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23-Ind-A1 Operations Research · December 2013

Question 8 of 10: Expected Value of Sample Information — The Coin-Toss Bet

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Notes on this paper

National Exams — December 2013 — 98-Ind-A1 Operations Research. Three-hour, open-book exam (any non-communicating calculator permitted); the paper totals 200 marks across 10 questions and only 100 marks are required, so a candidate would normally answer a subset — all ten are solved below for completeness.

Reference texts: Hillier & Lieberman, Introduction to Operations Research (11th ed., McGraw-Hill) — linear/integer programming, network optimization, dynamic programming, decision analysis and queueing theory; Nahmias, Production and Operations Analysis — inventory models with planned backorders.

Question 8: Expected Value of Sample Information — The Coin-Toss Bet (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Bet: pay $5 to play; heads → receive $100, tails → receive $0. Prior: $P(\text{double-tails coin})=0.8$, $P(\text{fair coin})=0.2$. One free-standing test toss is offered before the betting decision.

Find. The expected value of sample information (EVSI) of the pre-toss — the maximum amount worth paying for it.

Approach. Compute the EMV of betting with no information; then compute the EMV of the optimal bet/no-bet decision after each possible toss outcome (updating the coin-type probability by Bayes' rule), weight by the outcome probabilities, and subtract.

no test bet: EMV=0.8(−5)+0.2(45)=$5 pay for 1 toss H, P=0.1 → certainly fair T, P=0.9 → post P(fair)=1/9 bet, EMV=45 bet, EMV=5/9>0 EVSI of one pre-toss: with-info EMV = 0.1(45)+0.9(5/9) = $5 = without-info EMV
Every branch of the pre-toss tree still recommends betting, so the toss never changes the decision.
  1. EMV of betting, no information. $\text{EMV(bet)}=0.8(0-5)+0.2\!\left(\tfrac12(100)+\tfrac12(0)-5\right)=0.8(-5)+0.2(45)=-4+9=\boxed{\$5}$, which exceeds EMV(no bet)$=0$ — so without any test, the best action is already to bet, at $\$5$ expected profit.
  2. Toss-outcome probabilities. $P(H)=0.8(0)+0.2(0.5)=0.1$; $P(T)=0.8(1)+0.2(0.5)=0.9$ (a double-tails coin can never show heads).
  3. Optimal decision after Heads. Heads is impossible for the double-tails coin, so $P(\text{fair}\mid H)=1$ exactly: $\text{EMV(bet}\mid H)=\tfrac12(100)+\tfrac12(0)-5=\$45>0$ → bet.
  4. Optimal decision after Tails. By Bayes' rule, $P(\text{fair}\mid T)=\dfrac{0.2(0.5)}{0.9}=\dfrac19$, $P(\text{double-tails}\mid T)=\dfrac89$: $$\text{EMV(bet}\mid T)=\tfrac19(45)+\tfrac89(-5)=5-\tfrac{40}{9}=\boxed{\tfrac59\approx\$0.556}>0\ \Rightarrow\ \text{still bet.}$$
  5. Expected value with the toss, and EVSI. Since betting is optimal after either outcome, the toss never changes the decision: $$\text{EV(with info)}=P(H)\cdot45+P(T)\cdot\tfrac59=0.1(45)+0.9\!\left(\tfrac59\right)=4.5+0.5=\$5.$$ $$\boxed{\text{EVSI}=\text{EV(with info)}-\text{EMV(best action, no info)}=5-5=\$0.}$$
QuantityResult
EMV of betting, no information$5$
Optimal action after Headsbet (EMV = $45$)
Optimal action after Tailsbet (EMV = $5/9\approx0.56$)
EVSI of the pre-toss$0 — not worth paying anything for it