I'm trying to self-educated myself and I bought a probability book, which has this interesting question. It says not to look at any resources before you try it, but you may use a calculator.
In the card game bridge, each of $4$ players is dealt a hand of $13$ of the $52$ cards.
What is the probability that each player receives exactly one Ace?
I immediately thought that this was long-winded, but then I thought that it could be $\dfrac{1}{13} \times \dfrac{1}{13} \times \dfrac{1}{13} \times \dfrac{1}{13}$, although this in probably not correct. The reason I thought that was because $\dfrac{4}{52} \times \dfrac{3}{39} \times \dfrac{2}{26} \times \dfrac{1}{13}$, bu this is a very simple solution. Any help? Thanks a lot.
Best Answer
We need an ace in each hand of 13, how the rest of the cards go doesn't matter !
The first ace has to be in some group, each of the other aces have to fall in a different group, so the 2nd ace has 39 permissible slots out of 51, and so on.
Thus Pr = $\dfrac{39}{51}\cdot\dfrac{26}{50}\cdot\dfrac{13}{49} =\dfrac{2197}{20825}$