Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Also asked on stats.SE simultaneously – 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.

share|cite|improve this answer
Correct me if I am wrong, but \binom 6 1 = (6! / 1!(6-1)!) – Cameron Aziz Mar 22 '12 at 1:39
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 – Cameron Aziz Mar 22 '12 at 1:42
$\binom{6}{1}=6$ yes. And for $n=6$ I get 0.9845679. Some of your other binomials are incorrect – deinst Mar 22 '12 at 1:53
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? – Cameron Aziz Mar 22 '12 at 2:15
@CameronAziz Looks correct. – deinst Mar 22 '12 at 2:27

Not the answer you're looking for? Browse other questions tagged or ask your own question.