91Ó°ÊÓ

Given in the question that, a philanthropist writes a positive number $x$on a piece of red paper, shows the paper 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 the 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 your reward or to take the amount (either $200$or$50$) on the red paper .

We will show that the expected winning on the red paper is still greater than the expected winning on the blue paper. Observe that on the blue paper still is written $x$. On the other hand, the expected quantity of money on the red paper is

$\frac{1}{2}·\frac{x}{2}+\frac{1}{2}2x=\frac{5}{4}x$

So, whatever happens, it is always better to turn the blue paper first.

The expected quantity of money on the red paper is

$\frac{1}{2}·\frac{x}{2}+\frac{1}{2}2x=\frac{5}{4}x$

A philanthropist writes a positive number $x$on a piece of red paper, shows the paper 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 the 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 your reward or to take the amount (either $200$ or $50$) on the red paper.

Here we have two cases. If$xâ‰\yen y$, we stay with the amount on the blue paper, so the expected winning is simply

$E\left(R_y\left(x\right)\right)=x$

If $x<y$, we switch on the red paper. We know that on the red paper may be written $x/2$with the equal probability as $2x$, so the expected winning in this case is

$E\left(R_y\left(x\right)\right)=\frac{1}{2}·\frac{x}{2}+\frac{1}{2}2x=\frac{5}{4}x$

Expected winning

$E\left(R_y\left(x\right)\right)=\frac{1}{2}·\frac{x}{2}+\frac{1}{2}2x=\frac{5}{4}x$