Question

My question is this: If you roll a 6-sided die 6 times, what is the probability that you'll get k different numbers for k=(1,...,6). I can work out k=1,6 without a problem. The rest is where I'm confused. Is the only way to solve it though "manual" elimination of all the possibilities through inclusion-exclusion principle, or is there something more general and simple?

Answer 1

This is my idea for k=2. Am I making any sense? (I.e. we choose 2 numbers out of 6 and then there's 2^6 possibilities of sequencing them, except for those two where we'd just get on number) \[\frac{ \left(\begin{matrix}2 \\ 6\end{matrix}\right)*(2^6-2) }{ 6^6 }\]

Answer 2

The probabilities of obtaining a different number on each roll are as follows:
Roll 1: P(1) = 1
Roll 2: P(2) = 5/6
Roll 3: P(3) = 4/6
Roll 4: P(4) = 3/6
Roll 5: P(5) = 2/6
Roll 6: P(6) = 1/6
The probability of obtaining a different number on each of 6 rolls is:
P(1) * P(2) * P(3) * P(4) * P(5) * P(6) =
\[\frac{5!}{6^{5}}\]

Answer 3

Thanks, but that would be k=6, which, as I said, is fairly straightforward. It's the others I have a problem with.

Answer 4

the probability of 5 different numbers is found from the following:
\[P(5different)=\frac{6}{6}\times \frac{5}{6}\times \frac{4}{6}\times \frac{3}{6}\times \frac{2}{6}\times \frac{5}{6}\]
where 5/6 is the probability that the number on one roll will be the same as a number on one of the other five rolls.

Answer 5

Would I be then right to assume that
\[P(4)=\frac{ 6 }{ 6 }*\frac{ 5 }{ 6 }*\frac{ 4 }{ 6 }*\frac{ 3 }{ 6 }*(\frac{ 4 }{ 6 })^2\]
Sorry for being a bit slow, sadly, I haven't yet taken any probability course whatsoever and don't have much education in that field.

Answer 6

\[P(4different)=\frac{6}{6}\times \frac{5}{6}\times \frac{4}{6}\times \frac{3}{6}\times \frac{4}{6}\times \frac{1}{6}\]
where 1\6 is the probability that a throw will result in a number that is the same as the identical numbers that have resulted from two of the six throws.

Answer 7

Total Number of Combinations = 6^6 (includes repetitions)
Total Number of Combinations w/o repetition = 6!=6·5·4·3·2·1

The rest of the problem is easy...