# Combinatorics of Drawing Suits of card from a Stacked Deck?

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).

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?

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.

• For $m \le 104-n$, shouldn't the answer to part two should be the same as part one? – angryavian Apr 3 at 18:27
• Indeed. Many thanks for pointing this out. – Rahul Madhavan Apr 3 at 18:38
• 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. – SumNeuron Apr 5 at 7:31
• @RahulMadhavan if you have a few moments could you also maybe check out this – SumNeuron Apr 5 at 8:44
• Oh wait, I have a quick follow up. What happens if m > h? – SumNeuron Apr 5 at 14:26