A Fair Die Is Rolled N Times. What Is The Probability That At Least 1 Of The 6 Values Never Appears?

Answer :

I think via inclusion/exclusion the probability that at least one of the six values never appears after n rolls of the die would be:

$$p(n) = {6 \choose 1}({5 \over 6})^n - {6 \choose 2}({4 \over 6})^n + {6 \choose 3}({3 \over 6})^n - {6 \choose 4}({2 \over 6})^n + {6 \choose 5}({1 \over 6})^n$$

To understand, first just consider the probability of a 1 never showing up:

$$({5 \over 6})^n$$

Easy enough. Now what are the chances of either a 1 never showing up OR a 2 never showing up. To first order it's just twice the above, but by simply doubling the above, you've double-counted the events where neither a 1 nor a 2 show up, so you have to subtract that off to correct the double counting:

$$2({5 \over 6})^n - ({4 \over 6})^n$$

The final answer I gave above is just an extension of this where you first add the probability associated with the 6 ways of not rolling any particular number, then subtract off the probability of the ${6 \choose 2}$ ways of not rolling any two particular numbers, then add back in the probability of the ${6 \choose 3}$ ways of not rolling any three particular numbers, etc.

I made an A in probabilities about 25 years ago, but I haven't thought about this stuff much since, so there is a non-zero probability I'm totally wrong, but the results seem at least reasonable to me. I think it curious and nifty that the formula works for all $n \ge 1$. You pick an $n$ with $1 \le n \le 5$ and you get 1, but as soon as $n \ge 6$ the probability (appropriately) starts falling off:

$$p(1) = 1.00000000000000000000$$

$$p(2) = 1.00000000000000000000$$

$$p(3) = 1.00000000000000000000$$

$$p(4) = 1.00000000000000000000$$

$$p(5) = 1.00000000000000000000$$

$$p(6) = 0.98456790123456790136$$

$$p(7) = 0.94598765432098765444$$

$$p(8) = 0.88597393689986282585$$

$$...$$

$$p(100) = 0.00000007244804079771$$

Matt