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.

  • $\begingroup$ I tried to LaTeXify this and make it pretty, but I couldn't parse your solution attempt in a way that made sense. Could you rewrite it? $\endgroup$ – El'endia Starman Apr 15 '12 at 7:30
  • $\begingroup$ \binom{n}{m} is what you want $\endgroup$ – Graphth Apr 18 '12 at 13:24

$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$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.

  • $\begingroup$ I was trying to count out the possible outcomes of having at least one ace. $\endgroup$ – bizboy1 Apr 15 '12 at 0:00
  • $\begingroup$ Thanks. I'm understanding it better. $\endgroup$ – bizboy1 Apr 15 '12 at 0:04
  • $\begingroup$ @JyrkiLahtonen Yes I did something wrong there. Shame on me I wrote it way to fast. $\endgroup$ – Chi Apr 18 '12 at 13:18
  • $\begingroup$ @JyrkiLahtonen wrong answer is even worse than no answer for people who are seeking help. :p $\endgroup$ – Chi Apr 18 '12 at 13:23

The conditional probability setup has been done by Chi. The probability of $4$ aces is straightforward. There are $\binom{52}{13}$ hands, all equally likely. To make a $4$ Ace hand, we need to choose $9$ cards from the $48$ non-Aces, so the probability of a $4$ Ace hand is $\frac{\binom{48}{9}}{\binom{52}{13}}$.

To complete things we need the probability of at least one Ace. It looks as if this may be less clear, since the currently accepted answer is incorrect.

The probability of at least one Ace is $1$ minus the probability of no Aces. There are $\binom{48}{13}$ no Ace hands, so the probability of at least one Ace is $$1-\frac{\binom{48}{13}}{\binom{52}{13}}.\tag{$1$}$$ (Alternately, we could compute separately the probabilities of $1$, $2$, $3$, $4$ Aces and add up. More work.) So to get the final answer, divide $\frac{\binom{48}{9}}{\binom{52}{13}}$ by the expression $(1)$.

Remark: Suppose that (unlike any card game that I know), when we deal a card we put it back in the deck, shuffle, and do this $13$ times. Then the number of Aces we get to see has a Binomial Distribution, and ideas like the ones you described at the end of the post could become relevant. Hard to know, because what is written is not clear (formatting issues).

However, powers will not naturally be part of the answer when there is no replacement of the cards. The binomial distribution arises when there is independence between consecutive trials. That is not the case with dealing $13$ cards to North. If she get the $\spadesuit$ King on the $4$th card, it is certain she will not get it on the $8$-th card.


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