will be a card that matches suit, so you have a 12/51 chance that the second card will be "good".
11/50 chance that the third card will be "good". ....

So, you have 1x(12/51)x(11/50)x(10/49)x(9/48)x(8/47)x(7/46)=0.0000513

So you have 0.00513% chance of drawing a 7 card flush

  • You have 7/56 but I think that should be 7/46 - but also your math doesn't add up - I threw it into Excel and got .000051 - i.imgur.com/3Bs6Bgd.png
    – corsiKa
    Commented Apr 20, 2020 at 16:13
  • 1
    However despite the answer being off a bit, this is definitely the way I would solve the problem!
    – corsiKa
    Commented Apr 20, 2020 at 16:14
  • @corsiKa good catch. Had a typo in my formula in my first cell that I copied and pasted into the other cells.
    – Kevin
    Commented Apr 20, 2020 at 16:23

L. Scott Johnson is mostly correct except for 2 things:

  1. You specified it wasn't a straight flush (there are 7 8-card straight flushes per suit he should have removed.)
  2. You asked about the 7 card flush, which you were dealt before the draw, not the 8 card flush you drew into.

This gives us:

13C7 as the nCr we care about, giving us 1716 possible flushes of 7 cards.

Then we remove the 8 7-card straight flushes in each suit to give us 1708 possible flushes of 7 cards per suit.

We multiply that result by 4, since we want a flush of 7 cards, not specifically caring about suit giving us 6832 possible non-straight 7 card flushes.

Now we need 52C7, the number of combinations of 7 cards in 52 cards, which is 133,784,560.

This gives us 6832 out of 133,784,560 different 7 card hands that gives us odds of 1194444:61 against, or a 0.005107% chance of such a 7 card flush on the deal (or 0.001277% if you want a specific suit, or odds of 4777959:61 against)

If we do include the 7-card straight flushes, we have 6864 combinations in a deck of 7-card flushes, which gives us odds of 643162:33 against or 0.005131% of possible combinations.

  • 1
    In his particular scenario he didn't have a straight flush, but he asked "What are the odds of being dealt all 7 cards in the same suit?" Excluding a straight flush would lead to an incorrect answer to the question. Commented Apr 20, 2020 at 15:24
  • @GregSchmit Absolutely correct
    – Kevin
    Commented Apr 20, 2020 at 15:25
  • @GregSchmit then take a look at the last paragraph where I put the 32 straight flushes back in and give the odds on those.
    – Andrew
    Commented Apr 20, 2020 at 17:22
  • 1
    @Andrew No worries, I only mentioned that because you said that Scott was "mostly correct except...", indicating he was mistaken to include those. Commented Apr 20, 2020 at 17:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .