Recitation 22 Problem 3. The number of squares that a piece advances in one turn of the game Monopoly is determined as follows: Roll two dice, take the sum of the numbers that come up, and advance that number lares If you roll doubles(that is, the same number comes up on both dice), then you roll a second time, take the sum, and advance that number of additional squares If you roll doubles a second time, then you roll a third time, take the sum, and advance that number of additional squares (a)What is the expected sum of two dice, given that the same number comes up on Solution. There are six equally-probable sums: 2, 4, 6, 8, 10, and 12. Therefore, the expected sum is 1 6 6 (b)What is the expected sum of two dice, given that different numbers come up? (Use your previous answer and the Total Expectation Theorem Solution. Let the random variables D1 and D2 be the numbers that come up on the two dice. Let e be the event that they are equal. The Total Expectation Theorem Ex(D1+D2)=Ex(D1+ D2 I E). Pr(E)+Ex(D2+ D2 E). Pr(E Two dice are equal with probability Pr(e)=1/6, the expected sum of two indepen dent dice is 7, and we just showed that Ex(D1+ D2 E)=7. Substituting in these quantities and solving the equation, we find 7=7·+Ex(D2+D2|E) Ex(D2+D2|E)=7 (c) To simplify the analysis, suppose that we always roll the dice three times, but may ignore the second or third rolls if we didnt previously get doubles. Let the random variable X i be the sum of the dice on the i-th roll, and let ei be the event that the i-th roll is doubles. Write the expected number of squares a piece advances in these terms� � � � � � Recitation 22 7 Problem 3. The number of squares that a piece advances in one turn of the game Monopoly is determined as follows: • Roll two dice, take the sum of the numbers that come up, and advance that number of squares. • If you roll doubles (that is, the same number comes up on both dice), then you roll a second time, take the sum, and advance that number of additional squares. • If you roll doubles a second time, then you roll a third time, take the sum, and advance that number of additional squares. • However, as a special case, if you roll doubles a third time, then you go to jail. Regard this as advancing zero squares overall for the turn. (a) What is the expected sum of two dice, given that the same number comes up on both? Solution. There are six equally­probable sums: 2, 4, 6, 8, 10, and 12. Therefore, the expected sum is: 1 1 1 2 + 4 + . . . + 12 = 7 6 · 6 · 6 · (b) What is the expected sum of two dice, given that different numbers come up? (Use your previous answer and the Total Expectation Theorem.) Solution. Let the random variables D1 and D2 be the numbers that come up on the two dice. Let E be the event that they are equal. The Total Expectation Theorem says: Ex (D1 + D2) = Ex (D1 + D2 | E) · Pr (E) + Ex D2 + D2 | E · Pr E Two dice are equal with probability Pr (E) = 1/6, the expected sum of two indepen￾dent dice is 7, and we just showed that Ex (D1 + D2 | E) = 7. Substituting in these quantities and solving the equation, we find: 1 � � 5 7 = 7 · + Ex D2 + D2 | E 6 · 6 Ex D2 + D2 | E = 7 (c) To simplify the analysis, suppose that we always roll the dice three times, but may ignore the second or third rolls if we didn’t previously get doubles. Let the random variable Xi be the sum of the dice on the i­th roll, and let Ei be the event that the i­th roll is doubles. Write the expected number of squares a piece advances in these terms