[long] [images] More Dice Formulas (Help!)

[LazarusLong42]

Well, being both a programmer and a math-person (though amateur at both), I do understand both outlooks. The math-guy in me wants an elegant solution--i.e. one that does not require brute force to count everything.

Also, there remains the matter of efficiency. Getting 100d10, drop lowest, cannot take more than about 100 iterations with the equation I posted above, but would take 10^100--that is, one googol--iterations via a brute force method.

Dudes, you're making me ever happier and happier for my choice of using point buy.

See, this is the sort of comment which may well be non-productive, but only because I spent two minutes laughing at it. Very funny.

 

[frisbeet]

Attn: LazarusLong42

(bump)

Hey LL42,

Remind me again what those funny brackets over the summation symbols and in the "numerator" of the last combination factor mean, for your formula for the # ways of rolling an outcome R for N d-sided dice, keeping k and dropping 1.

Thanks a ton.

One of these days, probably after I pay the Government, I'm gonna work on this. Perhaps sooner. Not that anyone should hold their breath, of course.

 

[orsal]

(bump)

Hey LL42,

Remind me again what those funny brackets over the summation symbols and in the "numerator" of the last combination factor mean, for your formula for the # ways of rolling an outcome R for N d-sided dice, keeping k and dropping 1.

Thanks a ton.

One of these days, probably after I pay the Government, I'm gonna work on this. Perhaps sooner. Not that anyone should hold their breath, of course.

That's the "floor" operator -- round down to the nearest integer (but always down).

 

[ph0rk]

Dudes, you're making me ever happier and happier for my choice of using point buy.

A math major friend of mine and I were bored in the coffee shop one day and decided to figure out the difference between 6 rolld os 3d6 and 6 rolls, rerolling any 1's.

(tougher than it sounds)

As niether of us had any probability texts with us, we ended up going brute force.

I no longer have the results before me, but as you may know, the average score for 3d6 is 10.5

The average for 3d6 drop any 1's was around ~11.728

(by brute force we determined the distribution of final scores using the reroll 1's method and calculated average score based on this distribution)

In essence, it turned them into 5 sided dice, or very nearly so. (and the average score for 3 dice numbered 2-6 is 12)

where it gets interesting is when you stack multiple methods for eliminating low scores, like 4d6 reroll 1's drop the lowest, or 5d6, etc. Then your average approaches 13.5! (and will continue to raise by 1.5, give or take, for each effective number you eliminate)

you can overshoot the level of scores 32 point buy will give you quite easily if you combine 2 or more methods.

 

[CRGreathouse]

3d6 reroll 1s is the same as 3d5+3, so its average is (exactly) 12.

 

[ph0rk]

3d6 reroll 1s is the same as 3d5+3, so its average is (exactly) 12.

blargh, i meant 4d6 drop the lowest was 11.728 - not enough coffee today

 

[wocky]

Some time ago I did a program that calculates probabilities for the 4d6 drop-lowest method. The following are the results:

01 .000000
02 .000000
03 .000772
04 .003086
05 .007716
06 .016204
07 .029321
08 .047840
09 .070216
10 .094136
11 .114198
12 .128858
13 .132716
14 .123457
15 .101080
16 .072531
17 .041667
18 .016204

E(x) (mean) : 12.244599
V(x) (variance) : 8.104523
SD (standard deviation): 2.846844

Hope they are OK... I make no promises...

 

[WizarDru]

Brain......HURTING.


Seriously, guys. I mean....wow.

 

[Frostmarrow]

3d6 reroll 1s is the same as 3d5+3, so its average is (exactly) 12.

Now this even I can understand. Made it worthwhile to stay in the thread.