Anne, Ben and Feline play a game consisting of 32 distinct cards. The cards are shuffled at random. Anne receives 12 cards, Ben and Feline receive each 10. Anne’s hand contains precisely two Jacks, namely Jack of diamonds and Jack of hearts what is the probability that Ben and Feline have each one Jack?
This is how I approached the question, but I am not sure if this is the right way.
There are ${20}\choose{10}$${10}\choose{10}$ $=$ $184756$ ways to distribute 10 cards to each Ben and Feline.
There are 2! ways to distribute the jacks so that each person receives one and
${18}\choose{9}$${9}\choose{9}$ $=$ $48620$ ways to distribute the remaining cards so that each person receives 9 of them.
Hence, the desired probability is $\frac{2! * 48620}{184756}$ $=$ $\frac{10}{19}$
Imagine that 12 cards are dealt at random to A and none to B or F. Then A's hand is revealed showing 2 Jacks. The remaining 2 Jacks are still in the deck waiting to be dealt to either B or F.
Imagine that the next 10 cards are dealt at random to B, and the last 10 to F. Whether B and F get one Jack each depends on one Jack being in the next 10 cards and the other in the last 10 cards. The probability of this happening is $2.\frac{10}{20}.\frac{10}{19}=\frac{10}{19}$.