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I was looking at this question about "Expected time to roll all 1 through 6 on a die" where it was established that the average number of times needed to accomplish this would be 14.7

I understand that that is the average amount of time, but I don't understand how to expand that since the logic doesn't utilize things like combinations or permutations. How would you determine this variant?

"What is the chance that I roll every side on a six-sided dice at least once if I roll it 10 times?"

I'm not certain how to approach this, but I feel like I'm just missing something simple. Should I be using permutations? Or another approach?

RobPratt
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AwkwardBowman
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  • See Probability distribution in the coupon collector's problem – jlammy Jan 19 '21 at 22:50
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    You're not missing anything simple; this is a harder problem than it might look. The answer can be expressed in terms of Stirling numbers of the second kind, as explained in jlammy's link: https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind – Qiaochu Yuan Jan 19 '21 at 22:51
  • Linking to a general answer in terms of Stirling numbers is well-meaning, but I'm not convinced that it's helpful. – Misha Lavrov Jan 19 '21 at 23:08
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    @Misha: why? The argument you gave is just a special case of a general formula for the Stirling numbers. I mean to emphasize that the answer is not, say, just some binomial coefficient, so the OP doesn’t need to be worried that they don’t know how to solve it immediately. – Qiaochu Yuan Jan 20 '21 at 00:30
  • Between the links provided and the example below, I was able to create a thing to help me calculate it. I believe I understand it and was able to apply it to a larger example (which was the final goal). It's still a bit beyond my normal comprehension, but I can follow the steps for now at least. Thanks to both of you! – AwkwardBowman Jan 20 '21 at 00:33
  • @QiaochuYuan I guess that's fair. But what I mean is that saying "the answer can be expressed in terms of Stirling numbers" is not interesting when the Stirling numbers are just defined in terms of this problem; the hard part is computing them. – Misha Lavrov Jan 20 '21 at 00:37

1 Answers1

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For $i=1, \dots, 6$, let $A_i$ be the event "in the first $10$ rolls, we never roll $i$". Then $\Pr[A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5 \cup A_6]$ can be expanded out using the inclusion-exclusion principle as:

  • the sum of $\Pr[A_1]$ through $\Pr[A_6]$, each of which is $(\frac56)^{10}$;
  • minus the sum of all $\binom 62 = 15$ probabilities $\Pr[A_i \cap A_j]$, each of which is $(\frac46)^{10}$;
  • plus the sum of all $\binom 63 = 20$ probabilities $\Pr[A_i \cap A_j \cap A_k]$, each of which is $(\frac36)^{10}$;
  • minus the sum of all $15$ probabilities $\Pr[A_i \cap A_j \cap A_k \cap A_l]$, each of which is $(\frac26)^{10}$;
  • plus the sum of all $6$ probabilities $\Pr[A_i \cap A_j \cap A_k \cap A_l \cap A_m]$, each of which is $(\frac16)^{10}$: at this point we're down to cases where we roll only a single number.
  • minus the probabilitiy of $\Pr[A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5 \cap A_6]$, but that's $0$: at least one value has to be rolled.

This gives us a value of $$ 6 \left(\frac56\right)^{10} - 15 \left(\frac46\right)^{10} + 20 \left(\frac36\right)^{10} - 15 \left(\frac26\right)^{10} + 6 \left(\frac16\right)^{10} = \frac{101923}{139968} $$ for $\Pr[A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5 \cup A_6]$, which is the probability that we don't see all six outcomes. So the probability we do see all six outcomes is $1 - \frac{101923}{139968} = \frac{38045}{139968} \approx 0.272$.

It's a complete coincidence that this probability is within $0.006\%$ relative error of $\frac{e}{10}$.

Misha Lavrov
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