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if we choose a size $25$ subset of the set: $\{1,2,....100\}$ what is the expectancy of the number of sequential pairs in the subset? expectancy still confuses me, can anybody help?

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closed as off-topic by Namaste, user21820, RRL, Lord_Farin, Did Dec 19 '18 at 18:20

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Let $X_i$ be an indicator random variable that $i$ and $i+1$ ($i\in \{1,2,...,99\}$) are in a choosen subset.

Then $$ P(X_i = 1) = {{98\choose 23}\over {100\choose 25}}= {6\over 99}$$ Since $X= X_1+...+X_{99}$ we have $$ E(X) = E(X_1)+...+E(X_{99}) = 99{6\over 99} = 6$$

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  • $\begingroup$ Is that better @copper.hat? $\endgroup$ – Aqua Dec 5 '18 at 18:44
  • $\begingroup$ Yes thanks. ${}{}$ $\endgroup$ – copper.hat Dec 5 '18 at 18:49
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Let's assign a label $1,2 \dots 25$ (arbitrary order) to each of the selected numbers. Let $A_{i,j}$ with $1\le i<j \le 25$ be $1$ if elements with labels $i,j$ have sequential values, $0$ otherwise.

Then $X = \sum A_{i,j}$ counts the number of sequential pairs, which is what we want, and

$$E(X)=\sum A_{i,j}$$

But $E(X)=\sum E[A_{i,j}]= n_p P(A_{i,j} = 1) = n_p \frac{99}{\binom{100}{2}}$

Where $n_p = \binom{25}{2}=300$ is the number of pairs. Then

$$E(X) = \frac{300 \times 99} { \binom{100}{2}}=6$$

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