1. ## Find expectated payoff

Hi

There is a bag containing numbers from 1 to n. A number is drawn at random from the bag and put back.(let the number be "X") A second number is drawn after that. (let the number be "Y")

THe person gets 10$if X>y, he has to pay 5$ is x<y and gets 0$is x=y. What is the expected payoff of the game? (PS: I am not sure how to calculate the probabilities of x>y,x<y). The prob of x=y should be 1/n^2 most probably but not sure.. Any help is appreciated) Thanks 2. Originally Posted by champrock Hi There is a bag containing numbers from 1 to n. A number is drawn at random from the bag and put back.(let the number be "X") A second number is drawn after that. (let the number be "Y") THe person gets 10$ if X>y, he has to pay 5$is x<y and gets 0$ is x=y.

What is the expected payoff of the game?

(PS: I am not sure how to calculate the probabilities of x>y,x<y). The prob of x=y should be 1/n^2 most probably but not sure.. Any help is appreciated)

Thanks
By symmetry p(x>y)=p(x<y), p(x=y)=1/n and

p(x>y)+p(x<y)+p(x=y)=1

which will allow you to find the three probabilities, then the expected payoff is:

E=10p(x>Y)-5p(x<y)

CB

3. Hello, champrock!

There is a bag containing numbers from $\displaystyle 1\text{ to }n.$
A number is drawn at random from the bag and put back. (Let the number be $\displaystyle x$.)
A second number is drawn after that. (Let the number be $\displaystyle y$.)

The player: .$\displaystyle \begin{array}{cc}\text{wins \$10} & \text{ if }x > y \\ \text{pays \$5} & \text{if }x < y \\ \text{gets \$0}& \text{if }x = y \end{array} $What is the expected payoff of the game? There are: .$\displaystyle n^2$possible draws. In$\displaystyle n$cases, the numbers are equal. . .$\displaystyle P(x = y) \:=\:\frac{n}{n^2} \:=\:\frac{1}{n}$Among the other$\displaystyle n^2-n$cases, half of them have$\displaystyle x$greater than$\displaystyle y.$That is, in$\displaystyle \frac{n^2-n}{2}$cases,$\displaystyle x > y$. . Hence: .$\displaystyle P(x > y) \:=\:\frac{\frac{n^2-n}{2}}{n^2} \:=\:\frac{n-1}{2n} $Similarly: .$\displaystyle P(x < y) \:=\:\frac{n-1}{2n}$Therefore: .$\displaystyle EV \;=\;\left(\frac{1}{n}\right)(0) + \left(\frac{n-1}{2n}\right)(10) + \left(\frac{n-1}{2n}\right)(-5) \;=\;\boxed{\frac{5(n-1)}{2n}} $~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Translation: .$\displaystyle EV \:=\:\frac{5n-5}{2n} \:=\:\frac{5n}{2n} - \frac{5}{2n} \;=\;\frac{5}{2}-\frac{5}{2n}$On the average, you can expect to win slightly less than$2.50 per game.