Full question from actuarial exam practice problems:
The game of bridge is played by four players: north, south, east and west. Each of these players receive 13 cards. ... b) Consider a single hand of bridge. What is the probability for the hand to have exactly 3 Aces?
I know how to calculate the number of total possible outcomes, but can't figure out how to calculate possible outcomes with 3 Aces.
On
The probability that e.g. north will get $3$ aces is: $$\binom{13}{3}\binom{39}{1}\binom{52}{4}^{-1}$$
As explanation: put all card on a row and let the first $13$ cards be for north. If you only distinguish aces and non-aces then there are $\binom{52}{4}$ arrangements. Under the condition that $3$ of the aces belong to the first $13$ (hence are for north) there are $\binom{13}{3}\binom{39}{1}$ arrangements.
The probability that one of the players will get $3$ aces is: $$4\times\binom{13}{3}\binom{39}{1}\binom{52}{4}^{-1}$$
This because it concerns the union of $4$ disjoint events that each have probability $\binom{13}{3}\binom{39}{1}\binom{52}{4}^{-1}$.
The number of possible sets of aces is $\binom{4}{3}$. The number of possibilities for the $10$ remaining cards is $\binom{48}{10}$ (select $10$ cards that are not aces). Multiply these two to get the number of outcomes.