If seven balanced dice are rolled, what is the probability that each of the six different numbers will appear at least once?
My solution is $\frac{6 \cdot 7!}{6^7}$.
Is this correct? How would I implement a solution using multinomial coefficients?
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2 Answers
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4
There are $6^7$ sequences of $7$ tosses, all equally likely.
We now count the sequences in which all the numbers appear. The number that appears twice can be chosen in $6$ ways. For every such choice, where it appears can be chosen in $\binom{7}{2}$ ways. And the rest of the numbers can be put in the empty slots in $5!$ ways. So our probability is $$\frac{(6)\binom{7}{2}(5!)}{6^7}.$$
answered Dec 24, 2012 at 5:36
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$\begingroup$ What is wrong with assigning the seven slots $7!$ ways and multiplying it by 6 to represent a number that can be 6 different values? $\endgroup$ Dec 24, 2012 at 5:47
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4$\begingroup$ It double counts, for example, the sequence $1123456$. You might want to apply the same reasoning to a $2$-sided die (coin). Your procedure will give a probability greater than $1$. $\endgroup$ Dec 24, 2012 at 5:50
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$\begingroup$ What if the question asks for the probability of every possible number will appear? $\endgroup$– abuchayDec 16, 2015 at 20:12
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$\begingroup$ For this situation, that is another wording for the same question. $\endgroup$ Dec 16, 2015 at 20:45
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First the all tossing outcomes is $6^{7}$
Then we consider choose 6 tosses from 7 tosses, there are $\left(\begin{array}{c}7\\6\end{array}\right)$ ways, then assign them outcome value 1, 2, 3, 4, 5, 6 in order satisfying the conditions at least appear once. there are 6! ways. Then for the remaining last toss we can choose any value, there are 6 ways. Mutilplying them together result in $\left(\begin{array}{c}7\\6\end{array}\right)\times6!\times6$. However it's not the final result!!!
Consider the situation 1234561, in my ways of counting, '1' has been repeatedly counting 2! times. Therefore we have to divide total answer by 2!.Hence there are $\frac{\left(\begin{array}{c} 7\\ 6 \end{array}\right)\times6!\times6}{2!}$ ways.
So our probability is $\frac{\left(\begin{array}{c} 7\\ 6 \end{array}\right)\times6!\times6}{2!\times6^{7}}$
