# Probability with 8 number of dice

Eight identical dice are rolled simultaneously. In how many possible outcomes each of the six numbers appears at least once?

I got result as $\left(\dfrac 16\right)^8.$ And I think I missed some part. Could you please help me?

• It depends on what you choose as the set of outcomes. I would use the $6^8$ equally likely outcomes, by writing ID numbers on the dice. But the person asking may have a different notion of what the possible outcomes are. – André Nicolas Mar 4 '15 at 21:11
• @AndréNicolas I really do not get your answer completely.I think you want to say that my answer is correct dont you? – Nihad Azimli Mar 4 '15 at 21:14
• If we use the $6^8$ equally likely outcomes, your answer is not correct. There are many ways in which one can have all of $1$ to $6$ appearing. But as I mentioned earlier, the question asker may have a different sample space in mind. – André Nicolas Mar 4 '15 at 21:24
• Without an explicit specification of what we mean by possible outcome, I cannot produce an answer. I can produce an answer for the $6^8$ outcomes case, for then the possible distributions are of shape $3,1,1,1,1,1$ and $2,2,1,1,1,1$. But the asker may have different outcomes in mind. – André Nicolas Mar 4 '15 at 21:33

The overall number of possible outcomes is of course $6^8$. We want to subtract off the number of outcomes that have at most five different numbers.

Now let's start with an easier problem: How many outcomes are there with no sixes? That is obviously $5^8$. Similarly there are $5^8$ outcomes with no fives, with no fours, and so forth. So it looks like the answer will be $$6^8 - 6\cdot 5^8$$ But this is not quite right. Consider any roll with no sixes and no fives. We have subtracted that roll off twice -- once because it has no sixes, once because it has no fives. We should have subtracted it only once. So we will add those rolls back in. There are $\frac{6\cdot 5}{2}=15$ ways to lack 2 numbers, and each such pair leads to $4^8$ possible outcomes. So our answer so far is $$6^8 - 6\cdot 5^8 + 15 \cdot 4^8$$ But consider any roll that contains no sixes, no fives, and no fours. We will have subtract this roll three times, and we will have added it three times (once because it has no sixes or fives, once because it has no sixes or fours, once because it has no fives or fours). So this roll thus far has not been accounted for; we have to subtract it. Now there are $\frac{6\cdot 5 \cdot 4}{6}=20$ ways to choose $3$ numbers out of $6$. There are $3^8$ rolls for each combination of $3$ missing numbers, o our answer thus far is $$6^8 - 6\cdot 5^8 + 15 \cdot 4^8 - 20 \cdot 3^8$$ And now we come to rolls with no sixes, fives, fours, or threes. The same sort of reasoning shows that each of these was subtracted twice in total, so we need to add these back. And finally, consider wolls that are just six of the same number. We will have subtracted and added these the same number of times, so we will need to subtract these once more.

The final number of outcomes is $$6^8 - 6\cdot 5^8 + 15 \cdot 4^8 - 20 \cdot 3^8 + 15 \cdot 2^8 -6\cdot 1^8 = 191520$$

And the probability of rolling 8 dice and getting at least one of each number is $$\frac{191520}{6^8} = \frac{665}{5832} \approx 0.114$$

Thus the counterintuitive result is that you are likely to be missing at least one number when you roll 8 dice. In fact, in order to have better than a 50% chance of getting one of each number, you would have to roll 13 dice!

• Of course, the main problem with the question at hand, is that the user asked for the number of combinations, not for the probability of getting them. And as such, the answer depends on whether or not identical combinations are considered the same. For the sake of calculating probability, they are obviously not considered the same. But since the question is not about probability, it remains slightly unclear. I would guess that identical combinations are not considered the same, which makes your answer correct. – barak manos Mar 4 '15 at 21:43

To have all numbers appear at least once, you can have two pairs of the same number or one triplet. The number of ways to get a triplet is $6$ (choose the repeated number) $8 \choose 3$ (choose the positions for the triplet) $5!$ (order the remaining numbers)$=40320$ Can you do the two pairs?