Need a Programmer for Number-crunching

[resistor]

That's pretty obviously wrong. The probability of the highest stat being 3 is the same thing as the probability of all stats being 3s. And that's just (1/216)^6. Your value is 6 times as high.

You're right there. The three case is special in that the second probability is not 3-or-less, but equal-to-three. In this case, there's no choice of ordering so the multiplication by 6 is unneeded.

I stand corrected on that one, but I'll still pretty confident the others are correct.

 

[resistor]

I stand corrected on that one, but I'll still pretty confident the others are correct.

Actually, I take that back. The others are wrong too. They don't even sum to 1.0! Teach me to try to do probability late at night.

I suspect the problem with my approach is the multiplication by six. Because the second probability term is X-or-less, it may or may not be reorderable with the first term. So the six should really be some kind of weighted factor to represent the possible reorderings. Sounds infeasible offhand.

At any rate, I'm too tired to think about this more tonight. Brownie points to anyone whom gets the correct solution.

 

[resistor]

What you've got here is the answer for "what is the average number of 18s a character has." The problem is that some characters have multiple 18s, so the expected number of 18s is higher than the percentage of characters with 18s.

Ah, I see what you're saying now. Took a while to sink in properly. I suspect your approach will lead to the correct final solution.

Alright, REALLY going to bed now.

 

[arscott]

resistor, if you do the problem right, then order doesn't matter at all.

your chance of getting a 3 on a single roll is 1/216.
your chance of getting a 3 on every roll is (1/216)^6

your chance of getting a 3 or 4 on a single roll is 4/216
your chance of getting a 3 or 4 on every roll is (4/216)^6

And here's the important part:
The chance of your highest score being a 4 is simply the chance of getting all 3s and 4s minus the chance of getting all 3s. (That is, you've got at least one 4).
That's just (4/216)^6 - (1/216)^6.
And conveniently, we can factor out the denominator to get (4^6-1^6)/(216^6).

The chance of your highest score being a 5 is the chance of getting all 3s, 4s, and 5s [(10/216)^6] minus the chance of getting all 3s and 4s. And so on. Eventually you get:

 X  Prob. that X is the highest Ability
 3  (1^6)/(216)^6
 4  (4^6 - 1^6)/(216)^6
 5  (10^6 - 4^6)/(216)^6
 6  (20^6 - 10^6)/(216)^6
 7  (35^6 - 20^6)/(216)^6
 8  (56^6 - 35^6)/(216)^6
 9  (81^6 - 56^6)/(216)^6
10  (108^6 - 81^6)/(216)^6
11  (135^6 - 108^6)/(216)^6
12  (160^6 - 135^6)/(216)^6
13  (181^6 - 160^6)/(216)^6
14  (196^6 - 181^6)/(216)^6
15  (206^6 - 196^6)/(216)^6
16  (212^6 - 206^6)/(216)^6
17  (215^6 - 212^6)/(216)^6
18  (216^6 - 215^6)/(216)^6

Do the arithemetic, and you get the exact same numbers that Babomb came up with.

 

[Elric]

Hrm? No.

What I computed is the odds of having exactly one X and five X-or-below, multiplied by six because the X could be any of the six stats.

To put it another way, your odds of having a highest stat of 18 is the odds of rolling at least one 18 in six rolls of 3d6. Each roll is independent, with a chance on 1/216 on each. So the probability of rolling at least one 18 is 6/216.

Nope. You can see that this is wrong by considering what this would imply if a character had 217 ability scores. The odds of getting an 18 on each ability score is 1/216. So the odds of getting an 18 on one of all 217 ability scores is 217* (1/216)= over 100%! That's clearly not right. When you're looking for the chance to roll at least one 18, you can't simply add probabilities in this way.

 

[Flatus Maximus]

Each ability score has 6^3 possible rolled outcomes, and with 6 ability scores, that's 6^18, or 101,559,956,668,416 different possible sets of ability scores that can be rolled.

Minor nit-pick: There are 16^6 = 16,777,216 possible sets of ability scores using the "old skool" 3d6 method. (Each of the six scores takes one of the 16 values from {3, 4, ..., 18}.) There are, on the other hand, 6^18 possible outcomes on the eighteen d6 rolls before summing each 3d6. (You end up rolling eighteen d6, summing the first three to get the first score, summing the second three to get the second score, etc.)

 

[genshou]

Minor nit-pick: There are 16^6 = 16,777,216 possible sets of ability scores using the "old skool" 3d6 method. (Each of the six scores takes one of the 16 values from {3, 4, ..., 18}.) There are, on the other hand, 6^18 possible outcomes on the eighteen d6 rolls before summing each 3d6. (You end up rolling eighteen d6, summing the first three to get the first score, summing the second three to get the second score, etc.)

Yes, but you can't generate meaningful results out of 16,777,216. A significant part of the statistics here is that there is a bell curve distribution exists for the ability scores. Thus, exactly what dice were rolled to add up to 10 matters as much as the total of 10 itself.

Anyway, all this math talk isn't good for me at 3am. Time for bed.

 

[Flatus Maximus]

Yes, but you can't generate meaningful results out of 16,777,216. A significant part of the statistics here is that there is a bell curve distribution exists for the ability scores. Thus, exactly what dice were rolled to add up to 10 matters as much as the total of 10 itself.

Anyway, all this math talk isn't good for me at 3am. Time for bed.

Agreed, a minor nit-pick the mathematician in me just couldn't pass up.