Expected value of die rolling game.

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July 25th 2010, 09:57 AM
salohcin

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
July 25th 2010, 12:11 PM
lvleph

Looks correct to me.
July 25th 2010, 12:42 PM
Soroban

Hello, salohcin!

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

Quote:

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 $d.$
That is, if you roll four 3's in a row, then roll a one, you win $10d^4$ dollars.

Find the expected value of this game.

We have these two facts:

. . $\begin{array}{cccccccccccc}P(\$10) &=& P(\text{1 or 2}) &=& \frac{1}{3} & \Rightarrow & E_1 &=& \frac{1}{3}(10) &=& \dfrac{10}{3}\\ \\[-3mm]<br /> P(\$0) &=& P(\text{5 or 6}) &=& \frac{1}{3} & \Rightarrow & E_2 &=& \frac{1}{3}(0) &=& 0 \end{array}$

What about the third option?

Win in 2 rolls:
. . Roll a 3 or 4 on the first roll: $Prob = \frac{1}{3}$
. . The prize is $10d$ dollars.
. . Roll a 1 or 2 on the second roll: $Prob = \frac{1}{3}$
Hence: . $P(10d) \:=\:\frac{1}{3}\cdot\frac{1}{3} \:=\:\frac{1}{3^2}$

Win in 3 rolls:
. . Roll a 3 or 4 on the first roll: $Prob = \frac{1}{3}$
. . Roll a 3 or 4 on the second roll: $Prob = \frac{1}{3}$
. . The prize is $10d^2$ dollars.
. . Roll a 1 or 2 on the third roll: $Prob = \frac{1}{3}$
Hence: . $P(10d^2) \:=\:\frac{1}{3^3}$

. . and so on . . .

Hence: . $E_3 \;=\;\frac{1}{3^2}(10d) + \frac{1}{3^3}(10d^2) + \frac{1}{3^4}(10d^3) + \hdots$

. . . . . . $=\;\dfrac{10d}{3^2}\underbrace{\left[1 + \frac{d}{3} + \left(\frac{d}{3}\right)^2 + \left(\frac{d}{3}\right)^3 + \hdots \right]}_{\text{geometric series}}$

The geometrtic series has the sum: . $\dfrac{1}{1-\frac{d}{3}} \:=\:\dfrac{3}{3-d}$

Hence: . $E_3 \;=\;\dfrac{10d}{9}\cdot\dfrac{3}{3-d} \;=\;\dfrac{10d}{3(3-d)}$

Therefore: . $E \;=\;E_1 + E_2 + E_3 \;=\;\dfrac{10}{3} + 0 + \dfrac{10d}{3(3-d)} \;=\;\boxed{\frac{10}{3-d}}$