Math and the 4D6 Ability Score Rolls.

[Johntodo]

So I have a GM who wanted us to make 'Epic' characters. Very special beings and what not, yet still play by regular class rules.
One way we are special, is that instead of rolling 4D6, dropping the lowest like I have normally seen, he wants us to roll 4D6 and re-roll 1's and 2's (min 12, max 24).
I'm trying to math, but can't quite get the chart right. What I would really like to know, despite how difficult this would really be to figure out, is with 3's, 4's, 5's, and 6's, what is the percentage chance of getting each specific number?
Example, to get a 24, you NEED 4 6's. You have a 25% chance of getting a 6. To get 4 6's, you quarter 25 another 3 times. Rounded up that would be about a .4% chance (I probably didn't do that right..). Whats the chance of getting a 23? 22? 21? etc...
The middle numbers have a much higher chance, due to more combinations being able to reach it.
It's a lot to ask, I know. This is why you need to stay in school :)

P.S. Im (not-so) silently seething in rage at a player who I think used a 'random number generator' and got a 24 and multiple 22's. 5 out of 6 of his stats are above 20.
I happily rolled a 23 and 2 20's in front of the GM with legitimate dice rolling, and i was ecstatic.
That is kind of the purpose of this little side project.. and you know... Didn't something like this kill the cat?

[Johntodo]

I think my problem with not being able to figure this out is that the possible outcomes don't pay attention to order.
If order mattered, there would 256 possibilities (4*4*4*4).
There aren't that many in reality cause it doesn't matter which dice is first or last, and I don't know an equation for that.
SOOO I just wrote out all of the 35 possibilities (I think i got all of them..)
Obviously refer to 1's as 3's, 2's as 4's, etc. This was easier for me.
Now you just take 100% chance, divided by 35 possibilities (~2.86) and multiply that by however many are in each row.
Example - 15. 3 * 2.86 = 8.58% chance to get a 15.

12. 1111
13. 1112
14. 1113 1122
15. 1114 1123 1222
16. 1124 1133 2222 1223
17. 1134 1224 1233 2223
18. 1144 1234 2233 2224 1333
19. 2234 1244 2333 1334
20. 2244 1344 2334 3333
21. 2344 1444 3334
22. 2444 3344
23. 3444
24. 4444

Maybe this is the right way of doing it? I should have finished my math before posting *sigh*...

[Scrapper]

When I've gone with an epic stat roll, it's usually been 4d8 drop the lowest.

[Jerry Wright 307]

Try this site: Anydice.com.

Use the following formula: output 4d4 + 8

Set it for "Table" and "Normal"

Click "Calculate"

It will give you the odds.

I heartily recommend this site.

EDIT:
this kind of formula is hard to find on the site, so this is how you do 4d6, best 3:

output {1,2,3}@4d6

This means it will add the top three numbers of the 4d6 rolled, and drop the fourth.

If you use the "Roller" setting, it will let you generate numbers based on your formula.

[Matthew Downie]

Johntodo wrote:

I think my problem with not being able to figure this out is that the possible outcomes don't pay attention to order.

If order mattered, there would 256 possibilities (4*4*4*4).
There aren't that many in reality cause it doesn't matter which dice is first or last, and I don't know an equation for that.
SOOO I just wrote out all of the 35 possibilities (I think i got all of them..)
Obviously refer to 1's as 3's, 2's as 4's, etc. This was easier for me.
Now you just take 100% chance, divided by 35 possibilities (~2.86) and multiply that by however many are in each row.
Example - 15. 3 * 2.86 = 8.58% chance to get a 15.

12. 1111
13. 1112
14. 1113 1122
15. 1114 1123 1222
16. 1124 1133 2222 1223
17. 1134 1224 1233 2223
18. 1144 1234 2233 2224 1333
19. 2234 1244 2333 1334
20. 2244 1344 2334 3333
21. 2344 1444 3334
22. 2444 3344
23. 3444
24. 4444

Maybe this is the right way of doing it? I should have finished my math before posting *sigh*...

It's actually more complicated than that. A roll like '1234' is more common than '1111' because there are something like 24 different ways you can get 1234 (like 2134, 3124, 4321) but only one way to get a 1111.

To put it another way, if I want to get a 1234, it doesn't matter what the first dice is, the second dice I have a three in four chance of still being in with a shot... (Overall a 24/256 chance) But if I want a 1111 I have to get a 1 every single time (a 1/256 chance).

[Antimony]

The probabilities are:

12: 0.4%
13: 1.6%
14: 3.9%
15: 7.8%
16: 12.1%
17: 15.6%
18: 17.2%
19: 15.6%
20: 12.1%
21: 7.8%
22: 3.9%
23: 1.6%
24: 0.4%

[Johntodo]

Antimony wrote:

The probabilities are:

12: 0.4%
13: 1.6%
14: 3.9%
15: 7.8%
16: 12.1%
17: 15.6%
18: 17.2%
19: 15.6%
20: 12.1%
21: 7.8%
22: 3.9%
23: 1.6%
24: 0.4%

Thanks for the replies guys, and the probabilities Antimony!

Also, really cool site Jerry Wright 307, glad I know about that now.
Good point Matthew Downie, I knew it had to be more complicated than how I figured it.