D&D 5E 5e Champion - crit odds on multiple attacks?

[Jediking]

Hi all, I've tried to look up the stats but have not been able to. Any mathletes here that can help?
I have a Fighter (Champion) 5/Druid 4, and I switch between using a one-handed weapon (two attacks total) and two light weapons (3 attacks total) depending on if I am facing one tough opponent or a bunch of mooks.

I wanted to know the following crit chances (not looking to compare DPR, just the chance of a crit happening).

-Chance of rolling one critical for two attacks:
-Chance of rolling two criticals for two attacks:

-Chance of rolling one critical for three attacks:
-Chance of rolling two criticals for three attacks:
-Chance of rolling three criticals for three attacks:

Thanks for any help!

 

[Sacrosanct]

Maybe not exact numbers, but for a good idea (critical on a 19 or 20), off of the top of my head

-Chance of rolling one critical for two attacks: 1 in 5 (1 in 10 for normal crit of needing a natural 20)
-Chance of rolling two criticals for two attacks: 1 in 100 (1 in 400 for normal crits)

-Chance of rolling one critical for three attacks: 3 in 10 ( 3 in 20 for normal crits)
-Chance of rolling two criticals for three attacks: my brain is totally locking up on this one. Bleh.
-Chance of rolling three criticals for three attacks: 1 in 1000 (1 in 8000 for normal crits needing a natural 20)

*Edit added normal crits needing a natural 20 for comparison

 

[Stoutstien]

Hi all, I've tried to look up the stats but have not been able to. Any mathletes here that can help?
I have a Fighter (Champion) 5/Druid 4, and I switch between using a one-handed weapon (two attacks total) and two light weapons (3 attacks total) depending on if I am facing one tough opponent or a bunch of mooks.

I wanted to know the following crit chances (not looking to compare DPR, just the chance of a crit happening).

-Chance of rolling one critical for two attacks:
-Chance of rolling two criticals for two attacks:

-Chance of rolling one critical for three attacks:
-Chance of rolling two criticals for three attacks:
-Chance of rolling three criticals for three attacks:

Thanks for any help!

How often do you have advantage?

 

[DND_Reborn]

-Chance of rolling one critical for two attacks: 1 in 5 (close enough, actually 0.18)
-Chance of rolling two criticals for two attacks: 1 in 100 (0.1 x 0.1)

-Chance of rolling one critical for three attacks: about 1 in 4 (0.243)
-Chance of rolling two criticals for three attacks: about 1 in 37 (0.0247)
-Chance of rolling three criticals for three attacks: 1 in 1000 (0.1 x 0.1 x 0.1)

As @Stoutstien asks, advantage can change that a lot. Also, those are the discrete observations, if you want the probability of getting 1 OR more criticals, that is another question.

 

[NotAYakk]

Two attacks:

(.9+.1x)^2 = .81+.18x+.01x^2


Which means 18% one crit 1% two crits.

Three attacks:
(.9+.1x)^3 = .729 + .243x + .027x^2+.001x^3

Which means 72.9% 0 crits, 24.3% 1 crit, 2.7% 2 crits, 0.1% 3 crits.

With advantage:
(.81+.19x)^2 and (.81+.19x)^3

For 2 attacks:
65.61% 0
30.785% 1
3.61% 2

For 3:
53.1441% 0
37.3977% 1
8.7723% 2
0.6859% 3

Apologies if arithmetic errors. My in-head 2 digits cubing isn't reliable.

This is a trick called counting polynomials.

You can do crazy things with them, and do fun computation with bog standard polynomial multiplication.

For a fun thing you can do, take the derivative of those counting polynomials, and evaluate at 1; you get the average. Higher order moments can also be calculated (using nth derivative), but aren't quite as simple.

Spoiler

Or, another fun one, f(x) = x( 1+...+x^5 * f(x))/6 for an exploding d6.

Infinite polynomials can be represented via 1/(1-y) = 1+y+y^2+...

So the exploding case is 1/6(x+x^2+...+x^5)+ 1/36(x^7+x^8+...+x^11) + ...=(1/6+1/6^2*x^6+1/6^3*x^12+...)(x+...+x^5)
= 1/6 ( 1/(1-1/6 x^6) ) (x+...+x^5)

Which you can then take the derivative of or calculste its x^kth term or whatever.

 

[Jediking]

Thanks so much everyone!
Appreciate it