2 problems found
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:
Solution: \begin{align*} && \mathbb{E}(X^2) &= \frac12 \left (\frac16 \left ( 3(k -1)^2+2(2k-1)^2+(3k-1)^2 \right) +\frac16 \left ( 3(k -1)^2+2(k-2)^2+(k-3)^2 \right) \right) \\ &&&= \frac12 \left (\frac16 \left (20k^2-20k+6 \right) + \frac16 \left ( 6k^2-20k+20\right) \right) \\ &&&= \frac1{12} \left (26k^2-40k+ 26\right) \\ &&&= \frac{13}{6} (k^2+1) - \frac{10}{3}k \\ &&&= \frac{13}{6}(k-1)^2+k \end{align*} Since \(k\) a single digit positive number and \(\mathbb{E}(X^2)\) is an integer, \(6 \mid k-1 \Rightarrow k = 1, 7\). \begin{align*} \mathbb{E}(X | k=1) &= \frac12 \left (\frac16 \left ( 2+2 \right) +\frac16 \left ( 2+2 \right) \right) = \frac23 \not \in \mathbb{Z}\\ \mathbb{E}(X | k=7) &= \frac12 \left (\frac16 \left ( 3\cdot6+2\cdot13+20 \right) +\frac16 \left ( 3\cdot6+2\cdot5+4 \right) \right) = 8 \end{align*} Therefore \(k = 7\) The probability distribution is \begin{align*} && \mathbb{P}(X=4) = \frac1{12} \\ && \mathbb{P}(X=5) = \frac1{6} \\ && \mathbb{P}(X=6) = \frac12 \\ && \mathbb{P}(X=13) = \frac1{6} \\ && \mathbb{P}(X=20)= \frac1{12} \\ \end{align*} The only ways to score more than \(25\) are: \(20+6, 20+13, 20+20, 13+13\) The only ways to score exactly \(25\) are \(20+5\) \begin{align*} \mathbb{P}(>25) &= \frac1{12} \cdot\left(2\cdot \frac12+2\cdot\frac16+\frac1{12}\right) + \frac{1}{6^2} \\ &= \frac{7}{48} \\ \mathbb{P}(=25) &= \frac{2}{12 \cdot 6} = \frac{1}{36} \\ \\ \mathbb{E}(\text{payout}) &= \frac{7}{48}w + \frac{1}{36} = \frac{21w+4}{144} \end{align*} The casino needs \(\frac{21w+4}{144} < 1 \Rightarrow 21w< 140 \Rightarrow w < \frac{20}{3}\)
From the integers \(1, 2, \ldots , 52\), I choose seven (distinct) integers at random, all choices being equally likely. From these seven, I discard any pair that sum to 53. Let \(X\) be the random variable the value of which is the number of discarded pairs. Find the probability distribution of \(X\) and show that \(\E (X) = \frac 7 {17}\). Note: \(7\times 17 \times 47 =5593\).
Solution: There are \(\binom{26}3\binom{23}{1}2\) ways to obtain \(3\) pairs There are \(\binom{26}2 \binom{24}3 \cdot 2^3\) ways to obtain \(2\) pairs There are \(\binom{26}1 \binom{25}5 \cdot 2^5\) ways to obtain \(1\) pairs There are \(\binom{26}7 \cdot 2^7\) ways to obtain \(0\) pairs There are \(\binom{52}{7}\) ways to choose our integers, so \begin{align*} && \mathbb{P}(X = 3) &= \frac{\binom{26}{3} \cdot \binom{23}{1} \cdot 2}{\binom{52}{7}} \\ &&&= \frac{7! \cdot 26 \cdot 25 \cdot 24 \cdot 23 \cdot 2}{3! \cdot 52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 \cdot 47 \cdot 46} \\ &&&= \frac{7 \cdot 6 \cdot 5 \cdot 4 }{51 \cdot 2\cdot 49 \cdot 2\cdot 47 \cdot 2} \\ &&&= \frac{ 5 }{17\cdot 7\cdot 47} = \frac{5}{5593} \\ \\ && \mathbb{P}(X = 2) &= \frac{\binom{26}2 \binom{24}3 \cdot 2^3}{\binom{52}{7}} \\ &&&= \frac{220}{5593} \\ \\ && \mathbb{P}(X = 1) &= \frac{\binom{26}1 \binom{25}5 \cdot 2^5}{\binom{52}{7}} \\ &&&= \frac{1848}{5593} \\ \\ && \mathbb{P}(X = 0) &= \frac{\binom{26}7 \cdot 2^7}{\binom{52}{7}} \\ &&&= \frac{3520}{5593} \\ \\ && \mathbb{E}(X) &= \frac{1848}{5593} + 2 \cdot \frac{220}{5593} + 3 \cdot \frac{5}{5593} \\ &&&= \frac{2303}{5593} = \frac{7}{17} \end{align*} Notice we can find the expected value directly: Let \(X_i\) be the random variable the \(i\)th number is discarded. Notice that \(\mathbb{E}(X) = \mathbb{E}\left (\frac12 \left (X_1 +X_2 +X_3 +X_4 +X_5 +X_6 +X_7\right) \right)\) and also notice that each \(X_i\) has the same distribution (although not independent!). Then \begin{align*} &&\mathbb{E}(X) &= \frac72 \mathbb{E}(X_i) \\ &&&= \frac72 \cdot \left (1 - \frac{50}{51} \cdot \frac{49}{50} \cdots \frac{45}{46} \right) = \frac74 \left ( 1 - \frac{45}{51}\right) \\ &&&= \frac72 \cdot \frac{6}{51} \\ &&&= \frac7{17} \end{align*}