In this case, you have an exercise in permutations.
Let us look at how many permuations there are (a permutation is where rolling 1 then 2 then 3 then 4 is different from 1 then 2 then 4 then 3 - if they're considered the same, that's a combination).
You have 20^4 permuations (each die can come up in 20 different ways and there are four dice), or 160,000 permutations.
How many of those permutations contain at least two 1's? Easy.
1-1-X-X has 400 possible permutations (the two X dice can come up in 20 different ways - including more 1's - and there are two of them, so it's 20^2).
1-X-1-X has another 400 (800 total)
1-X-X-1 has another 400 (1200 total)
X-1-1-X has another 400 (1600 total)
X-1-X-1 has another 400 (2000 total)
X-X-1-1 has another 400 (2400 total).
Note that we have already included combinations with 3 and/or 4 ones in the above total. You would have X coming up in 19 different ways (instead of 20) if you wanted EXACTLY 2 rolls of 1 (instead of AT LEAST 2 rolls of 1) since in that case you would want only rolls of 2 through 20. In that case you would have instead of 6*(20^2) the total of 6*(19^2) or 2166 total possible permutations.
Our final fraction for AT LEAST two 1's is then 2400/160,000 or 0.015 (corresponding to 1.5%).
Our final fraction for EXACTLY two 1's is 2166/160,00 or 0.0135375 (which is about 1.35%).
--The Sigil
