Probability Distribution for Product of two Random Numbers? If I multiply two random numbers from 1 to 10 together I will get an outcome from 1 to 100. Over a large enough sample size the outcomes will tend to cluster closer to 1 than to 100 (only 21 of the 100 possible outcomes are greater than 50, for example).


*

*How would the probability distribution of the outcomes be described?

*Suppose that I have a percentage figure, say 60% in mind; how would I find the number below which I could expect 60% of all outcomes to fall? (For example - 79% of all outcomes fall below 50).
Please keep in mind that I am looking for a solution that is easily programmable (ie not reliant on graphs or tables etc).
 A: 
How would the probability distribution of the outcomes be described?

You would describe it precisely using a probability mass function. This simply lists the probabilities of each outcome individually:
$$\begin{eqnarray*}
\rm{Pr}(X=1) &=& 0.01 \\
\rm{Pr}(X=2) &=& 0.02 \\
\rm{Pr}(X=3) &=& 0.02 \\
\rm{Pr}(X=4) &=& 0.03 \\
\ldots && \\
\rm{Pr}(X=100) &=& 0.01
\end{eqnarray*}$$
The sum of all the probabilities has to add up to one, of course.
$$\sum_{x=1}^{100}\rm{Pr}(X=x) = 1$$
Descriptive statistics can also be used. E.g. a graphical description using a histogram, left, or a boxplot, right.

Or, you can quantitatively summarize the distribution. The median is 24, the mean is 30.25, the standard deviation (a measure of spread) is 23.94. The statistical term used for "Over a large enough sample size the outcomes will tend to cluster closer to 1 than to 100" is that the distribution is right-skewed, or colloquially, has a right tail (this can be expressed quantitatively using skew).

Suppose that I have a percentage figure, say 60% in mind; how would I find the number below which I could expect 60% of all outcomes to fall? (For example - 79% of all outcomes fall below 50).

Mathematically, what you are doing here is using the quantile function, $Q$. This is the inverse of the cumulative distribution function $F$, which is defined as follows:
$$F(x) = \rm{Pr}(X\le x)$$
So, $Q = F^{-1}$, e.g. to find the median $m$, we solve $\rm{Pr}(X\le m) = 0.5$.

Please keep in mind that I am looking for a solution that is easily programmable (ie not reliant on graphs or tables etc).

For this particular problem, I would say the easiest solution is to list all the possible outcomes ($1\times 1, 1\times 2, 1\times 3 \ldots 10\times 10$), and sort the list. Then the $n^{\rm{th}}$ value in the list gives $\rm{Pr}(X\le n/100)$, because all outcomes are equally likely. In a language such as C, C++ or Java you could write a nested for loop to write the outcomes into an array and then use built-in functions to sort the array.
In a language like R, you can make use of built-in vector operations like so. The R code below confirms your assertion that "79% of all outcomes fall below 50".
> x=sort(outer(1:10,1:10))
> x[79]
[1] 49
> x[80]
[1] 50

A: I suppose you mean the random variables to be independent. Then you should simply program a simulation. As an example, following yours, suppose ¤X¤ and ¤Y¤ are two independent random variables, each with a uniform distribution on $\{1,2,3,4,5,6,7,8,9,10\}$. The do this in R:
repl <- 100000
X <- sample(1:10, repl, replace=TRUE)
Y <- sample(1:10, repl, replace=TRUE)
Z <- X*Y
hist(Z)

or you can do any other calculations you want with your sample $Z$
to describe its distribution.
