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In a first round, $n > 1$ pairwise distinct playing cards are handed out to $n$ people, where each person receives one card. After people have studied their own cards, the cards are collected, permuted uniformly at random and redistributed. For $1 ≤ i ≤ n$, let $A_i$ be the event that person $i$ now receives the same card as in the first round.

$(i)$ Find $P(A_i)$ for $1 ≤ i ≤ n$

$(ii)$ Let $X$ be the total number of people who receive the same card in both rounds. Find E[$X$].

I need help with part $i$. I think I can then try solving the second part myself after that but I will ask if I get stuck. I am assuming that player $1$ receives the first card, then player $2$, all the way up to player $n$.

I am confused on part $i$. Consider $A_1$. So person $1$ receives their same card but the remaining players $2-n$ can have any card except from the one player $1$ receives. This gives $$P(A_1)=\frac{1}{n} \cdot \frac{n-1}{n-1} \ldots \cdot \frac{1}{1} = \frac{1}{n}. $$

Now, $P(A_2) = \frac{n-1}{n} \cdot \frac{1}{n-1} \cdot \frac{n-2}{n-2}\ldots \cdot \frac{1}{1} = \frac{1}{n}$

So I think $P(A_i) = \frac{1}{n}$ for all $1 \leq i \leq n$.

TShiong
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    Yes, $P(A_i)=\frac{1}{n}$ regardless the value of $i$. For (ii) just add these up. – JMoravitz May 30 '23 at 13:39
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    If you want to solve part (ii) without invoking linearity of expectation, see the extra explanation in the linked answer. – JMoravitz May 30 '23 at 13:42
  • @JMoravitz So my $(i)$ is correct ? There is no factorial ? –  May 30 '23 at 13:44
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    Yes. All factorials and such cancel each other out and the answer for (i) is that $\Pr(A_i)=\frac{1}{n}$ for every $i$. – JMoravitz May 30 '23 at 13:58
  • @JMoravitz understood. Can you show me how to solve ($ii$) using linearity of expectation. I know I can write $X = X_1 + .... +X_n$ where $X_i$ is when $i$ people receive same card. –  May 30 '23 at 14:04
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    $E[X]=E[X_1+\dots+X_n]=E[X_1]+\dots+E[X_n]=\frac{1}{n}+\frac{1}{n}+\dots+\frac{1}{n}=n\times \frac{1}{n}=1$ – JMoravitz May 30 '23 at 14:45

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