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
$$P(\text{3ofakind}) =\frac{C^{39}_{1} \cdot (C^{4}_{1})^3 \cdot C^{36}_{18} \cdot (C^{4}_{1})^{18}}{ C^{156}_{21}}$$
It may be easier to find the probability that a hand of 21 cards does not include three of a kind.
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 $$1-p = \boxed{0.107529}$$