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

[Thanee]

Average of 4d6 drop lowest should be about 12.

Bye
Thanee

 

[hong]

Computers being what they are these days, it's faster and easier to do 10,000 trials and find the average (about 12.24), than to work out the analytical solution.

[edit: you'll notice that Nathan is still working on the answer. ]

Hong "trust me, I'm a statistician" Ooi

 

[Nathan]

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...

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

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

 

[hong]

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

This is why I'm a statistician, not a mathematician.

 

[Storminator]

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

I was advocating using the computer to count all permutations of dice rolls, not roll a bunch of dice and post the results.

Personally, I find actually working out such solutions tedious. As soon as you do one someone asks "what happens if you roll 8 stats and keep 6?" Then you start over.

PS

 

[Zappo]

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. It allows you to plan exactly what will happen at 4th and 8th level when you'll get those free points, as well as requirements for feats and prestige classes, in the most efficient way, while with the random generation you risk being screwed and you'll likely waste some points. It's the same reason for which the fixed HP optional rule in the DMG suggests giving half max hit dice instead of average (eg, 6 for a barb instead of 6.5). Safety is an advantage in itself.

 

[Nathan]

I was advocating using the computer to count all permutations of dice rolls, not roll a bunch of dice and post the results.

Of course, everything else doesn't make much sense. However, if you have a large number of dice with many sides (i.e. many possible dice rolls), you have to stick to the random procedure of rolling a bunch of dice.

Personally, I find actually working out such solutions tedious. As soon as you do one someone asks "what happens if you roll 8 stats and keep 6?" Then you start over.

PS

It just takes a few seconds to calculate the answer with my formula:

24.88681341...

 

[Storminator]

It just takes a few seconds to calculate the answer with my formula:

24.88681341...

Hey now! that's not what I meant! I meant you do 4d6 drop low and generate 8 stats, then keep the six best stats. Now what's the average stat?

PS

 

[Nathan]

Okay, I haven't read your message carefully enough. Sorry.

I can't give you quickly a solution to the problem you've posted because my formula at the moment is only usable for the question of n s-sided dice where n-k are dropped.

Perhaps I will spend some time on answering your question and post the formula.

 

[Undead Pete]

Ain't it wonderful being geeky??