Monday, February 1, 2010

Monday Probability 4: Rolling dice thrice

A dice is rolled and the number which comes can be taken as the prize money or you can ask for another roll. The new number is your prize money or you can ask for the final roll. In all cases the final roll is your prize money. What is the expected return for an optimal strategy?

Answer:14/3

2 comments:

  1. Let the strategy be represented by 2 integers (t1,t2)
    This means that whenever you get a number lower than t1 in the first throw, you ask for another throw. Whenever you get a number lower than t2 on the second throw, you ask for the third throw.
    Clearly the expected prize money is
    E=(t1-1)/6*E2+1/6*(t1+...+6)
    where E2 is the expected prize money obtained from second throw onwards
    Similarly
    E2=(t2-1)/6*E3+1/6*(t2+...+6)
    where E3=(1+2+3+4+5+6)/6=7/2

    Hence E=-1/12*t1^2 + (-1/72*t2^2 + 1/9*t2 + 41/72)*t1 + (1/72*t2^2 - 1/9*t2 + 217/72)

    which is maximized when t1=5 and t2=4 and the optimal value is 14/3~4.667

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