Fair Value of a Dice Game 1 die. Up to 3 rolls. Your winnings are equal to the value of a single roll. You can stop after roll 1 or roll 2. If you proceed to the next roll (i.e the 2nd or 3rd), you forfeit the previous value (no memory). Therefore, if you roll 3 times, the value of the 3rd roll is what you win.
What is the fair price of this game?

Strategy
Since the expected value of a single roll is 3.5, I will stop on any roll if I get a 4, 5, or 6.
If Fair Price = Expected Value
And Expected Value = P(i_th roll) x E(i_th roll) i=1 to 3
Then Fair Value = P(1 roll) E(1st roll) + P(2 rolls) E(2nd roll) + P(3 rolls) * E(3rd roll)
E(i_th roll) = 3.5 for any roll
P(only 1 roll) = 1/2 :: must roll 4|5|6
P(only 2 rolls) = 1/2 * 1/2 :: must roll 1|2|3 then 4|5|6
P(3 rolls) = 1/2 * 1/2 :: must roll 1|2|3 then 1|2|3
$\frac{1}{2} * 3.5 + \frac{1}{4} * 3.5 + \frac{1}{4} * 3.5 = $
$\frac{1}{2}*\frac{7}{2} + \frac{1}{4}*\frac{7}{2} + \frac{1}{4}*\frac{7}{2} = $
$\frac{7}{4} + \frac{7}{8} + \frac{7}{8} = $
$\frac{14}{8} + \frac{7}{8} + \frac{7}{8} = $
$\frac{28}{8} = 3.50$
Whn trying to solve this before, I think I made a mistake and end up with something like $4\frac{5}{8} = 4.625$
Should 3.5 be the answer, or should the 1st and 2nd rolls be "wieghted" by what values you must roll to move on (1,2 or 3) instead of the general 3.5 expected value for a single roll?

The value of a single roll is $3.5$
With two rolls, keep the first if it is more than $3.5$, so the value is $$\frac12\times5+\frac12\times3.5=4.25$$
With three rolls, keep the first if it is more than $4.25$
