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We have $n$, 6-face fair dice. At each time $n$ dice are thrown independently. I want to calculate the probability of number of times we should throw dices before having at least one 6 from each of the dice.
We know that for one die to have one 6, the probability of $k$ failures obeys a Geometric Distribution with $p=1/6$ and expectation of 6.

How should I extend this to $n$ dices?

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1 Answer 1

up vote 2 down vote accepted

Label the dice $1$ to $n$. Let $X_1$ be the number of throws of the first die until we get a $6$. Define random variables $X_2, X_3,\dots, X_n$ analogously.

Let $Y=\max(X_1,X_2,\dots,X_n)$. Your question asks for an expression for $\Pr(Y=y)$.

It is easier to find $\Pr(Y\le y)$. That will do it, because $$\Pr(Y=y)=\Pr(Y\le y)-\Pr(Y\le y-1).\tag{$1$}$$

Now we go after $\Pr(Y\le y)$. We have $Y\le y$ iff we have at least one $6$ on all the dice by time $y$.

The probability that there was at least one $6$ on die $1$ by time $y$ is $1$ minus the probability of no $6$, so it is
$$1-\left(\frac{5}{6}\right)^y.$$ For $n$ dice, take the $n$-th power. We get $$\Pr(Y\le y)= \left(1-\left(\frac{5}{6}\right)^y\right)^n.$$ Now use $(1)$.

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Thank you @Andre. This seems correct. I tried using Negative Binomial and calculating P(X1 + X2 + ... + Xn > N). Actually your method is much more simpler! –  HamedKhan Dec 5 '12 at 18:43
    
I think the sum of my $X_i$ does not lead to a solution of the problem. –  André Nicolas Dec 5 '12 at 19:06

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