1,000,000/20 = 50,000
50,000/20 = 2500
2500/20 = 125
125/5 = 25
25/5 = 5
5/5 = 1

so, you need 6 dice
the first three need to roll 1
the second three need to roll from 1 to 5

Roll a 19 or 20 six times in a row.

>>98185879
>the second three need to roll from 1 to 5
from 1 to 4, pardon me
because from 1 to 4 is one fifth of the d20.

Rolled 3, 6, 1, 5, 5, 7 = 27 (6d10)

>>98185879
>so, you need 6 dice
yeah, 6 D10s

>>98185889
Fair enough. That requires all 6 roll 10's too. Or at least all the same.

>>98185856
First, roll a 20. Then roll three more 20s. Then roll 8 on a D8. Thats pretty close. Otherwise, if you need exactly a million, roll a 19+ six times.

Otherwise, get a critical hit through disadvantage, and deal 34 damage on 4d6+d2.

>>98185993
I think 19+ 6 times in a row is sufficient.

So 5 skill checks with a 19+ difficulty followed by a to-hit of 19+. I'm going to instantly kill whatever is on the receiving end in heroic panache. But it has to take exactly 6 19+ rolls.

Rolled 7, 16, 5, 2, 20, 20 = 70 (6d20)

>inb4 failure

>>98186032
Do note that the chance that anyone would ever roll these exact numbers in this sequence is 1 in 64,000,000.

Roll a 20 then 1 to 4, three times in a row.

Now some maths since you asked how.

Chance to roll 1 on d10 is 1/10

The chance to roll a 1 then a 1 is would be the same as rolling two dice at the same time and both of them show 1. That's equivalent to rolling 11 on d100. Intuitively that's a 1/100 chance.

The chance to roll three 1 in a row is the same as rolling three dice at the same time all showing 1, which is a result of exactly 111 on d1000. There's only one way to do this so chances are 1 in 1000.

Keep going until six 1 in a row is equivalent to 111 111 on d1 000 000.

From another viewpoint, we want a way to generate 1 million possible outcomes. Happily, the number 1 million can easily be decomposed into factors like this:
1 000 000 = 10 * 10 * 10 * 10 * 10 * 10
If we had a die that had 10 faces on it we could roll it to generate 10 outcomes. If we did that six times in order we could get 1 000 000 distinct outcomes. We choose any one of those possible outcomes as the target, that gives us 1 in a million chance.

The easiest way to do that is to choose a six digit number consisting of that digit repeated, then roll a single d10 six times in a row to get that digit. You could also roll six dice, each one in its own box, with box one for the first digit, etc. You could use six distinguishable dice such as being different colours and define before hand the order in which the colours will be read.

You don't have to use a repeated digit, you could say the target number is 123 456, that's just as likely as 111 111. Some people would use 000 000 as that is to d1 000 000 the same as 00 is to d100.

Using equations, the chance to roll 1 on d10 is
P(1) = 1/10 = 0.1
To roll two 1 in a row, multiply the probabilities: P(1,1) = P(1) * P(1) = 0.1 * 0.1 = 0.01
And so on to
P(1,1,1,1,1,1) = P(1) * ... * P(1) = 0.1 * ... * 0.1 = 0.000 001 = P(1,2,3,4,5,6)

The particular number doesn't matter.

>cont

>>98189637
The chance to roll 123 456 is the same as the chance to roll 111 111 which is why I tagged that bit on the end.

The anon who replied roll 19 or 20 six times chose the simple route of using 10 * 10 * 10 * 10 * 10 * 10 = 1 000 000. I chose the much more fun route of 20 * 5 * 20 * 5 * 20 * 5 = 1 000 000.

That anon is interested in choosing numbers to have a 1 in 10 chance, for example he chose 19 and 20 on the d20 you requested.
P(19 or 20 on d20) = 2/20 = 1/10
Roll 19 or 20 six times in a row, or roll 19 or 20 or six distinguishable dice in whose order has been predetermined.

So I needed to use a d20 roll to get a 1 in 20 result, and a d20 roll to get a 1 in 5 result.

This is possible because 20 is a factor of 20, and 5 is also a factor of 20. Both a 1 in 20 result and a 1 in 5 result can be generated directly using d20.

To get a 1 in 20 result on d20, roll any single chose value, for example 1.
P(1) = 1/20

To get a 1 in 5 result, partition the d20 into five equal groups for example {1,2,3,4}, {5,6,7,8}, {9,10,11,12}, {13,14,15,16}, {17,18,19,20} and then choose any one of them.
P(1 or 2 or 3 or 4) = 4/20 = 1/5

You don't have to partition the die so that the values are consecutive. A partition such as {3, 7, 11, 19} is valid and will give you the same chance to roll.

You don't have to use the same partition each time, so long as you use four distinct values and establish the target before knowing the value of the roll (or to complicate things needlessly, you need to do it at least independently, such as by using another random number generator).

Because 1/20 * 1/5 = 1/100 = 1/10 * 1/10 you can substitute any 1/20 * 1/5 pair with a 1/10 * 1/10 pair.

Ultimately, to get a 1 in 1 000 000 result from d20 you need to roll in order

one of {3, 6, 13, 17}
then 7
then either 4 or 13
then 9 or 12
then one of {2, 16, 14, 19}
then end with an 8 because it's Discworld.
Those are all randomly generated, none are my favourite Discworld novels modulo 20.

>>98185856
>What combination of rolls can lead exactly to a million to 1 chance? Specifically, a D20
use it as d10, roll exactly one chosen result on 6d10