Find expectated payoff

champrock · #1 $Old$ July 6th, 2009, 07:06 AM

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

CaptainBlack · #2 $Old$ July 6th, 2009, 07:57 AM

Quote:

Originally Posted by champrock $View Post$

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

Soroban · #3 $Old$ July 6th, 2009, 08:17 AM

Hello, champrock!

Quote:

There is a bag containing numbers from $1\text{ to }n.$
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 player: . $\begin{array}{cc}\text{wins \$10} & \text{ if }x > y \\ \text{pays \$5} & \text{if }x < y \\ \text{gets \$0}& \text{if }x = y \end{array}$

What is the expected payoff of the game?

There are: . $n^2$ possible draws.

In $n$ cases, the numbers are equal.
. . $P(x = y) \:=\:\frac{n}{n^2} \:=\:\frac{1}{n}$

Among the other $n^2-n$ cases, half of them have $x$ greater than $y.$
That is, in $\frac{n^2-n}{2}$ cases, $x > y$
. . Hence: . $P(x > y) \:=\:\frac{\frac{n^2-n}{2}}{n^2} \:=\:\frac{n-1}{2n}$

Similarly: . $P(x < y) \:=\:\frac{n-1}{2n}$

Therefore: . $EV \;=\;\left(\frac{1}{n}\right)(0) + \left(\frac{n-1}{2n}\right)(10) + \left(\frac{n-1}{2n}\right)(-5) \;=\;\boxed{\frac{5(n-1)}{2n}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Translation: . $EV \:=\:\frac{5n-5}{2n} \:=\:\frac{5n}{2n} - \frac{5}{2n} \;=\;\frac{5}{2}-\frac{5}{2n}$

On the average, you can expect to win slightly less than $2.50 per game.