I have been working working on the following probability problem:
Alice has a set of 52 perfectly shuffled cards, hands the first 2 cards to Bob, Alice takes the next 2 cards. Let:
B:= Bob has a pair
A:= Alice has a pair
Calculate: $\text{Pr}[A]$, $\text{Pr}[B]$, $\text{Pr}[A\cap B]$
$\text{Pr}[B]$ was straight forward, $\frac{3}{51}$.
$\text{Pr}[A]$ took some thinking, but as far as I understand, the point is that handing cards to Bob and not looking at them is equal to not drawing them in the first place. Hence, $\text{Pr}[A]=\frac{3}{51}$
I am now stuck with $\text{Pr}[A\cap B]$. The way I approached it was to use conditional probability: $\text{Pr}[A\mid B]\cdot \text{Pr}[B]$, splitting $\text{Pr}[A\mid B]$
- Alice draws the same pair as Bob
- Alice draws a different pair
For (1), I calculated: $\frac{2}{52} \cdot \frac{1}{52}$, because Bob has already drawn the first 2 cards, so Alice has to get exactly the remaining 2
For(2), I calculated: $12 \cdot \frac{4}{52} \cdot \frac{3}{51}$ because there are 12 other possible pairs to make s.t. it is different from Bob's pair.
Hence, I am left with: $$\text{Pr}[A\mid B]\cdot \text{Pr}[B]=\left(\frac{2}{52} \cdot \frac{1}{52} + 12 \cdot \frac{4}{52} \cdot \frac{3}{51}\right) \cdot \frac{3}{51}.$$
Unfortunately, this does not agree with the solution of $\frac{73}{20825}$.
I simply cannot find the mistake, help is greatly appreciated!
Thanks!
It should be $$\text{Pr}(A\cap B)=\frac{\overbrace{13\cdot 12\cdot \binom{4}{2}^2}^{\text{different pairs values}}+\overbrace{13\cdot \binom{4}{2}}^{\text{same pair value}}}{\binom{52}{2}\binom{50}{2}}=\frac{73}{20825}.$$
P.S. You may also modify your evaluation as $$\left(\frac{2}{\color{red}{50}} \cdot \frac{1}{\color{red}{49}} + 12 \cdot \frac{4}{\color{red}{50}} \cdot \frac{3}{\color{red}{49}}\right) \cdot \frac{3}{51}=\frac{73}{20825}.$$