If you were to roll 7 dice what are the chances that exactly

>>226974
Possible rolls=xxxxxxx where x can be [1,2,3,4,5,6]; number of possible rolls = 6*6*6*6*6*6*6=279963

Accepted rolls:

44xxxxx
4x4xxxx
... (4+5+4+3+2 lines omitted)
xxxxx44

Where x can be [1,2,3,5,6]
(6+5+4+3+2+1)*(5*5*5*5*5)=21*3125=65625

Chances of getting exactly two 4s = 65625/279963=23%

>>226986
(6+5+4+3+2+1)*(5*5*5*5*5)=21*3125=65625
Could you explain this line anon?

Two ways to do it:

************* 1 ****************
(like previous answer, but it's 279936, not 279963)
Count number of acceptable ways and divide by total number of ways.

Acceptable: Choose 2 dice from 7 possibilities = 7C2=7!/2!/(7-2)!=21
For each of the two dice with a 4, you have only one possibility (getting a 4). For other dice, you have 5 possibilities (not a 4):
Total possible ways: 21*(1*1*5*5*5*5*5)=65625

Total number of ways is 6*6*6*6*6*6*6=279936.

Result is 65625 / 279936 = 21875 / 93312 = 0.2344 = 23.44%

***************** 2 ****************
With probabilities:
To get 44xxxxx, the odds are:
1/6 * 1/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6

But you have 7C2=21 different ways to pick two dice showing 4.

Result is then 21 * 1/6 * 1/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 = 65625 / 279936 = 21875 / 93312 = 0.2344 = 23.44%

>>227006
When you're looking for the condition that exactly 2 dice out of 7 come up 4, then there are 5 dice left that can come up in any number of ways that do not include 4. The total number of combinations is 5 (6 sides minus 1 because we're excluding 4) raised to the 5th power (5 dice), so 5*5*5*5*5.

The (6+5+4+3+2+1) comes from the fact that there are that many ways for exactly 2 fours to come up in a generic roll ("generic" being numbers replaced by a generic x, as >>226986 did).

So, having two dice, I just tried this out for myself. I only had one six sided one, the other was a twelve, but it still worked for the purposes of my experiment.

In two rolls, I got just one four, even though I didn't expect to get any fours at all. Funny, math is cool and weird.

I had to reroll the first time because it fell off the table, but I didn't look at what number it landed on. Will this skew the results? Should I try it again?