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.
Best Answer
$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$this is the formula for conditional probability.
In this problem, A=Player 1 has 4 aces; B=Player 1 has at least one aces.
We need to figure out what AnB is: it is easy in this case, since if player 1 has 4 aces, then she must has at least one. Event A is clearly "narrower", so $A\cap B$ is A. $P(A\cap B)=P(A)=\frac{\binom{4}{4}\binom{48}{9}}{\binom{52}{13}}$, since we need to choose 4 cards from 4 aces, and 13 cards from all other cards. P(B)=$1-P(B^{c})=1-\frac{\binom{48}{13}}{\binom{52}{13}}$ (Thank Jyrki Lahtonen for pointing me out the original error), it should be 1-P(No Ace at all). $\frac{P(A\cap B)}{P(B)}$is the correct answer, hope you can get the intuition behind this from my sloppy explanation.
By the way, can you explain to me what you were trying to do in your attempt? I am afraid that I didn't get your idea and couldn't tell where you went wrong.