Answer: 6
let Expected number of throws be $ E[X]$
$E[X] = \sum E[x_{i}] * P_{i}$
$E[X] = $ (case when first output = 2 or 4 or 6) + (case when first output = 1 or 3 or 5)
$E[X] = $ (case when first output = 2 or 4 or 6) *(case when second output = 2 or 4 or 6)
+ (case when first output = 2 or 4 or 6)* (case when second output = 1 or 3 or 5)
+ (case when first output = 1 or 3 or 5)
reference image :
writing it as recursive equation , since we are back to original state (back to starting state)
i.e, say after k trails
$E[X_{i}] = P_{i} * (k + E[X])$
solving the equation:
$E[X] = $$3*(2 * \frac{1}{6} * \frac{3}{6})$ + $3*((2 + E[X] )* \frac{1}{6} * \frac{3}{6})$ + $((1 + E[X] )* \frac{3}{6} )$
$E[X] = $$\frac{6 * 3}{6 * 6}$ + $((2 + E[X] )* \frac{3*3}{6*6})$ + $((1 + E[X] )* \frac{3}{6} )$
$E[X] = $$\frac{1}{2}$ + $((2 + E[X] )* \frac{1}{4})$ + $((1 + E[X] )* \frac{1}{2} )$
$E[X] = $$\frac{2}{4}$ + $((2 + E[X] )* \frac{1}{4})$ + $((1 + E[X] )* \frac{2}{4} )$
4* $E[X] = $ $2 + (2+ E[X] ) + 2*(1+ E[X])$
4* $E[X] = $ $2 + 2+ E[X] + 2+ 2E[X]$
we get, $E[X]$ = expected number of throws = $6$