2

I am trying to work out the probability of seeing all 6 numbers on a fair die at least once after n throws, where n > 6.

So I found a related question: Probability of rolling a dice 8 times before all numbers are shown. and a part of the provided answer seems to work for n = 7, but the problem I have is I don't know how to generalize this to work with all n > 6.

Would be nice if someone could explain the general approach to such a problem.

Community
  • 1
Peter
  • 123
  • Welcome to math stack exchange! This problem is known as the coupon-collector-problem. It has been asked multiple times here. As far as I remember, you need stirling numbers (I do not remember whether of first or of second kind). – Peter Jul 26 '16 at 08:12
  • You can also use the inclusion-exclusion-method. To determine the probability that numbers are missing calculate the probability that one fixed number is missing, that two fixed numbers are missing and so on and use the inclusion-exclusion-formula ($p_1-p_2+p_3-p_4+p_5-p_6$ ) – Peter Jul 26 '16 at 08:15

2 Answers2

2

First, use inclusion/exclusion principle in order to count the number of combinations:

  • Include the number of combinations with at most $\color\red{6}$ values showing: $\binom{6}{\color\red{6}}\cdot\color\red{6}^n$
  • Exclude the number of combinations with at most $\color\red{5}$ values showing: $\binom{6}{\color\red{5}}\cdot\color\red{5}^n$
  • Include the number of combinations with at most $\color\red{4}$ values showing: $\binom{6}{\color\red{4}}\cdot\color\red{4}^n$
  • Exclude the number of combinations with at most $\color\red{3}$ values showing: $\binom{6}{\color\red{3}}\cdot\color\red{3}^n$
  • Include the number of combinations with at most $\color\red{2}$ values showing: $\binom{6}{\color\red{2}}\cdot\color\red{2}^n$
  • Exclude the number of combinations with at most $\color\red{1}$ values showing: $\binom{6}{\color\red{1}}\cdot\color\red{1}^n$

Then, divide by the total number of combinations in order to compute the probability:

$$\frac{\sum\limits_{k=0}^{6-1}(-1)^k\cdot\binom{6}{6-k}\cdot(6-k)^n}{6^n}$$

barak manos
  • 43,109
0

Roll the die $n$ times and for $i\in\{1,\dots,6\}$ let $E_i$ denote the event that number $i$ does not show up.

Then you are looking for: $$1-\Pr(E_1\cup E_2\cup E_3\cup E_4\cup E_5\cup E_6)$$

Now attack this using inclusion/exclusion and symmetry.

drhab
  • 151,093