Which Strategy Maximizes Expected Reward in a Paper-Picking Game?

[e(ho0n3]

[SOLVED] Which Paper Do I Pick?

Homework Statement
A philanthropist writes a positive number x on a piece of paper, shows it to an impartial observer, and then turns it face down on the table. The observer then flips a fair coin. If it shows heads, she writes the value 2x, and, if tails, the value x/2, on a piece of blue paper which she then turns face down on the table. Without knowing either the value x or the result of the coin flip, you have the option of turning over either red or the blue piece of paper. After doing so, and observing the number written on that paper, you may elect to receive as a reward either that amount or the (unknown) amount written on the other piece of paper. For instance, if you elect to turn over the blue paper and observe the value 100, then you can elect either to accept 100 as a reward or to take the amount (either 200 or 50) on the red paper. Suppose that you would like your expected reward to be large.

(a) Argue that there is no reason to turn over the red paper first because if you do so, then no matter what value you observe, it is always better to switch to the blue paper.

(b) Let y be a fixed nonnegative value, and consider the following strategy. Turn over the blue paper and if its value is at least y, then accept that amount. If it is less than y, then switch to the red paper. Let R_y(x) denote the reward obtained if the philanthropist writes the amount x and you employ this strategy. Find E[R_y(x)]. Note that E[R_0(x)] is the expected reward if the philanthropist writes the amount x when you employ the strategy of always choosing the blue paper.

The attempt at a solution
(a) If I look at the blue paper and observe 100, then either 200 or 50 is written on the red paper. If I look at the red paper and observe 100, then either 200 or 50 is written on the blue paper. It doesn't matter which paper I look at first then. If I decide to receive the amount in the unknown piece of paper, then I will obtain twice the value I saw 50% of the time right? But this means the value I observe will be twice that written on the other piece of paper 50% of the time. So it doesn't matter what decision I make.

Of course, if I just want to obtain twice what the philanthropist wrote, I will always pick the blue paper. So, under this criteria, there is no need to look at the red paper.

(b) E[R_0(x)] = 2x * 0.5 + x/2 * 0.5 = x + x/4 = 5/4 x right? For arbitrary y, if x < y, then R_y(x) = x. If x >= y, then R_y(x) = 2x or x/2. P{R_y(x) = x} = P{R_y(x) = x|x < y}P{x < y} + P{R_y(x) = x|x >= y}P{x >= y} which simplifies to P{x < y}. What is this though?

 

[D H]

Of course, if I just want to obtain twice what the philanthropist wrote, I will always pick the blue paper. So, under this criteria, there is no need to look at the red paper.

You don't know that! The blue paper might have half what the philanthropist wrote on the red paper.

This problem is a variant of the http://en.wikipedia.org/wiki/Exchange_paradox" .

If by positive number you mean positive integer, then there is a definite twist to this problem as compared to the standard versions of the exchange paradox. If you pick the blue paper first and see an odd amount written on it then you should choose the value on the red paper instead.

 

[e(ho0n3]

You don't know that! The blue paper might have half what the philanthropist wrote on the red paper.

I know, but I'll have to risk it if I want to get twice what the philanthropist wrote.

If by positive number you mean positive integer, then there is a definite twist to this problem as compared to the standard versions of the exchange paradox. If you pick the blue paper first and see an odd amount written on it then you should choose the value on the red paper instead.

I think "positive number" means positive real, so x/2 is given exactly (no rounding).

 

[e(ho0n3]

If x is real, I don't see any possible advantage of turning over one of the papers. If I employ the strategy of always selecting as reward the amount written on the red paper, then my expected reward is x. If I employ the strategy of always selecting as reward the amount written on the blue paper, then my expected reward is 5/4x. Thus, if I want my expected value to be large, I must employ the latter strategy.

If x is an integer, then the situation is different: Suppose x is odd and x/2 was written on the blue paper. I choose to look at the blue paper and observe that the number ends in .5. I immediately conclude that x must be odd and so I choose the red paper. It is thus advantageous to look at the value on the blue paper.

But what is my expected reward in this situtation? Assume that x is odd 0.5 of the time. If I employ the strategy of picking as reward the value written on the red paper even if I see that the value on the blue paper does not end in .5, my expected reward is x. If I employ the strategy of picking as reward the value on the blue paper if that value does not end in .5 and the value on the red paper otherwise, then my expected reward is 0.5 * x + 0.25 * 2x + 0.25 * x/2 = 9/8x. If I employ the strategy of just picking as reward the value on the blue paper regardless of what I see, then my expected reward is 5/4x. It's counterintuitive that the latter strategy will actually yield a higher expected reward that the strategy mentioned before it. Hmm...

So by this analysis, it is always better to pick the value on the blue paper. It's not even necessary to turn over one of the papers.

 

[e(ho0n3]

The above post takes care of problem (a). The solution to (b) would depend, I think, on the distribution the philanthropist is using for choosing his x's. Am I right?

 

[e(ho0n3]

I had a look at the answer in the book. I got (a) right but (b) wrong.

Concerning (b), I don't have to worry about the distribution the philanthropist is using for choosing x. All I needed to do was to consider these three cases: y < x/2, x/2 <= y < 2x and y >= 2x.

Given y < x/2, then I would always pick the blue paper and so my expected value is 5/4x. Given x/2 <= y < 2x, then I would pick the value on the red paper half of the time and the value on the blue paper the rest of the time. The expected value in this situation is 0.5 * 2x + 0.5 * x = 3/2x. Given y >= 2x, I will always end up picking the value on the red paper and so the expected value is x.

Sigh. I can't believe it was this easy. That's what I get for overcomplicating.