Question

Suppose we are interested in bidding on a piece of land and we know one other bidder is interested. The seller announced that the highest bid in excess of $10,200 will be accepted. Assume that the competitor's bid x is a random variable that is uniformly distributed between $10,200 and $14,600.

a. Suppose you bid $12,000. What is the probability that your bid will be accepted (to 2 decimals)?

b. Suppose you bid $14,000. What is the probability that your bid will be accepted (to 2 decimals)?

c. What amount should you bid to maximize the probability that you get the property (in dollars)?

d. What is the expected profit for this bid (in dollars)?

Answer 1

a.

probability that your bid will be accepted = P(x < 12,000)

= (12,000 - 10,200) / (14,600 -  10,200)

= 0.4090909

b.

probability that your bid will be accepted = P(x < 14,000)

= (14,000 - 10,200) / (14,600 -  10,200)

= 0.8636364

c.

Let Y be the amount you bid.

probability that your bid will be accepted = P(x < y)

= (y - 10,200) / (14,600 -  10,200)

=  (y - 10,200) / 4400

The maximum probability that you get the property, will be for the maximum value of y.

We know that the probability lies between 0 and 1.

Thus,

0 <  (y - 10,200) / 4400 < 1

0 < y - 10,200 < 4400

10,200 < y < 14600

The maximum value of y is 14600.

Thus, you should bid for $14,600 to maximize the probability that you get the property.

d.

When you bid for $14,600,  the probability that you get the property is 1.

If you know someone is willing to pay you $A for the property.

Expected profit for this bid = probability that you get the property * ($A - $14,600)

= 1 * $(A - 14,600)

= $(A - 14,600)