Solution. Let \(X_i\) be the score on the \(i\)th die. Each takes the values \(1, \dots , 6\) with probability \(\frac 16\), so \(E(X_i) = \frac {1 + 2 + \cdots + 6}{6} = \frac 72\). The total is \(X = X_1 + X_2 + X_3\), and \[E(X) = E(X_1) + E(X_2) + E(X_3) = 3\cdot \frac 72 = 10.5 .\]

Solution. Let \(X_i = 1\) if person \(i\) picks their own hat and \(0\) otherwise, and \(X = \sum _{i=1}^N X_i\). Each person is equally likely to get any of the \(N\) hats, so \(P(X_i = 1) = \frac 1N\) and \[E(X) = \sum _{i=1}^N E(X_i) = N\cdot \frac 1N = 1 .\] For the variance, \(\var (X) = \sum _i\var (X_i) + 2\sum _{i<j}\cov (X_i, X_j)\). Each \(X_i\) is Bernoulli with \(p = \frac 1N\), so \(\var (X_i) = \frac 1N\left (1 - \frac 1N\right ) = \frac {N - 1}{N^2}\). For \(i\ne j\), both get their own hats with probability \(P(X_i = 1)P(X_j = 1\mid X_i = 1) = \frac 1N\cdot \frac {1}{N - 1}\), so \[\cov (X_i, X_j) = E(X_iX_j) - E(X_i)E(X_j) = \frac {1}{N(N - 1)} - \frac {1}{N^2} = \frac {1}{N^2(N - 1)} .\] There are \(\binom N2\) pairs \(i<j\), so \[\var (X) = N\cdot \frac {N - 1}{N^2} + 2\binom N2\frac {1}{N^2(N - 1)} = \frac {N - 1}{N} + \frac {1}{N} = 1 .\] Both answers are free of \(N\). For large \(N\) the number of matches is approximately Poisson with mean \(1\), and the probability that nobody gets their own hat tends to \(e^{-1} = 0.368\).

Solution. Let \(X_i = 1\) if type \(i\) appears at least once among the \(10\) coupons. A given coupon is not of type \(i\) with probability \(\frac {24}{25}\), so \(P(X_i = 0) = \left (\frac {24}{25}\right )^{10}\) and \(E(X_i) = 1 - \left (\frac {24}{25}\right )^{10}\). The number of distinct types is \(X = \sum _{i=1}^{25}X_i\), so \[E(X) = 25\left [1 - \left (\frac {24}{25}\right )^{10}\right ] = 25(1 - 0.6648) = 8.38 .\] The indicators are dependent (if one type is missing the others are a little more likely to appear), but that does not affect the expectation.