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All Odds Before Even

Asked atCitadelIMC TradingSusquehanna International Group

You roll a fair six-sided die again and again. What is the probability that all three odd numbers ( $1$ , $3$ , $5$ ) have appeared before any even number ( $2$ , $4$ , $6$ ) appears?

Solution

Try it yourself first. Open a hint if you get stuck.

Hint 1

Repeated rolls don't matter. Only the order in which each face shows up for the first time matters.

Hint 2

That order of first appearances is a random arrangement of $1, \dots, 6$ , with every arrangement equally likely.

Show solution

Ignore the repeats. Rolling a face you have already seen changes nothing, so we only care about the order in which the six faces show up for the first time. That order is an arrangement of $1, 2, \dots, 6$ , and by symmetry every one of the $6! = 720$ arrangements is equally likely.

What we need. All odd faces before any even face means the first three faces to appear are $1, 3, 5$ in some order, and the last three are $2, 4, 6$ in some order.

Count. The odd faces can be ordered in $3! = 6$ ways and the even faces in $3! = 6$ ways, so there are $36$ good arrangements:

P = \frac{3! \cdot 3!}{6!} = \frac{36}{720} = \frac{1}{20}.

Another way to see it: the three odd faces have to take the first three of six places. Choosing which $3$ of the $6$ places are odd can be done in $\binom{6}{3} = 20$ equally likely ways, and only one of them is "the first three". So $P = 1/20$ .

Answer: $0.05000$ .

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