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Hard rolling dice problem

  1. Mar 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Five dice are rolled. You win a dollar for each number other than a five that is rolled, but you don't win extra for duplicate numbers; for example if [3,5,3,2,1] is the result, then you win three dollars. What should you pay to play this game (what is expected value)?

    2. Relevant equations

    Can't think of any specific ones

    3. The attempt at a solution

    It's hard for me to explain what I did because I don't know how to draw a tree diagram on a computer.

    First I did (5/6)*(4/6)*(3/6)*(2/6) but apparently that's wrong. Then I drew a tree diagram so I had 5/6 chance of rolling 1,2,3,4,6. If I rolled a 1, then I have a 2/3 chance of rolling a 2,3,4,6 but not 1,5. If I rolled a 3 let's say then 1/3 chance of rolling one of the remaining numbers namely 4,6. If I rolled a 4 then 1/6 chance of rolling a 6 and lastly multiplied by random variable 5 for five tosses. Thus my equation is this starting from bottom of the tree:

    5(1/6)2(1/3)4(1/2)5(2/3)(1/6)=1.85 The 5,2,4,5 I added in to account for the fact that I only considered certain numbers and left some out.

    Is this anywhere near correct? Thanks
     
    Last edited: Mar 4, 2009
  2. jcsd
  3. Mar 4, 2009 #2

    tiny-tim

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    Homework Helper

    Hi leumas614! :smile:

    You need to be systematic …

    how many ways are there of winning $1?
    $2?

    $5? :wink:
     
  4. Mar 4, 2009 #3
    # ways of rolling anything is 5[tex]^{5}[/tex]

    # of ways of winning $0: only 1 (rolling 5,5,5,5,5)

    # of ways of winning $5:

    1,2,3,4,6 in any order so 5!

    # of ways of winning $1

    any combination of 1 and 5, 2 and 5 ... 4 and 5. The prob is the binomial distribution:
    4[tex]\sum[/tex]from n=1 to 5([tex]^{5}_{n}[/tex])(1/6)[tex]^{n}[/tex](1/6)[tex]^{5-n}[/tex]=4(1/6)[tex]^{5}[/tex](1+5+10+5)=0.01080

    There has to be a way to solve this without enumerating all possible answers. What about this:

    odds of rolling only two numbers out of 6 are 5[(2/6)[tex]^{5}[/tex]-(1/6)[tex]^{5}[/tex]] It's multiplied by 5 because you can make 5 pairs of two numbers (1,5 2,5...6,5). It's minus (1/6)[tex]^{5}[/tex] to account for the overlap. You are counting one number twice. For the other numbers:

    C(5,1)[(2/6)^5-(1/6)^5] +C(5,2)[(3/6)^5-(2/6)^5]+....+C(5,5)[(1-(5/6)^5]=3.2447

    This includes rolling 3 numbers, 4 numbers, 5 numbers
     
  5. Mar 4, 2009 #4
    Never mind I found out the answer.

    X = I1+I2+I3+I4+I6


    I1 is the indicator of getting a 1. That is you either get a 1 or you don't. So X is getting a number other than 5 and no repeats which equals [tex]\sum[/tex]In

    Take expected value of both sides so E(X) = P(1)+P(2)+P(3)+P(4)+P(6) which is just 5P(1).

    P(1) is the odds of getting at least one 1 which is just 1-(5/6)5

    So E(X)=5P(1)=5(1-(5/6)5)=2.99
     
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