1. I have three dice. What is the chance that when rolling all three at once, the highest two will equal 10, 11, or 12?
2. Two dice rolled at once, what's the chance of the total of the two being more than 9?
If you're wondering, this represents how often the Black Knights in my Vampire Counts army will successfully run down and destroy a fleeing enemy, assuming the Knights have defeated them in close combat and they autobreak.
I think 2 is 1 over 4. I'm good with math, but on a 10th grade level.
For #2, just count how many dice combinations of the 36 you can get a total of 10, 11 or 12; so you have (3+2+1)/36 = 1/6
For #1, I think you multiply the solution from #2 by 3C2, so you get 1/2. (not quite sure if I'm repeating some results here though)
I threw dice 50 times for each one, and got 50% for the first one, and 20% for the second... so it seems that this banner upgrade will be very worth it. :O
Lets see if you take 3+2+1 and divide it by 36 then times it by pi then divide it by Sine you get
a brotha that has no clue what hes talkin' about.....
2+2=4? >_>
LOL laughs at everyone.
A die has 6 sides to get 10, 11 or 12 with 2 you need a 4, 5 or 6 that is 3/6 sides or 1/2 of the die. Yet if you get two 4's or a 4 and a 5 you won't get it. One die is 50% to get one of these numbers. However with 3 to get 2 dies to = 10, 11 or 12 the chances are a bit less.
6!
----------
3!(6−3)!
0.5^3(1−0.5)^6−3=
6×5×4×3×2
----------------
(3×2)(3×2)
(0.125×0.125)=0.3125
So the % is 31 1/4% ^_^
1. 42
2. 42
3. 42