Results 1 to 3 of 3

Math Help - Expected value of die rolling game.

  1. #1
    Junior Member
    Joined
    Apr 2010
    Posts
    40

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Mar 2009
    Posts
    378
    Looks correct to me.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,749
    Thanks
    649
    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 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}}

    Last edited by Soroban; July 25th 2010 at 01:04 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. probability of winning dice rolling game
    Posted in the Advanced Statistics Forum
    Replies: 7
    Last Post: July 17th 2013, 06:34 AM
  2. Replies: 1
    Last Post: January 9th 2012, 11:48 AM
  3. Replies: 5
    Last Post: January 4th 2012, 01:49 AM
  4. 2 player game rolling a die
    Posted in the Advanced Statistics Forum
    Replies: 8
    Last Post: September 9th 2009, 05:04 AM
  5. Sequential-form game (game tree diagram)
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: October 5th 2008, 06:51 PM

Search Tags


/mathhelpforum @mathhelpforum