Ok math geeks I need help

[drothgery]

I'll usually just write a quick program to roll the dice a few thousand times and map out the distribution; I took the probability and stats in college to learn the 'right' way to get the correct answer, but Monte Carlo-style estimations are a lot easier to do when your day job is writing web server code ...

 

[tzor]

Dice should be easy because they are independent. (The roll of one die doesn’t impact the roll of the other.) Given that the general way to figure the odds of A or B is P(A)+P(B)-P(A and B) (the later is to eliminate the overlap). The odds of A and B is P(A) * P(B).

Now let’s assume A & B are the same. Let’s just call P(A) P for now. If you have two dice the odds are 2P-P^2. What happens when you want to add another P? (2P-P^2)+P-(2P-P^2)*P) = (3P-(P^2+P))-(2P^2-P^3))

OK this is getting confusing. So let’s use a little spreadsheet magic.
Assume that P is the original percentage.
P’ = 2P-P^2
P” = P+P’ – (P*P’)
P’” = P+P” – (P*P”)

Rolling “under 4” is rolling a 1-2-3. Using this and going to 4 dice I get 75.99%
Rolling “under-6” is rolling a 1-2-3-4-5. Using this and going 2 dice I get 75.000%

So there you have it. 4d10 under 4 is a tad better than 2d10 under 6.

 

[Storminator]

I just make out charts in excel and count out the number of desired responses. Now i feel like a math idiot.

No job is too big for brute force and ignorance!

PS