In MapleFlow lowerscript L is not easily available.  It should be accessible with the CRTL+space after typing L.

Or in a list of common symbols (that pallete is not available in Maple Flow)

I had to to use maple to discover how to enter it.

`ℓ produced the correct result and then deleting the quote I could achieve the desired result.

I got the proportional symbol to work once, typing "proportional" + CRTL + Space.  Went for wlak came back and could not get it to work at all.

Does it actually work or am I imagining things?

I have this image on a vase and I want to flatten it using Maple to see how it would look on a flat piece of paper.  How can we go about it. 

The image is attached.

Image1-Greek.zip

So, every year our family picks names for Christmas, but this year seemed odd.  My wife pointed out that one family always just happens to randomly pick another families name.  That is the majority of the time most of one family is always trading with another, the only rule is that each family member can not trade with a member of their own family. 

I was wondering if someone could come up with a way to calculate just how random the choice actually was from the previous year or years using whatever means necessary .. graph theory came to mind.  The draw was done behind closed doors, so my wife questioned how failrly that was done, and so brought me here with the question ...
Was the name choosing really random or was it actually a fixed draw?

In our group there are total of 9 adults and 9 children, but actually the 2 youngest children swap gifts so it's really 9 adults and 7 children who can swap presents.  But I will breakdown the families less the two trading children.

Family A - 2 adults 1 child
Family B - 2 adults 2 children
Family C - 2 adults 2 children
Family D - 2 adults 2 children
Family E - 1 adult

Oh, a child can trade with another child or a parent and similarily a parent can trade with another parent or a child.  

Something to be aware of.  I realized you can't use rand in an equation with other rand's since the calculation affects the other call to it. 

Here is an example:

A simulation of 3 six sided die rolls added together for 50 trials. 

with(Statistics):
a:=[]:
for i from 1 to 50 do
  b:=rand(1..6)()+rand(1..6)()+rand(1..6)():
  a:=[op(a),b]:
end do:
Histogram(a,frequencyscale=absolute,discrete=true)

You'll notice the rand calls have affected the other calls to rand, and has eliminated some possible results.  

[added by moderator, from a Reply by the OP]

The idea was to generate values from 1 to 20 (I just chose 20 as an example) using 6 sided dice, and one that which produces as uniform a distrubution as possible. I appreciate the explanations.  I chose 4 as the smallest number of rolls needed. In the physical world as n approaches large numbers rolling that many becomes increasingly more difficult and time consuming.

So just wondering what the best solution would be for a uniform distribution in a range from 1 to n using only six sided dice?