Puzzle 6 of 26 · Medium

Evens before odds

Your answer

You roll a fair die until every face has come up at least once. What’s the chance that 2, 4 and 6 have all appeared before you see any odd number?

The evens need to show up before any odd.

Right answers go through as you type.

Show the solution

Answer: 1/20 = 5%

Evens first

\displaystyle 3!

Then odds

\displaystyle 3!

\displaystyle 36

Step 1 · Only the order faces first appear matters
$\displaystyle a \text{ random order of the } 6 \text{ faces}$
Repeats don’t change anything. The first new face, then the next new face and so on form an order of 1 to 6, and every order is equally likely.
Step 2 · Orders in all
$\displaystyle 6! = 720$
6 choices for the first new face, 5 for the next, and so on.
Step 3 · Orders with the 3 evens first
$\displaystyle 3! \times 3! = 36$
Arrange the evens in the first three spots and the odds in the last three.
Step 4 · Chance
$\displaystyle 36 \div 720 = \frac{1}{20}$
The good orders out of all the equally likely ones.

Answer $\displaystyle \frac{1}{20} = 5\%$

Step 4 of 4