This is a one-player game. The goal of the game is to have N points.
You start with zero points.
You take a turn by rolling a six-sided die.
On a 1, you gain 1 point.
On a 6, you lose all points.
On a 2, 3, 4, or 5, nothing happens.

On average, how many turns will it take you to reach N points?
6*((2^N)-1)



f(cur, n): the expected number of turns to the end of the game

if cur == n: 0
else: (
        1 + f(cur+1, n) +
        1 + f(cur, n) +
        1 + f(cur, n) +
        1 + f(cur, n) +
        1 + f(cur, n) +
        1 + f(0, n)
    ) / 6
    == (6 + f(cur+1, n) + 4*f(cur, n) + f(0,n)) / 6


let X = f(cur, n) when cur != n
X = (6 + f(cur+1, n) + 4*X + f(0,n)) / 6
6*X = 6 + f(cur+1, n) + 4*X + f(0,n)
2*X = 6 + f(cur+1, n) + f(0,n)
X = (6 + f(cur+1, n) + f(0,n)) / 2
f(cur, n) = (6 + f(cur+1, n) + f(0,n)) / 2


Let n == 4.
Let Z = f(0,4)
f(cur, n) = (Z + 6 + f(cur+1, n)) / 2

f(4,4) = 0.
f(3,4) = (Z + 6) / 2 = Z/2 + 3
f(2,4) = (Z + 6 + Z/2 + 3) / 2 = (3/4)*Z + 9/2
f(1,4) = (Z + 6 + (3/4)*Z + 9/2) / 2 = ((7/4)*Z + 21/2) / 2  = ((7/8)*Z + 21/4)
f(0,4) = Z = (Z + 6 + (7/8)*Z + 21/4) / 2
2*Z = Z + 6 + (7/8)*Z + 21/4
Z = 6 + (7/8)*Z + 21/4
(1/8)*Z = 6 + 21/4
Z = (6 + 21/4)*8 = 48 + 42 = 90

WLOG, f(0, n) = uhhhhhhhh