Fair price to play a dice game We play a game with a pot and a single die. The pot starts off empty.

*

*if the die roll is  1, 2 or 3, I put 1 pound in the pot, and the die is thrown again

*if its 4 or 5, the game finishes, and you win whatever is in the pot

*if its 6, you leave with nothing

What is the fair price for you to play this game?

*

*1/2

*1/3

*3/2

*2/3


This question is equivalent to asking the expected value of the price in the pot when the game ends.
There’s a $\frac{1}{2}\cdot \frac{1}{3}$ chance the total is 1 (1, 2, or 3 and then 4 or 5), $\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{3}$ chance the total is 2 (1, 2, or 3 and then 1, 2, or 3 and then 4 or 5), and so on, such that we want to find $\frac{1}{3}(1\cdot (\frac{1}{2})^1+2\cdot (\frac{1}{2})^2+3\cdot (\frac{1}{2})^3+\dots).$
Let $S = \frac{1}{3}(1\cdot (\frac{1}{2})^1+2\cdot (\frac{1}{2})^2+3\cdot (\frac{1}{2})^3+\dots)$
Then $\frac{S}{2} = \frac{1}{3} (1\cdot (\frac{1}{2})^2+2\cdot (\frac{1}{2})^3 + \dots)$
Subtracting the two equations gets $S-\frac{S}{2} = \frac{1}{3}(1\cdot (\frac{1}{2})^1 + 1\cdot (\frac{1}{2})^2+1\cdot (\frac{1}{2})^3+\dots).$ The parentheses is just equal to $1,$ so $\frac{S}{2} = \frac{1}{3}\cdot 1,$ so $S = \boxed{\frac{2}{3}.}$
