Coin Problem

Problem:

Suppose you can bid on a coin of an unknown value between 0 and 100. If your bid placed is greater than the value of the coin, then you win the coin. Furthermore, if you win the coin, then you are guaranteed to be able to sell it to your friend, a coin collector, for 1.5 times the coin's value. How much should you bid for the coin?

Solution:

Let X be the value of the coin, where X ~ unif(0,100).

Let Y be your bid.

Your return, R, is 1.5X - Y if Y > X, or -Y if Y ≤ X

Your expected return, E(R), is given by the following formula:

E(R) = E(1.5X - Y | Y > X)*P(Y > X) + E(-Y)*P(Y ≤ X)

E(R) = -Y + 1.5*E(X | Y > X)*P(Y > X)

To gain some intuition into how these terms work, here are some numerical examples:

Bid Y = 0

In this case, P(Y > X) is 0, because 0 is the lower bound of the coin's value.

So E(R) = 0.

 

Bid Y = 1

E(R) = -1 + 1.5*E(X | X < 1)*P(X < 1)

The probability of X < 1 is 1/100. If we know that X < 1, then the random variable X | X < 1 is distributed uniformly between 0 and 1. Thus, E(X | X < 1) = 1/2.

So E(R) = -1 + 1.5*0.5*0.01 = -0.9925

 

Bid Y = 2

E(R) = -2 + 1.5*E(X | X < 2)*P(X < 2)

The probability of X < 2 is 2/100. If we know that X < 2, then the random variable X | X < 2 is distributed uniformly between 0 and 2. Thus, E(X | X < 1) = 1.

So E(R) = -2 + 1.5*1*0.02 = -1.97

 

Bid Y = 10

E(R) = -10 + 1.5*E(X | X < 10)*P(X < 10)

Following a similar pattern, E(R) = -10 + 1.5*5*0.1 = -9.25

 

Bid Y = 50

E(R) = -50 + 1.5*E(X | X < 50)*P(X < 50)

Following a similar pattern, E(R) = -50 + 1.5*25*0.5 = -31.25

 

Bid Y = 80

E(R) = -80 + 1.5*E(X | X < 80)*P(X < 80)

Following a similar pattern, E(R) = -80 + 1.5*40*0.8 = -32

 

Bid Y = 100

E(R) = -100 + 1.5*E(X | X < 100)*P(X < 100)

Following a similar pattern, E(R) = -100 + 1.5*50*1 = -25

This leads us to the finding that E(X | Y > X) = Y / 2; additionally, P(Y > X) = Y / 100.

Therefore, our general formula for our expected return is:

E(R) = -Y + 1.5*(Y/2)*(Y/100), or:

E(R) = 0.0075Y² - Y

It turns out that under the rules of this game, there is no bid that returns a positive expected return. To see this, we can find the optimum of the return function:

E(R)' = 0.015Y - 1 = 0

Y* = 1 / 0.015 = 66.66667

However, this optimum Y* is a local minimum rather than a local maximum, because the expected return function is convex; E(R)'' = 0.015 > 0. Indeed, E(R) when Y = Y* is -33.33333.

In our numerical examples above, we checked the boundary conditions of Y = 0 and Y = 100, and thus have confirmed that for this game, Y = 0, or not placing a bet at all, yields the highest return.

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Problem:

Suppose the return multiplier on the resale value of the coin differs from 1.5. What multipliers would cause you to change your bidding strategy?

Solution:

Let a represent the multiplier, and assume a > 0. Our expected return formula is modified to the following:

E(R) = -Y + a*(Y/2)*(Y/100)

E(R) = (a/200)*Y² - Y

We can first identify the values of a under which we would expect any positive returns; or, solve for a such that:

(a/200)*Y² - Y > 0

Y((a/200)*Y - 1) > 0

a > 200/Y

In our most extreme bid of Y=100, this inequality indicates that a > 2 is the requirement for this game to yield any positive return. This explains why a = 1.5 indicated that the best strategy was not to place a bid.

Now, assuming a > 2, what is the optimal Y to bid? Once again, we consider the optimum Y* of the return function:

E(R)' = (a/100)*Y - 1 = 0

Y* = 100/a

However, once again, this is a local minimum, since E(R)'' = a/100 > 0. So we must check the boundary conditions.

When Y = 0, E(R) = 0. When Y = 100, E(R) = 50a - 100. With a > 2, this indicates that bidding 100 is always the strategy that yields the highest expected return.

Intuitively, this makes sense, as a bid of 100 guarantees that you will win the coin and that you will receive the payout. As the payout multiplier increases above 2, the expected payout dwarfs the fixed cost of placing the bid.