Statistically, is rolling a six sided die $n$ times equal to choosing a random number between $n$ and $6n$? Statistically, is rolling a six-sided die $n$ times equal to choosing a random number between $n$ and $6n$?
Thank you
 A: The possible numbers are $n$ through $6n,$ not $1$ through $6n.$
Does "random" mean uniformly distributed? Not in the usage of probabilists, but many people, including many mathematicians, sometimes use the term that way.
Notice that if $n=2$ and $X$ is the sum of the two resulting numbers, then $\Pr(X=2) = \Pr(X=12) = 1/36,$ and $\Pr(X=3) = \Pr(X=11) = 2/36,$ and $\Pr(X=4) = \Pr(X=10) = 3/36,$ and so on, so these are not uniformly distributed.
For the sum of the numbers shown by $n$ dice, the mean is $\frac 7 2 n$ and the variance is $\sigma^2/n$ where $n$ is the result from a single die, and that variance is much smaller than the variance of a uniform distribution on the set $\{n,n+1,n+2,\ldots,6n\}.$ So it's nowhere near uniformly distributed. You would only rarely get a result as small as $n$ or as large as $6n,$ and far more often get a number that's near $\frac 7 2 n.$
A: If you sum the number you get then the answer is no, since


*

*there is no way you can achieve $1$ (the least possible value is $2$)

*$P(X+Y = 2) = \frac{1}{36}$ while $P(X+Y = 7) = \frac{1}{6}$ as you can easily check, where $X$ is a random variable representing the result of one dice roll

