SOLUTION TO PROBLEM H-9

If he attempts question 1 first, then he will win

$$\begin{array}{rcl} 0 &\mbox{with probability}& 1 - P_1\\ V_1... ...)\\ V_1 + V_2 & \mbox{with probability} & P_1P_2. \end{array}$$

Hence his expected winnings in this case will be

V₁P₁(1-P₂) +(V₁+V₂)P₁P₂.

On the other hand, if he attempts question 2 first, his expected winnings will be V₂P₂(1-P₁) + (V₁+V₂)P₁P₂. Therefore it is better to try question 1 first if $V_1P_1(1-P_2) \ge V_2P_2(1-P_1)$ or, equivalently, if

$$\frac{V_1P_1}{1-P_1} \ge \frac{V_2P_2}{1-P_2}$$

Thus, for instance, if he is $60\%$ certain of answering question 1, worth $\$ 200$, correctly and he is $80\%$ certain of answering question 2, worth $\$ 100$, correctly, then he should attempt question 2 first because

$$400 = \frac{(100)(.8)}{.2} > \frac{(200)(.6)}{.4} = 300$$

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Thu Mar 14 19:12:53 CST 2002