13 cards randomly dealt to 4 players. What's the probability of 1 player being dealt quads/four-of-a-kind? Inspired by the Vietnamese game Tiến lên where you are dealt $13$ cards. One specific rule states if you are dealt quads 2 (four 2's), you automatically win that round. I want to find the probability of that happening.
My method:
$\frac{4}{52} \times \frac{3}{51} \times \frac{2}{50} \times \frac{1}{49} \times {}^{13}C_4 = 0.264\%$
The first 4 terms are the probability of being dealt quads 2 in 1 particular order. We have 13c4 of these equivalent orders, so I multiply it by 13c4.
2 questions:

*

*Is my answer correct?

*I tried to use another approach: P(I have quads 2) = 1 - P(3 other players have no 2's). I tried to calculate the subtracted probability by doing 48/52 * 47/51 * ... * 11/15 * 10/14 but it seems that it's wrong. How would I go about doing this?

Thanks!
 A: If the question is about the probability that the $13$ cards dealt to you contain the $4$ twos and $9$ other cards, then the probability would be $$\frac{\,^{4}C_4 \,^{48}C_{9}}{{\,^{52}C_{13}}} \approx 0.002641$$
This is the same as your answer of $\dfrac{\,^{13}C_4 }{{\,^{52}C_{4}}}$, as can be shown if you expand into factorials
Your alternative approach of considering the other $39$ cards also gives the same answer, though you should not subtract from $1$: $$\frac{48}{52}\times \frac{47}{51}\times \cdots\times \frac{10}{14} = \frac{ \,^{48}C_{39}}{{\,^{52}C_{39}}}= \frac{ \,^{48}C_{9}}{{\,^{52}C_{13}}}$$
A: This is correct for the probability that a given player has all four twos (in a random hand of $13$ cards). There is perhaps a simpler way to get the same answer, as follows. There are ${}^{48}C_9$ hands which contain all four twos (since each hand contains $9$ other cards chosen from the remaining $48$ cards), and there are ${}^{52}C_{13}$ hands in total, so the probability is $\frac{{}^{48}C_9}{{}^{52}C_{13}}$, and this simplifies to $\frac{13\times12\times11\times 10}{52\times51\times50\times 49}$.
If you want the probability that one of the four players has all the twos, then since these are disjoint events you can just add the probabilities, i.e. multiply the above by $4$.
