Any Math Geeks out there that like to mess with Dice averages?

[Chacal]

Up to mistakes I've made so far:

the exact solution to the problem

> average of 4d6, drop lowest <

is:

15869/1296

That is

12.24459877...

Thanks !

My general formula is a little bit complicated, at least now. I'm trying to simplify it so it can be posted.

There's no problem with a complicated formula, if you explain a bit the reasoning behind it.

Could you post it and enlighten us ?

By the way, knowing an exact answer (i.e. an analytic one) is more fun than having run a computer program 10.000 times... [/B]

Agreed !
Plus it helps to determine which parameters will have which effects, instead of testing them all by programming.


Chacal

 

[Christian]

I think the reason for which the 25 point buy is lower than 4d6 drop lowest in terms of total modifiers is that by using point buy you have an inherent (not too small) advantage in being able to choose exactly your stats.

Depends how you count it. For the record: 4d6 drop the lowest gives an average of ~12.25. So, the total of the six stats will average ~73.5. Point buy gives a base of eight in each stat (a total of 48 for the six), and (up to 14) costs one building point per stat point. So, barring munchkins who just have to buy up some high stats , you'll have 48+25 = 73 total stat points on 25-point buy. If you weight the average rolls by their point-buy cost, you'll get a higher number in the rolling method, though.

My old 486 is still grinding away at an exact solution to the original question (hey-I have to do real work on my primary machine!), but a ~5,000,000 sample run in Excel gave an average of 12.91 with the 'seven rolls, drop the lowest' method. This is approximately equivalent to 29-point buy.

BTW, I use this system too. It only slightly increases the average and the chance of getting really high rolls, but it significantly decreases the chances of really low rolls, virtually eliminating the incidence of characters with two or more scores under 8.

 

[Elric]

I also used the 4d6 drop the lowest, roll 7 times and keep the highest six method. Of course, one player rolled an 88 (at least he didn't roll an 18!) and the rest rolled 80-82. Then the last person rolled very badly by comparison (like 70-72) so I let him reroll and he ended up in the 79-80 range.

Nathan- a generalized formula will probably be very, very complex. I think that I figured out the exact formula for the average when you roll N S-sided dice and keep only the highest number rolled. However, it involves Bernoulli numbers and it was very hard to derive even the simplest case (roll 2, keep 1) by hand.

 

[hong]

My old 486 is still grinding away at an exact solution to the original question (hey-I have to do real work on my primary machine!), but a ~5,000,000 sample run in Excel gave an average of 12.91 with the 'seven rolls, drop the lowest' method. This is approximately equivalent to 29-point buy.

Good heavens, there's no need to use that huge a sample. By comparison, the 60,000 trials I did to get my answer of 12.24 to the original question had a standard error of 0.011. This effectively means I'm within +/- 0.02 of the true result.


Hong "this is why I went to uni" Ooi

 

[tarchon]

There is an compact way to write the formula as an iterative sum if you neglect the "hopeless character" rule, but you're still going to need a computer to calculate it. In the original problem, we wanted to deal with the standard PHB rules, so you pretty much have to do that by some kind of direct simulation method. For this problem, computers are fast enough now that you can sort the entire event space without having to resort to Monte Carlo though. There are only 2^28 outcomes if you treat each 4d6-drop-lowest as a single event (the PMF for it is easy to calculate), and 30 million iterations of a couple comparisons and float multiplications isn't that big of a deal in most compiled languages. The other nice thing about simulation code is that you can easily introduce variations into it to see how they affect the outcome, though that holds for either Monte Carlo or a full census of the event space.

 

[The Sigil]

Ugh.

The "average of 4d6, drop the lowest" is easy. Laughably so.

There are 6^4 permutations, since each die can throw a 1-6 and you're throwing 4 dice.

That means there are 1296 permutations. With a little work in MS Excel, you can quickly count the number of permutations that yield each 3-18 result.

3 -> 1
4 -> 4
5 -> 10
6 -> 21
7 -> 38
8 -> 62
9 -> 91
10 -> 122
11 -> 148
12 -> 167
13 -> 172
14 -> 160
15 -> 131
16 -> 94
17 -> 54
18 -> 21

To get the average, you simply sum R*N/T for all R where R is the roll (3-18), N is the number of permutations that yield that roll, and T is the total number of permutations.

This yields 15869/1296 or 12.24459877 (to 10 significant digits)

However, this does not answer the question as to "what is the mathematical average if I make this roll 7 times and drop the lowest of those 7?" That one's MUCH tougher because there are 16^7 possible permutations... (16 different possible answers from 3-18 and 7 throws gives 16^7) or 2^28 (over a quarter of a billion) AND the odds of coming up with each given permutation are weighted thanks to the fact that it's not equally likely that a given die throw will result in a 3 as an 18. IOW, it gets really ugly really fast. For instance, the odds of throwing a 3-3-3-3-3-3-3 are 1 in 2^28 times 1 in 1296 or 1 in 2^32*3^4 or about one in 348 billion.

That's right, essentially you have to count up 348 billion permutations and figure out how they work to yield a result. Hence, it's MUCH faster to use a computer and have it "average stuff."

Simply put, 348 billion total permutations sucks.

I'm trying to find a mathematical shortcut, but have been unsuccessful thus far.

--The Sigil

 

[CRGreathouse]

Up to mistakes I've made so far:

the exact solution to the problem

> average of 4d6, drop lowest <

is:

15869/1296

I'm impressed. This is exactly correct! Usually, people on the boards aren't that good with probability...

 

[Christian]

Good heavens, there's no need to use that huge a sample. By comparison, the 60,000 trials I did to get my answer of 12.24 to the original question had a standard error of 0.011. This effectively means I'm within +/- 0.02 of the true result.

But you see, I was trying to replace an exact analytical answer. If you want to be picky, I determined the average to be 12.91152 +/- .00005 . And it only took a few minutes for Excel to pump that out while I was doing other things with that same PC (the Pentium 4, not the 486), so what the hey.

 

[Christian]

Simply put, 348 billion total permutations sucks.

I'm trying to find a mathematical shortcut, but have been unsuccessful thus far.

I've got my 486 doing the former, calculating the weighted total of each of the possible permutations. I think I broke it, though ... the hard drive has stopped whirring and is now making a whimpering noise instead.

Good luck with the shortcut.

 

[Storminator]

Full descriptive stats are a bitch here, but you can still get some info.

Frex, what's the chance of getting at least one 18? Count up all ways to NOT get an 18, which is (1-.0162)^7=89%, and subtract that from 100% In other words, nearly 11%

What's the chance of getting a 3? That's the odds of rolling TWO 3s, (since you can drop one) =(1/1296)^2 <= 1/1,000,000

What's the odds of having a low stat of say 7? For that you need one 7, and one score 7 or lower. So odds of a 7 = 38/1296, odds of one score 7 or lower = 74/1296, odds of 5 scores 7 or higher (1260/1296)^5 multiply it all together and get ~0.15%

PS