Question: A deck of cards is shuffled and dealt to four players, with each receiving 13 cards. Find: The probability that the first player holds all the aces given that she holds at least one.
Attempt at solution: P(Player 1 has 4 aces | at least one)=P(4 aces)/(1-P(none)) =binomial(4,4)*binomial(48,9)/binomial(52,13)/(binomial(52,13)/binomial(52,13)-binomial(4,0)*binomial(48,13)/binomial(52,13))=bimonial(48,9)/(binomial(52,13)-binomial(48,13))=5/1318=.003794.
This is the right answer, according to the text. But I'm not sure why P(Player 1 has 4 aces | at least one)=P(4 aces)/(1-P(none)). I really struggle with these type of equations. Can someone teach me how to solve P(A|A,B,C,D,...)?
Also, I'm curious why I couldn't solve it this way. I get a wrong answer:
AP^3*q^9 + AAP^2*q^9 + AAAp*q^9 + AAAAq^9
=binomial(12,3)*3*2*1/(51*50*49) + binomial(11,2)*2*1/(50*49) + binomial(10,1)*1/49= 25% (where A stands for ace).
Thanks in advance!!! This is for review, not homework.