# 8 dice, 4 identical odd numbers, 4 identical even numbers

8 dice are rolled, what is the probability that you roll 4 identical even numbers and 4 identical odd numbers.

First Odd/Even Roll 6C1 = 6

Next 4 Rolls All Odd/Even 8C4 = 70

Opposite Odd/Even Roll 3C1 = 3

Last 4 Rolls Opposite Odd/Even 4C4 = 1

All Possible Rolls 6^8 = 1679616

$P = (6*70*3*1)/1679616 = 0.00075$

Have I done this problem correctly?

• Imagine that the dice are thrown in sequence. Pick which four locations in the sequence will be used for the even numbers: $\binom{8}{4}$ ways. Pick which even number it specificially is: $3$ ways. Now pick which odd number it specifically is for the others: $3$ ways. This gives a total of $70\cdot 3\cdot 3$ possible sequences of dice rolls with four identical even numbers and four identical odd numbers, making your answer off by a factor of two. – JMoravitz Feb 13 '18 at 18:36
• If you want to modify your own approach, begin by picking the first number in the sequence: $6$ ways. Then pick which three more of the remaining seven locations match the first: $\binom{7}{3}=35$ ways (we do not do 8C4 here since the first term in the sequence has already been decided by something else). Then pick the value for the remaining dice in $3$ ways, giving $6\cdot 35\cdot 3$, same answer as I gave in previous comment. – JMoravitz Feb 13 '18 at 18:39
• Thank you very much! Both ways you described make sense to me and that was very helpful – Byron Feb 13 '18 at 18:43

To add to JMoravitz's comments, a way of thinking of this is that there are $3$ choices for the even face and $3$ choices for the odd face.
Then, out of the $8!$ ways to order the $8$ dice, since the even and odd faces are indistinguishable, we have $$\frac{8!}{4!4!}$$ different orderings. Hence there are $$3\cdot3\cdot\frac{8!}{4!4!} = 630$$ ways for this to happen, and so a probability of $$\frac{630}{6^8} = \frac{35}{93312} \approx 0.00037508573.$$