Recitation 22 Solution. From the total expectation formula, we get Ex(advance)=Ex(X1| E1). Pr(E1 +Ex(X1+X2|E1∩E)P(E1∩E) +Ex(X1+X2+X3E1∩E2∩E3)Pr(E1∩E2∩E3) (0|E1∩E2∩E3)·Pr(E1∩E2∩E3) Then using linearity of (conditional) expectation, we refine this to Ex(advance) Ex(X1|E1. Pr(E1) +(Ex(X1E1∩E2)+Ex(X2|E1∩E2)·Pr(E1∩E) +(Ex(X1E1∩E2∩E3)+Ex(X2|E1∩E2∩E)+Ex(X8|E1∩E2∩E3) (E1∩E2∩E3) +0. Using mutual independence of the rolls, we simplify this to (X1l E1) Pr(E1) +(Ex(X1 l E1)+Ex(X2 1 E2). Pr (E1).Pr(E2 +(Ex(X1 E1)+Ex(X2 E2)+Ex(X3 E3).Pr(E1). Pr(E2). Pr(E3 (d)What is the expected number of squares that a piece advances in Monopoly? Solution. We plug the values from parts(a)and(b) into equation(4) 5 )=7·2+(7+7) 15 115 66 (7+7+7) 66� � � � � � � � � � | | � � � � Recitation 22 8 Solution. From the total expectation formula, we get: Ex (advance) = Ex X1 | E1 Pr E1 � · � � � + Ex X1 + X2 | E1 ∩ E2 Pr E1 ∩ E2 � · � � � + Ex X1 + X2 + X3 | E1 ∩ E2 ∩ E3 · Pr E1 ∩ E2 ∩ E3 + Ex (0 | E1 ∩ E2 ∩ E3) · Pr (E1 ∩ E2 ∩ E3) Then using linearity of (conditional) expectation, we refine this to Ex (advance) = Ex X1 E1 Pr E1 � � | · � � �� � � + Ex X1 | E1 ∩ E2 + Ex X2 E1 ∩ E2 Pr E1 ∩ E2 � � � | � · � � �� + Ex X1 | E1 ∩ E2 ∩ E3 + Ex X2 E1 ∩ E2 ∩ E3 + Ex X3 E1 ∩ E2 ∩ E3 · Pr E1 ∩ E2 ∩ E3 + 0. Using mutual independence of the rolls, we simplify this to Ex (advance) = Ex X1 | E1 Pr E1 (4) � · � �� � � + Ex (X1 | E1) + Ex X2 E2 · Pr (E1) Pr E2 � | � · �� � � + Ex (X1 | E1) + Ex (X2 E2) + Ex X3 E3 | | · Pr (E1) · Pr (E2) Pr · E3 (d) What is the expected number of squares that a piece advances in Monopoly? Solution. We plug the values from parts (a) and (b) into equation (4): 5 1 5 1 1 5 Ex (advance) = 7 + (7 + 7) + (7 + 7 + 7) · 6 · 6 · 6 · 6 · 6 · 6 19 = 8 72