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What is the probability that a five card hand contains exactly two aces, given that we know it contains at least one ace?

$P(A|B) = \frac{P(A \cap B)}{P(B)}$ where $A$ is contains exactly two aces and $B$ is contains at least one ace. The book is telling me that $P(A \cap B) = P(A)$, but why would this be? Wouldn't it make more sense that $B \subseteq A$ because $B$ is considering $1$ OR more aces?

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    $\begingroup$ $A\subseteq B$ means: if a hand contains exactly two aces, then it contains at least one ace. $B\subseteq A$ means: if a hand contains at least one ace, then it contains exactly two aces. Now, which statement makes sense to you? $\endgroup$ – bof Sep 6 '16 at 2:47
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    $\begingroup$ It's at this point I now realize I need to sleep. I forgot to check the logic of what $A \subset B$ meant concerning sets. Everything in $A$ is in $B$ because $B$ contains all subsets sets with one ace and two aces and ect. That makes sense. $\endgroup$ – Oliver G Sep 6 '16 at 2:56
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Here is a simulation to consider when you wake up. Also, some hints.

A million poker hands are fairly 'dealt' using R statistical software. Simulated numbers of aces in each are counted. Then approximated and exact hypergeometric probabilities of relevant events are found. [Perhaps look at 'hypergeometric distribution' in your text or at the Wikipedia article.]

m = 10^6;  nr.aces=numeric(m)
for (i in 1:m) {
   hand = sample(1:52, 5)         # aces = 1,2,3,4
   nr.aces[i] = sum(hand <= 4) }  # nr aces in hand
cond = (nr.aces >= 1)
mean(nr.aces[cond] == 2)          # read '[...]' as 'such that ...'
## 0.1175526                      # aprx P(exactly 2 | at least 1)
table(nr.aces)/m  # simulated distribution of nr of aces in poker hand
nr.aces
       0        1        2        3        4 
0.658740 0.299346 0.040116 0.001774 0.000024 
round(dhyper(0:4, 4, 48, 5), 5)  # exact hypergeometric distribution
## 0.65884 0.29947 0.03993 0.00174 0.00002

The histogram below shows the simulated distribution of the number of aces, and the blue dots atop the bars show exact hypergeometric probabilities.

enter image description here

Let $X$ be the number of aces in a fairly dealt poker hand. Then you should verify exact values of $P(X = 0),\, P(X \ge 1),\, P(X = 2),$ and $P(X = 2 | X \ge 1)$ for yourself. The first and the third are the only ones that require much arithmetic. Notice that $\{X = 2\} \subseteq \{X \ge 1\}.$

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A = 2 aces in hand B = >0 ace in hand

P(A|B) = P(A AND B) / P(B)

P(B) = 1 - P(no ace dealt) $= 1 - (48/52)(47/51)(46/50)(45/49)(44/48) = 1 - (48! / 43!) / (52! / 47!)$

no hands that have 2 aces = (ways to choose 2 of 4 aces) x (ways to choose 3 of 48 non aces) = 103776 hands possible = 2598960

P(A) = P(A and B) = 103776 / 2598960

P(A|B) = P(A AND B) / P(B) = (103776 / 2598960) / (1 - (48! / 43!) / (52! / 47!)) = 0.117

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