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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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    What are A and B?2012-09-29
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    P(A) represent the odds of a 52 card deck same for P(B)2012-09-29
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    In a probability question, the experiment has to be described in detail, else there can be several different interpretations, with several quite different numerical answers.2012-09-29
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    @AndréNicolas understood. But doesn't the title describe the problem quite well?2012-09-29
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    @Haraldo: Not really. Are the decks kept separate, with Alice and Bob turning up the top card? Do they only do turn up the top cards, or do they go through the entire deck, as in the children's game of War? Or are the decks shuffled together and the top two cards are turned up? Or $\dots$.2012-09-29
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    @AndréNicolas You are right - fair enough. Your first assumption is right though!2012-09-29
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    Then the probability of equality is $1/52$. Whatever Alice turns up, the probability Bob's matches is $1/52$.2012-09-29
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    @AndréNicolas correct! you win a cookie :)2012-09-29
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    You guys are WAY to smart for me! This question popped up randomly on Stack Exchange home page, and I thought, I'd take a peek! Awesome! If only this existed when I was in school, I would have got some better grades! I have a question, the original answer, (which I thought was correct btw!) must be the answer to a question, but what question? Is it the odds of Alice and Bob both cutting the deck to the same card?2012-09-29
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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

2 Answers 2

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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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    I would vote you up but I don't have enough rep!2012-09-29
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    Does the dot represent something after the 1/52 answer?2012-09-29
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    The end of a sentence :).2012-09-29
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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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    not really, it should be $1\over 103$, as two deck is combined so there will be 103 card left. Your answer is fine with 2 separate deck2012-09-29
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    Your second point is what I am actualy looking for...2012-09-29
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    Oh, I understood the 2 decks are separate..2012-09-29
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    @Mathematics Thanks, so its a 1/103 chance of both turning up an Ace of spades?2012-09-29
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    @Berci Yes the decks are seperate. Two people turning cards over. What are the odds of them both turning an Ace of spades at the same time.2012-09-29
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    No. If the decks are mixed, then $2/104 \cdot 1/103$, the first card can be out of $2$, and the second then is unique.2012-09-29
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    @Haraldo then your original calculation was right.2012-09-29
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    @Berci it just seems very high odds. If I go through this process 1 or two times(e.g. work through a whole pack with someone else as well doing the same). We often turn up the exact same card. Anyway maths is never wrong!2012-09-29
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    If you work through a whole pack, the chance of getting the same card at least one time is .6321205588.2012-09-29
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    Yes, this answer is valid only for the *first play*. Or, if you fully randomly mix each decks before the turns, then it is valid for all turns.. But, if you keep playing on, only rotating your own deck, then it becomes much more often..2012-09-29
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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