So my question has two parts and I have managed to confused myself into lack of comprehension.

Suppose you are handed a shuffled, stacked deck of playing cards i.e. instead of 52 cards there are 104 as the I included an extra copy of each card for the heart suit (two 2 of hearts, two 3 of hearts, etc) and four copies of each spades (four 2 of spades, four 3 of spades, etc).

(P(hearts) = 1/4, P(spades)=1/2, P(clubs)=P(diamond)=1/8)

Pick a number, n. What is the likelihood that in the top n cards there are h hearts, s spades, and c clubs? where h + s + c + d = n (where d is the number of diamonds)

A bit harder: Now pick two numbers m and n. First remove the first m cards from the randomized deck. Now what is the likelihood that in the top n cards there are h hearts, y spades, and c clubs, h + s + c + d = n (where d is the number of diamonds).

So for the first part (without removing m cards at random) is clear to me that the top n cards of the stacked deck of cards is 104Cn an the number of ways the specific combination of (h, s, c, d) is the multinomial coefficient (n!/(h!*s!*c!*d!)). Where I am confusing myself is, is this just the (n C h,s,c,d / 104Cn * 100) as 104Cn inherently handles the uneven distributions in the suits of cards? or do I need to weight the multinomial coefficient somehow?


1 Answer 1


For the first part, let the probability equal $p_1$. Then, $$p_1 = \dfrac{\dfrac{n!}{h!\ s!\ c!\ d!} \cdot \dfrac{104-n!}{(26-h)!\ (52-s)!\ (13-c)!\ (13-d)!}}{\dfrac{104!}{26!\ 52!\ 13!\ 13!}} = \dfrac{{26 \choose h}\cdot {52 \choose s}\cdot {13 \choose c}\cdot {13 \choose d}}{{104 \choose n}}$$

The numerator of the RHS in the above expression is more indicative of the choices favorable to the problem. The denominator is the total number of ways of choosing n cards from 104.

For the second part, let m be restricted to the case where $m \leq 104-n$ (as otherwise the probability is 0).

We are dividing the pile into 3 parts of size $m,n$ and $104-m-n$. The total number of ways of doing this are $\frac{104!}{m!\cdot n!\cdot (104-m-n)!} = \binom{104}{m,n,104-m-n}$

For the pile of size n, we have the probability of favorable cases as ${26 \choose h}\cdot {52 \choose s}\cdot {13 \choose c}\cdot {13 \choose d}$. The remaining $104-n$ need to be split into $m$ and $(104-m-n)$, which can be done in ${104-n \choose m}$ ways

Let $p_2$ be the required probability for the second part. Then:

$$p_2 = \dfrac{{26 \choose h}\cdot {52 \choose s}\cdot {13 \choose c}\cdot {13 \choose d}\cdot {104-n\choose m}}{\frac{104!}{m!\cdot n!\cdot (104-m-n)!}} = \dfrac{{26 \choose h}\cdot {52 \choose s}\cdot {13 \choose c}\cdot {13 \choose d}}{\binom{104}{n}}$$

which is same as the first answer. Many thanks to @angryavian for the correction in second part.

  • 1
    $\begingroup$ For $m \le 104-n$, shouldn't the answer to part two should be the same as part one? $\endgroup$
    – angryavian
    Apr 3, 2021 at 18:27
  • $\begingroup$ Indeed. Many thanks for pointing this out. $\endgroup$ Apr 3, 2021 at 18:38
  • 1
    $\begingroup$ thank you for taking the time to answer and clarify. I see where I confused myself and how the second case (removing m then revealing n) is identical to the first. $\endgroup$
    – SumNeuron
    Apr 5, 2021 at 7:31
  • $\begingroup$ @RahulMadhavan if you have a few moments could you also maybe check out this $\endgroup$
    – SumNeuron
    Apr 5, 2021 at 8:44
  • $\begingroup$ Oh wait, I have a quick follow up. What happens if m > h? $\endgroup$
    – SumNeuron
    Apr 5, 2021 at 14:26

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