Cripple Mr Onion is a card game that involves 8 suits (104 cards in toto), with 2-7 players being dealt 10 cards each (without replacement).

I've tried several different websites for hypergeometric distribution, but really don't know what I'm doing with them (I've no idea whether the results are correct or not, since I can't tell if I input the appropriate data).

I've also tried it by hand, but can't wrap my head around the proper method. When the suit doesn't matter, then it's easy enough; when the suit matters, it becomes rather more complicated.

How does one calculate the probability of hands in a card game, when the deck involved has 8 suits?

Taking two basic examples, I would hope that I could then work out the rest for myself:

  1. I imagine that calculating the probability of my hand containing two aces (of any suits) is the simplest starting point for this game's rules.
  2. For comparison and contrast, I then also calculate the probability of my hand containing {4, 5, 7} in only one or two specific suits (not the other six possible suits).

There is an excellent work-up regarding Yu-Gi-Oh! in another post's answer, but I don't know how to modify it properly for this problem (which would probably be the best solution possible).

  • How good a programmer are you? Writing a small program that generates a large number of hands, will get you very good approximations.
    – John
    Commented Jul 8, 2018 at 15:57
  • 1
    @John There is no need to write a program to do that as you can use math to calculate the exact values, just need to know the correct formulas.
    – Joe W
    Commented Jul 8, 2018 at 15:58
  • I'm math-oriented (which is frustrating, since card probabilities shouldn't present any difficulties whatsoever... yet they do). Knowing the exact values isn't necessary, since this is only for my own curiosity, but I'm too OCD to rely upon a program's generated aggregation. Commented Jul 8, 2018 at 16:06
  • 1
    @mmathis, several of us are capable of answering this question. I have an answer ready, once/if it gets reopened.
    – John
    Commented Jul 9, 2018 at 14:46
  • 5
    Simple math questions on games are on topic.
    – John
    Commented Jul 9, 2018 at 15:28

2 Answers 2


I find hypergeometric series nice for computing, but poor for explaining. For teaching purposes, I much prefer expanding these out using binomial coefficients. Some books, and Wolfram Alpha call this "choose."

The key here is to realize that order does not matter. The total number of possible hands is choose(104, 10) = 26 trillion.

Now, to count how many hands have 2 aces, you need to figure out how many ways there are to choose 2 aces, and multiply that by the number of ways to get 8 cards with no aces. Writing it out in detail we have:

                    # of hands w 2 aces    
P(2 aces in hand) = -------------------
                        # of hands

     # ways to choose 2 aces * # ways to choose non aces
  = ----------------------------------------------------

     choose(8,2) * choose(96,8)
  = ----------------------------  = 0.14

Here's the Wolfram Alpha expression for all the accuracy you could ever want.

Your second question, how to find {4,5,7} in one of two suits is ambiguous. Do all three have to be in the same suit? Or different suits? There's another issue. Let's call the suits ABCDEFGHIJ, and say you want the 4,5,7 to be in suits A or B. Do you care if there is a 4 in suit F?

P(4,5,7 in either of 2 suits, don't care about other suits) = 

     # ways to choose {4,5,7} * number of ways to choose other cards
  = ----------------------------------------------------

  = ----------------------------  = 0.004

Or, the related: P(4,5,7 in either of 2 suits, no other 4,5,7 in hand) =

     # ways to choose {4,5,7} * number of ways to choose other cards
  = ----------------------------------------------------

  = ----------------------------  = 0.001

You might reasonably ask, why are these numbers so much lower than for aces? This is because you are specifying three cards in the hand, and the aces are only specifying two cards in the hand.

Keep in mind, that in most games, this isn't the full story. For instance, the way I've computed things, I've included a pair of aces and three kings. In poker, that's a full house, and so needs to be removed from this probability. If that's what you want, then things get more complicated.

  • 1
    On the P(2 aces in hand) part ... you wrote choose(10,2) for the # ways to choose 2 aces... however, as there are 8 suits, there are 8 aces in the game. So, shouldn't it be chose(8,2) instead? ... seems that the prob is correct (0.12) but you just confused the 8 with 10 (i.e.: the calculation was made with 8, but you wrote 10).
    – DarkCygnus
    Commented Jul 9, 2018 at 23:01
  • @DarkCygnus, you are correct.
    – John
    Commented Jul 10, 2018 at 0:33

How does one calculate the probability of hands in a card game, when the deck involved has 8 suits?

This depends on the query you seek, and the specific restrictions it might have. Answering your example:

I imagine that calculating the probability of my hand containing two aces (of any suits) is the simplest starting point for this game's rules.

First, I'm taking this as the probability of your hand containing two, and exactly two, Aces.

You have 8 suits, 104 cards in total, which gives 27 cards per suit.

We know that each suit has only one copy of an Ace. That is, there is only one ace of Hearts, etc. Thus, there are 8/104 cards that are aces. This gives us a 8/104 chance to pull out any Ace. From this we also have a 96/104 chance of not pulling an Ace.

So, a hand consists of 5 cards 10 cards. Based on the previous probabilities, and considering that when you pull a card the total number of cards is reduced by one we have:

Prob. of a hand with exactly 2 aces = (8/104)x(7/103)x(96/102)x(95/101)x(94/100)x(93/99)x(92/98)x(91/97)x(90/96)x(89/95) = 0.003161

Do note that the chances would be higher if the query where containing at least two aces.

The terms being multiplied are: The first two the chances of pulling Aces, and the rest the chance of not pulling an ace. All factors consider that cards pulled out are not returned to the pool.

Edit: If we consider the a priori probabilities instead, that is without accounting for cards drawn and without any specific order, the calculation changes to:

(8/104)2x(96/104)8 = 0.003119

  • You need to multiply your result by binomial(10,2). You are computing the probability the first two cards of the 10 are aces, but the order of the draw is unimportant in this case.
    – John
    Commented Jul 9, 2018 at 22:17
  • @John so, you say that probabilities should be considered as a whole, without counting the reduced deck size per each draw? That would give instead: (8/104)**2 x (96/104)**8... or is that combination probability to include to compensate for the cards drawn?
    – DarkCygnus
    Commented Jul 9, 2018 at 22:29
  • I don't understand your question. You used permutations (instead of binomial coefficients) to compute your probability. But that forces order to matter. Your proposed computation in the comments is appropriate for a dice rolling scenario, but as these are samples without replacement, we need an "urn" model. Not sure I'm helping. So, I wrote up a complete answer. Hope that makes it clearer.
    – John
    Commented Jul 9, 2018 at 22:53
  • @John perhaps yours is more clean it seems :)
    – DarkCygnus
    Commented Jul 9, 2018 at 22:59

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