Optimal Bidders I

Hard

Statistics Expectations

Problem

Carter has come into contact with a bounty of gold. He takes it to an auction shop. The shop tells him that every bidder will place a bid uniformly between $500 an d$ 1000. They also say they can recruit bidders for $5 per person.

Carter needs to figure out the optimal number of bidders to maximize his expected payout. What is this maximum payout?

Original Problem Link: Click here

Solution

Step 1: Modeling the Maximum Bid

Each bidder bids randomly between 500 and 1000. Now, if there are $n$ bidders, we can assume that the bids will be evenly spaced between 500 and 1000 on average.

The expected spacing between bids is:

\frac{500}{n + 1}

Thus, the expected maximum bid will be:

M (n) = 1000 - \frac{500}{n + 1}

Step 2: Writing the Expected Payout

Now, Carter’s expected earnings are simply:

Max bid - Cost of bidders

Which gives us:

E (n) = M (n) - 5 n

Substituting $M (n)$ :

E (n) = 1000 - \frac{500}{n + 1} - 5 n

Now, we need to maximize this function.

Step 3: Finding the Optimal $n$

We differentiate $E (n)$ and set it to 0:

\frac{d}{d n} (1000 - \frac{500}{n + 1} - 5 n) = 0

\frac{500}{( n + 1 ) ^{2}} = 5

Solving:

500 = 5 (n + 1)^{2}

100 = (n + 1)^{2}

n + 1 = 10 \Rightarrow n = 9

Thus, the optimal number of bidders is $n = 9$ .

Step 4: Calculating the Maximum Expected Payout

Substituting $n = 9$ :

E (9) = 1000 - \frac{500}{10} - 5 (9)

= 1000 - 50 - 45

= 905