[BruceZ]

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You roll a fair 6 sided die and are given an amount of money equal to the number you roll, 1=$1 2=$2...etc. The caveat is that if you wish to reroll you may, but would then be stuck with whatever you get the second time.

The question is, how much would you be willing to pay to play this game? read: what is the EV

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The average for 1 die is 3.5, so keep 4,5,6 (avg = 5) and roll 1,2,3 again. The EV is

(1/2)*5 + (1/2)*3.5 = $4.25 for 2 dice.


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A slightly less trival extension, what if you get up to 6 rolls, meaning you can choose to reroll 5 times before you are stuck with the 6th.

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EV(n) = EV for n rolls.
Work backward from 2 remaining rolls.
Since EV(2) = 4.25, keep 5+ (avg = 5.5) and roll 1,2,3,4:

2 rolls left, keep 5+, EV(3) = (1/3)*5.5 + (2/3)*4.25 = 4 + 2/3

3 rolls left, keep 5+, EV(4) = (1/3)*5.5 + (2/3)*(4 2/3) = 4 + 17/18

4 rolls left, keep 5+, EV(5) = (1/3)*5.5 + (2/3)*(4 + 17/18) = 5 + 7/54

5 rolls left, keep 6, EV(6) = (1/6)*6 + (5/6)*(5 + 7/54) = 5 + 84/325 =~ $5.27 for 6 dice.


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Finally and most interestingly, how about generalizing it for n rolls...just for kicks

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For n > 6, EV(n) = (1/6)*6 + (5/6)*EV(n-1)

recursion: EV(n) = 1 + (5/6)*EV(n-1)

EV(6) = 5.27
EV(7) = 1 + (5/6)*5.27
EV(8) = 1 + 5/6 + (5/6)^2 * 5.27
EV(9) = 1 + 5/6 + (5/6)^2 + (5/6)^3 * 5.27
...
EV(n) = sum [i = 0 to n-7](5/6)^i + (5/6)^(n-6) * 5.27

from finite sum geometric series:

EV(n) = [1 - (5/6)^(n-6)]/(1 - 5/6) + (5/6)^(n-6) * 5.27

EV(n) = 6 - 6*(5/6)^(n-6) + 5.27 * (5/6)^(n-6)

EV(n) = 6 - 0.73*(5/6)^(n-6), for n > 6 dice.