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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 $\text{Pr}[A\mid B]$

  1. Alice draws the same pair as Bob
  2. 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!

Robert Z
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iMrFelix
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  • Robert Z is correct, but let me add that where you went wrong is thinking that in (1) and (2) Alice was still picking from 52 cards. In (1) she is picking from 50 cards (she can't pick the two Bob is holding). In (2) she is picking from 48 cards, because she can't pick the two Bob is holding, nor the other two cards of the same number. – Paul Sinclair Aug 11 '19 at 16:46
  • @PaulSinclair I see what you mean and that's how I first thought of it, but then I didn't get why Alice's probability is the same as Bob's. In a similar question link, the second comment by Robert Isreal suggests that having drawn a card previously does not matter as you do not look at it and therefore have no extra information – iMrFelix Aug 13 '19 at 06:21
  • I was talking about your calculations in (1) and (2). There you are trying to calculate $P(A\mid B)$, not $P(A)$. $P(A)$ is indeed the same as $P(B)$, as Robert Israel said. But $P(A\mid B)$ is something different. In $P(A)$, Bob's cards are not known - not just to Alice, but to the entire concept of this probability, which does not take them into account. Bob's cards are just 2 of the 50 cards other than Alice's. But the difference between $P(A)$ and $P(A\mid B)$ is that now we do take Bob's cards into account. They are now two fixed cards that Alice cannot get. – Paul Sinclair Aug 13 '19 at 16:33

1 Answers1

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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 $$\left(\frac{2}{\color{red}{50}} \cdot \frac{1}{\color{red}{49}} + 12 \cdot \frac{4}{\color{red}{50}} \cdot \frac{3}{\color{red}{49}}\right) \cdot \frac{3}{51}=\frac{73}{20825}.$$

Robert Z
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