I guess you probably have lost interest in this by now, but my bed is currently occupied by an agressive animal, and I have nothing else to do...
There are 24 possible combinations you can roll a D6 and a D4, and in 18 of those combinations the D6 will be higher. That means the chance of success is 75%.
There are 192 combinations you can roll a D6, D4 and D8. The D8 will be higher than the D6 and the D6 will be higher than the D4 in 88 of those I think. The Reason is that in one case all 8 results on a D8 will result in sucess, in two cases 7 results will be sucess etc. A table will look like this:
D4|D6 6_5_4_3_2_1
1 ____ 3+4+5+6+7+8
2 ____ 3+4+5+6+7
3 ____ 3+4+5+6
4 ____ 3+4+5
The number in the table is the number of successful results on the D8 for a certain result on the D6 and D4. It all adds up to 88, wich means 88/192, or about 46% of the results are successful so far.
When it comes to the D12 I am not sure how to make this visible, but basically you can imagine a diagram like the above, only with 3 dimensions instead of two. The number of total combinations is by now 2304 so its gets hard to do calculation without having an effective way of doing it, but after laborating a bit i think the number of successful combinations are 499. That would mean about 22% chance.
Im afraid the D20 makes too many combinations for using the previous method though.
I am not good at statistics, but im quite sure I've got the first two right. The third I think is correct too, but I wouldnt bet anything expensive on it. To check I made a small program on my calculator that tries to solve the problem empirically. (Dont have any compiler of any kind on this computer
) The result was as follows:
first check: 78%
first and second: 45%
first, second and third: 27%
all checks: 18%
So the result seems resonable. The calculator is very slow though, so this is only about 700 test, wich is not very much.