Possible Duplicate:
Expected time to roll all 1 through 6 on a die
Probability of picking all elements in a set

Im pretty new to the stackexchange, and posted this is statistics, and then discovered this site, and thought it was much more appropriate, so here I go again:

It is fairly easy to figure out what is the average rolls it would take to roll all faces of a die [1 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1 = 14.7], but that got me thinking of a seemingly more complicated problem.

Say you roll a die 1-5 times, the is the odds of ALL faces showing, is obviously 0. If you roll a die 6 times, the odds of all faces showing can easily be calculated like so:

1 * (5/6) * (4/6) * (3/6) * (2/6) * (1/6) = .0154 or 1.54%

Now is where I get stuck. How to do 7, or more times, and calculate it with n.

Any tips is helpful!


marked as duplicate by joriki, Sasha, Henning Makholm, user940, t.b. Mar 22 '12 at 11:12

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  • 2
    $\begingroup$ Also asked on stats.SE simultaneously $\endgroup$ – Dilip Sarwate Mar 22 '12 at 1:03

The probability of not rolling a 1 in $n$ rolls is $(5/6)^n$, similarly for not rolling a $2,\ldots,6$. Now, $6(5/6)^n$ would be the probability that we are not rolling a $1,\ldots,6$, but we would be double counting the rolls where we do not roll both a 1 or a 2. The probability of not rolling two specified numbers in $n$ rolls is $(4/6)^n$ and there are $\binom{6}{2}$ pairs of numbers. But if we subtract these out we undercount the rolls that avoid three numbers. This generalizes to the inclusion-exclusion principle, giving us the probability of missing any number as $$\binom{6}{1}(5/6)^n-\binom{6}{2}(4/6)^n+\binom{6}{3}(3/6)^n-\binom{6}{4}(2/6)^n+\binom{6}{5}(1/6)^n$$ as the probability of missing at least one number in $n$ rolls. The probability of rolling all of them is just 1 minus this probability.

  • $\begingroup$ Correct me if I am wrong, but \binom 6 1 = (6! / 1!(6-1)!) $\endgroup$ – Cameron Aziz Mar 22 '12 at 1:39
  • $\begingroup$ With that, I somehow come up with 1.197 for n = 6 [(720/120)(5/6)^n] - [(720/48)(4/6)^n] - [(720/18)(3/6)^n] - [(720/8)(2/6)^n] - [(720/5)(1/6)^n] = 1.197 $\endgroup$ – Cameron Aziz Mar 22 '12 at 1:42
  • $\begingroup$ $\binom{6}{1}=6$ yes. And for $n=6$ I get 0.9845679. Some of your other binomials are incorrect $\endgroup$ – deinst Mar 22 '12 at 1:53
  • $\begingroup$ got it! (720/120)(5/6)^n] - [(720/48)(4/6)^n] - [(720/36)(3/6)^n] - [(720/48)(2/6)^n] - [(720/120)(1/6)^n] eh? $\endgroup$ – Cameron Aziz Mar 22 '12 at 2:15
  • $\begingroup$ @CameronAziz Looks correct. $\endgroup$ – deinst Mar 22 '12 at 2:27

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