Problems

The score shown on a biased \(n\)-sided die is represented by the random variable \(X\) which has distribution \(\mathrm{P}(X = i) = \dfrac{1}{n} + \varepsilon_i\) for \(i = 1, 2, \ldots, n\), where not all the \(\varepsilon_i\) are equal to \(0\).

1. Find the probability that, when the die is rolled twice, the same score is shown on both rolls. Hence determine whether it is more likely for a fair die or a biased die to show the same score on two successive rolls.
2. Use part (i) to prove that, for any set of \(n\) positive numbers \(x_i\) (\(i = 1, 2, \ldots, n\)), \[\sum_{i=2}^{n}\sum_{j=1}^{i-1} x_i x_j \leqslant \frac{n-1}{2n}\left(\sum_{i=1}^{n} x_i\right)^2.\]
3. Determine, with justification, whether it is more likely for a fair die or a biased die to show the same score on three successive rolls.

A game in a casino is played with a fair coin and an unbiased cubical die whose faces are labelled \(1, 1, 1, 2, 2\) and \(3.\) In each round of the game, the die is rolled once and the coin is tossed once. The outcome of the round is a random variable \(X\). The value, \(x\), of \(X\) is determined as follows. If the result of the toss is heads then \(x= \vert ks -1\vert\), and if the result of the toss is tails then \(x=\vert k-s\vert\), where \(s\) is the number on the die and \(k\) is a given number. Show that \(\mathbb{E}(X^2) = k +13(k-1)^2 /6\). Given that both \(\mathbb{E}(X^2)\) and \(\mathbb{E}(X)\) are positive integers, and that \(k\) is a single-digit positive integer, determine the value of \(k\), and write down the probability distribution of \(X\). A gambler pays \(\pounds 1\) to play the game, which consists of two rounds. The gambler is paid:

• \(\pounds w\), where \(w\) is an integer, if the sum of the outcomes of the two rounds exceeds \(25\);
• \(\pounds 1\) if the sum of the outcomes equals \(25\);
• nothing if the sum of the outcomes is less that \(25\).

The probability of throwing a 6 with a biased die is \(p\,\). It is known that \(p\) is equal to one or other of the numbers \(A\) and \(B\) where \(0 < A < B < 1 \,\). Accordingly the following statistical test of the hypothesis \(H_0: \,p=B\) against the alternative hypothesis \(H_1: \,p=A\) is performed. The die is thrown repeatedly until a 6 is obtained. Then if \(X\) is the total number of throws, \(H_0\) is accepted if \(X \le M\,\), where \(M\) is a given positive integer; otherwise \(H_1\) is accepted. Let \({\alpha}\) be the probability that \(H_1\) is accepted if \(H_0\) is true, and let \({\beta}\) be the probability that \(H_0\) is accepted if \(H_1\) is true. Show that \({\beta} = 1- {\alpha}^K,\) where \(K\) is independent of \(M\) and is to be determined in terms of \(A\) and \(B\,\). Sketch the graph of \({\beta}\) against \({\alpha}\,\).

Show Solution

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Solution: \(X \sim Geo(p)\). \(\alpha = \mathbb{P}(X > M | p = B) = (1-B)^{M}\) \(\beta = \mathbb{P}(X \leq M | p = A) = 1 - \mathbb{P}(X > M | p = A) = 1 - (1-A)^{M}\) \begin{align*} \ln \alpha &= M \ln(1-B) \\ \ln (1-\beta) &= M \ln(1-A) \\ \frac{\ln \alpha}{\ln (1-\beta)} &= \frac{\ln(1-B)}{\ln(1-A)} \\ \ln(1-\beta) &= \ln \alpha \frac{\ln (1-A)}{\ln(1-B)} \\ \beta &= 1- \alpha^{ \frac{\ln (1-A)}{\ln(1-B)} } \end{align*} and \(K = \frac{\ln (1-A)}{\ln(1-B)} \) Since \(0 < A < B < 1\) we must have that \(0 < 1 - B < 1-A < 1\) and \(\ln(1-B) < \ln(1-A) < 0\) so \(0 < K < 1\)

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