obediah said:
janjetina said:
To roll 18, you need to roll 666X, where X is any number from 1 to 6 (6 possibilities).
Any out of 4 dice can be the 'X' one, so there are 4*6=24 outcomes.
You've counted 6666 4 times.
To roll 17, you need to roll 665x. Dice 5 and dice X can be any two out of four dice, so the number of outcomes is (4 2) * 6 = 36
x can not be 6, so you need (4 2) * 5 = 30.
And the answer should be 54. I'm almost but not quite desperate enough to avoid work to spend the morning figuring this out.
To roll 16, you need to roll 664x or 655x. Following from the last case, you get (4 2) * 6 from both possibilities, so the subtotal is 2 * (4 2) * 6 = 72.
This should be 94.
EDIT: The take home message is that it's easier to write a simulation than to do the math.
It's even more complicated than that.
As you said, to get 18, you need 666x, out of 4*6 possibilities, 6666 is counted 4 times, so there are 4*6-3 = 21 favorable outcomes.
17 is more complicated. In 665x, There is a total of 4*3*6 possibilities (order matters, so it's not (4 2) * 6, but 4*3*6, but e.g. each 6656 is counted 3 times, with 4 places to choose for a '5', so you have to subtract 4*2=8 outcomes, each 5566 is counted twice, with (4 2) places to put fives in, so (4 2)*1 = 6 has to be subtracted for the total of:
4*3*6 - (4 1) * 2 - (4 2) * 1 = 72-8-6=58 outcomes.
For 16, you have 664x and 655x, each essentially the same as 665x, so you can get a 16 in 58*2=116 ways.
I wrote a computer program and checked the results - they are correct.
The problem is, many 655x, 665x, 666x and 664x outcomes are overlapping as well, so to get a number of >=16 outcomes you have to subtract the number of overlappings of (655x, 665x), (655x,666x), then add the number of cases where three out of four overlap.
In short, just use a computer program looping through the possibilities, as suggested.