# Thread: Expected value of die rolling game.

1. ## Expected value of die rolling game.

Hello all,

I was told that this is how to solve the following problem, but I just can't totally buy it and get my head around it. So if any of you can verify that this is correct, I would be indebted to you.

You've got a fair die, and I propose the following game. If you roll a 1 or a 2, I will give you 10 dollars (and the game ends). If you roll a 3 or a 4, you get another roll but the prize is discounted by d. That is, if you roll four 3's in a row, then roll a one, I give you 10d^4. If you roll a 5 or a 6, the game is over and you get nothing.

I want to find the expected value of this game, lets call this expected value G. The way I was told to solve is the following. There is a 1/3 chance you a 10 dollar prize, a 1/3 chance you get another roll (discounted) and a 1/3 chance you get nothing. Hence....

G=1/3(10)+1/3(dG)

solving for G we get...

10/(3-d)

Is this correct? Thank you kindly for any advice.

Sincerely,

Nick

2. Looks correct to me.

3. Hello, salohcin!

I cranked this out . . . using my usual baby-talk.

You roll a fair die.

If you roll a 1 or a 2, you win $10 dollars. If you roll a 5 or 6, you get nothing. If you roll a 3 or a 4, you get another roll but the prize is discounted by$\displaystyle d.$That is, if you roll four 3's in a row, then roll a one, you win$\displaystyle 10d^4$dollars. Find the expected value of this game. We have these two facts: . .$\displaystyle \begin{array}{cccccccccccc}P(\$10) &=& P(\text{1 or 2}) &=& \frac{1}{3} & \Rightarrow & E_1 &=& \frac{1}{3}(10) &=& \dfrac{10}{3}\\ \\[-3mm] P(\$0) &=& P(\text{5 or 6}) &=& \frac{1}{3} & \Rightarrow & E_2 &=& \frac{1}{3}(0) &=& 0 \end{array}$What about the third option? Win in 2 rolls: . . Roll a 3 or 4 on the first roll:$\displaystyle Prob = \frac{1}{3}$. . The prize is$\displaystyle 10d$dollars. . . Roll a 1 or 2 on the second roll:$\displaystyle Prob = \frac{1}{3}$Hence: .$\displaystyle P(10d) \:=\:\frac{1}{3}\cdot\frac{1}{3} \:=\:\frac{1}{3^2}$Win in 3 rolls: . . Roll a 3 or 4 on the first roll:$\displaystyle Prob = \frac{1}{3}$. . Roll a 3 or 4 on the second roll:$\displaystyle Prob = \frac{1}{3}$. . The prize is$\displaystyle 10d^2$dollars. . . Roll a 1 or 2 on the third roll:$\displaystyle Prob = \frac{1}{3}$Hence: .$\displaystyle P(10d^2) \:=\:\frac{1}{3^3} $. . and so on . . . Hence: .$\displaystyle E_3 \;=\;\frac{1}{3^2}(10d) + \frac{1}{3^3}(10d^2) + \frac{1}{3^4}(10d^3) + \hdots $. . . . . .$\displaystyle =\;\dfrac{10d}{3^2}\underbrace{\left[1 + \frac{d}{3} + \left(\frac{d}{3}\right)^2 + \left(\frac{d}{3}\right)^3 + \hdots \right]}_{\text{geometric series}} $The geometrtic series has the sum: .$\displaystyle \dfrac{1}{1-\frac{d}{3}} \:=\:\dfrac{3}{3-d}$Hence: .$\displaystyle E_3 \;=\;\dfrac{10d}{9}\cdot\dfrac{3}{3-d} \;=\;\dfrac{10d}{3(3-d)}$Therefore: .$\displaystyle E \;=\;E_1 + E_2 + E_3 \;=\;\dfrac{10}{3} + 0 + \dfrac{10d}{3(3-d)} \;=\;\boxed{\frac{10}{3-d}} \$