Question

Tom is trying to enter the used car business. He knows that Jean-Ralphio will sell him a car that needs repairs. Once repaired, Tom can sell it for $100 more than he spent to purchase it. Further, he knows that each car has an 80% chance of being a good car, and a 20% chance of being a bad car. Good cars only cost $20 to repair, but bad cars cost $200 to repair.

Mona-Lisa decides to sweeten the deal by offering Tom a warranty. Tom can pay her $40 and in return, she will pay half of the repair costs, up to $80. Therefore, Tom’s choices are to buy the car, buy the car and the warranty, or not buy anything and stick to his day job wih the Parks Department.

1. (a) Draw the decision making chart to help Tom. Be sure to include his actions, the states of the car, and his resulting payouts.

2. (b) Based on the Expected Monetary Value criterion, what should Tom do?

3. (c) Now assume that you are uncertain of the probability that the car is good. De- termine how much the probability can change before the optimal option changes under the Expected Monetary Value criterion.

4. (d) Return to the original probabilities of 0.8 and 0.2 from the original problem statement. Assume the repair cost for a good car is unknown. How much could this value change before the optimal option changes under the Expected Monetary Value criterion?

Answer 1

Tom’s choices are to

1. buy the car
   1. If good car (with a probability of P(G)=0.80), make a profit of  $100-$20=$80, after spending $20 on repair
   2. If bad car (with a probability of P(B)=0.20), make a profit of  $100-$200=-$100, after spending $200 on repair
2. buy the car and the warranty (cost of warranty is $40)
   1. If good car (with a probability of P(G)=0.80), make a profit of  $100+10-$40-$20=$50, after spending $20 on repair, $40 on warranty and warranty paying the half of the repair cost= $20/2=$10
   2. If bad car (with a probability of P(B)=0.20), make a profit of  $100+$80-$40-$200=-$60, after spending $200 on repair, $40 on warranty and warranty paying the half of the repair cost upto $80= min($200/2,80) = $80
3. not buy anything and stick to his day job with the Parks Department.
   1. Payoff is $0

a) The decision tree is given below

b) At chance node 2: The expected monetary value of buying the car without warranty

At chance node 3: The expected monetary value of buying the car with a warranty

At the decision node 1, Tom’s choices are

• to buy the car with an expected payout of $44
• buy the car and the warranty with an expected payout of $28
• not buy anything and stick to his day job with the Parks, with an expected payoff of $0

Based on the Expected Monetary Value criterion Tom should buy the car.

c) Let P(G) be the probability of car being a good car and P(B)=1-P(G) be the probability of car being a bad car.

The expected monetary value of buying a car (node 2, EMV) is

The expected monetary value of buying a car with a warranty (node 3, EMV) is

We will find the value of P(G) at which the decisions change, that is EMV(2)<EMV(3)

This means that EMV(2)< EMV(3) for P(G)<0.5714

Hence if the probability of good car falls below 0.5714, then it makes sense to buy the car with a warranty.

But what about sticking with the day job, that is payoff of zero?

We will see at what probability EMV(3)<0

That is if the probability of good car falls below 0.5455, it makes sense to not buy the car (with or without warranty)

ans: When the probability of good car is above 0.5714 buy the car, when between 0.5455 to 0.5714 buy the car with warranty, when less than  0.5455 do not buy the car.

d) Let the repair cost for the good car be R.

The expected monetary values when buying the car without a warranty is

The expected monetary values when buying the car without a warranty has 2 cases

When R is less than 160, the warranty pays R/2, When R is greater than 160, the warranty pays $80.

The EMV, when R is less than 160 is

At R=$20 ENV(2)> EMV(3). the decision changes if EMV(2)<EMV(3)

That means if the repair cost for good car is more than $60, then it makes sense to buy the car with warranty

What about sticking with the day job?

Let us see when does EMV(3)<0

That means if the repair cost for good car is more than $90, then it makes sense to not buy the car with/without warranty

This means for when R is more than 160, the optimal choice would still be not to buy the car. Hence we will stop at this

ans: If the repair cost is less than $60, then buy the car, if between $60 to $90 buy then the car with warranty, if greater than $90 then do not buy the car.