So, in my family we play a card game called Continental, on each round one card is added to the initial cards until you are dealt 13 cards (the rules of the game are not important as my question is at the time the cards are dealt) we usually combine 3 decks (1 deck = 13 cards per suit x 4 suits + 2 jokers). The other day while playing, my brother got all 3 heart-suit Aces available on this 13 cards stage, so we asked ourselves, what are the odds? and I've been trying to figure it out but my combinatorics are very basic. My attempt to resolve it was:
$$\frac{12×\binom{13}{1}}{\binom{158}{3}}$$
Any help with this will be very much appreciated.
As I understand it, you want to find the probability of having three $A\heartsuit$ in a thirteen card hand drawn from three standard decks. The number of such hands would be $\binom{159}{10}\approx 2.1\times 10^{15}$, since any hand with those three aces would be completed by $10$ cards from the remaining $3\times 54-3=159$ cards. The total number of possible hands is clearly $\binom{162}{13}\approx 5.2\times 10^{18}$. Therefore, the probability of drawing such a hand is $$\frac{\binom{159}{10}}{\binom{162}{13}}\approx 0.000411203\approx 0.04\%,$$ which is roughly similar to the odds of getting drafted to the NBA... Good luck to your brother!