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Ok I think this is very easy. In fact I think it may be the following equation:

P(A) = 1/52 P(B) = 1/52 P(A and B) = P(A) * P(B) = 1/2704 

However it doesn't feel right. If you play this out for real the odds seem a lot better than 1 out of 2704.

Could someone enlighten me?

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    @MisterITGuru the original answer (@mlqxxxx) was correct but for the wrong reason. At least that's the way I interprete it. So I'm not sure there is a question afterall.2012-09-29

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If $A$ and $B$ represent the cards turned over in the first and second decks, then $ \begin{eqnarray} P[A=x] = P[B=x] &=& \frac{1}{52} \\ P[A=x\wedge B=x]&=&\frac{1}{52^2} \end{eqnarray} $ for any particular card $x$. The probability that $A$ and $B$ are equal to each other is then $ P[A=B]=\sum_{x}P[A=x\wedge B=x]=52\times\frac{1}{52^2}=\frac{1}{52}.$

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    lol - I thought it might be some maths notation ;)2012-09-29
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No. The first card doesn't matter what, it can be 52 out of 52, and only $B$ is restricted then. So, it is $\displaystyle\frac 1{52}$.

What you answered is the probability that both $A$ and $B$ are the ace of spades, for example.

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    @ByronSchmuland: In fact the expected number of matches if you go through the whole pack(s) is exactly 1 (aka. the expected number of fix points in a random permutation is 1).2012-09-29