# Probability and cards

In Nepal, we play a card game mixing $$3$$ deck of cards. We deal $$21$$ cards to each player. What is the probability of getting $$3$$ of a kind (both number and suit)? I got $$2.8\%$$ but it's dubious.

Sample space: $$C^{156}_{21}$$

Choice card: $$C^{39}_{1}$$

Suite of choice: $$(C^{4}_{1})^3$$

Remaining cards: $$C^{36}_{18}$$

Suite choice: $$(C^{4}_{1})^{18}$$

so

$$P(\text{3ofakind}) =\frac{C^{39}_{1} \cdot (C^{4}_{1})^3 \cdot C^{36}_{18} \cdot (C^{4}_{1})^{18}}{ C^{156}_{21}}$$

• Hi Anthony, it would help if you posted your method for coming to that answer, so you can get some fruitful feedback, rather than having someone just post a solution. – user445909 Feb 25 at 11:54
• Would you care to share your calculations with us? – Vinyl_cape_jawa Feb 25 at 11:54
• I heavily edited your question, plese check if I done everything right. – Vinyl_cape_jawa Feb 25 at 12:22
• By "three of a kind", do you mean exactly one three of a kind, or at least one three of a kind? – awkward Feb 25 at 12:43
• I don't understand the logic of the calculation. Let's say you meant "exactly one". Then there are $52$ choices for the triple. Having made that choice you now need to populate the rest of the hand. To do that, you'll have some doubles and some singletons. You should specify the number of doubles and count accordingly, summing over the possible numbers of doubles. I don;'t understand what your formula computes. – lulu Feb 25 at 12:49

There are $$\binom{156}{21}$$ 21-card hands possible, all of which are equally likely. We would like to count the number of hands which do not include three of a kind. More generally, we might consider $$r$$-card hands. The generating function for the number of $$r$$-card hands which do not contain three of a kind is $$f(x) = \left[1 + \binom{3}{1} x + \binom{3}{2} x^2 \right]^{52}$$ The prospect of expanding this polynomial by hand is disheartening, but it's easy if we use a computer algebra system. We are really only interested in the coefficient of $$x^{21}$$, which turns out to be (courtesy of Mathematica) $$[x^{21}]f(x) = 4.85069 \times 10^{25}$$ So the probability that a 21-card hand does not contain three of a kind is $$p = \frac{[x^{21}]f(x)}{\binom{156}{21}} = 0.892471$$ The answer to the original problem, the probability that a 21-card hand includes at least one three of a kind, is $$1-p = \boxed{0.107529}$$