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Completely shuffle and then lay out a full deck of $52$ cards in a single row (i.e. single row $52$ columns). Do the same thing with a second deck of playing cards and place this deck in the second row(also has $52$ columns). What is the expected total number of matches in all the columns?

A match means that the top card and the bottom card in the column are the same (same rank and suit).

I was trying to solve it by brute force but seems like it's not that easy to me.

Min Gao
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1 Answers1

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For $i=1$ to $52$, let $X_i=1$ if the cards in column $i$ match, and let $X_i=0$ otherwise.

Then the number $Y$ of matches is given by $Y=X_1+X_2+\cdots+X_{52}$. It follows by the linearity of expectation that $$E(Y)=E(X_1)+E(X_2)+\cdots +E(X_{52}).$$ I expect you can find $E(X_i)$.

Remark: One could find the distribution of the number of matches. However, that is a much more complicated path to the expectation. The method of indicator random variables can be very useful.

André Nicolas
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  • just one more question I'm feeling a little bit confused, are all E(X_i) s independent of each other? – Min Gao Jan 02 '15 at 07:47
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    The random variables are (fairly clearly) not independent. The linearity of expectation does not require independence, it holds unconditionally. – André Nicolas Jan 02 '15 at 07:49
  • The distribution is that based on rencontres numbers, though with a large number of cards such as $52$ then this will be close to the Poisson distribution with the expectation as calculated here. – Henry Jan 02 '15 at 08:20
  • I was wondering, isn't the probability for a card to match $\frac{1}{52}$? And as there are 52 events, shouldn't we expect only 1 card to match? @AndréNicolas – Gummy bears Jan 02 '15 at 11:22
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    Yes, if by match you mean exact match, then $\Pr(X=1)=\frac{1}{52}$ and therefore $E(X_i)=\frac{1}{52}$ and therefore $E(Y)=1$. If you are satisfied with partial match (we care only about kind, not suit) same analysis works, but $\Pr(X_i=1)=\frac{4}{52}$. – André Nicolas Jan 02 '15 at 11:32
  • Are you sure this is correct? This model assumes that it is possible to have 51 cards match. However, this isn't possible, because if one card does not match, it must have displaced at least one other card. – brainmurphy1 Apr 07 '17 at 02:35