Even Steven

Solution

Answer: 0.05

A very clean way to think about it is to ignore repeated rolls and only look at the order in which the faces appear for the first time. Eventually, as you keep rolling, the numbers \[ 1,2,3,4,5,6 \] will each show up for the first time in some order. For example, you might first see them in the order \[ 4,2,6,1,5,3. \] That gives a permutation of the six faces. Now, you win exactly when all three even numbers \[ 2,4,6 \] have appeared before the first odd number appears. In terms of this first-appearance order, that means the first three positions must all be the evens. So the first three spots must be \[ 2,4,6 \] in any order, which gives \[ 3! \] possibilities. Then the last three spots must be the odds \[ 1,3,5 \] in any order, which gives another \[ 3! \] possibilities. So the number of favorable first-appearance orders is \[ 3!\cdot 3!. \] Since every ordering of first appearances is equally likely, the total number of possible orders is \[ 6!. \] So the probability is \[ \frac{3!\cdot 3!}{6!} = \frac{6\cdot 6}{720} = \frac{36}{720} = \frac{1}{20}. \] So the chance is \[ \frac{1}{20} = 5\%. \]