It's a Wand! It's a Crossbow Bolt! It's a Floor Wax!

[Stalker0]

ninja, batman, star trek

 

[maggot]

How about five dice and all of them d10's? So wands will be measured in terms of number of one's you can roll before the wand expires. When built, the wand will expire on its 5th one and be worth 100% of full cost. After a single one is rolled, the wand is worth 80% and will expire after 4 more ones. Etc.

This keeps the costs linear, and involves no more bookkeeping than keeping track of which die you are on in the d20-d12-d8-d6-d4 system.

 

[Anax]

I have to say that I find the idea of a 1/20 chance of 96 charges to be a bit freaky... But this is certainly an interesting system, overall.

Spatz: There’s no way to argue that my math as wrong, either. :> The probability of rolling on the d20 a+1 times, the d12 b+1 times, the d8 c+1 times, the d6 d+1 times, and the d4 e+1 times is:

p(a,b,c,d,e) = (1/20)(19/20)^a(1/12)(11/12)^b(1/8)(7/8)^c(1/6)(5/6)^d(1/4)(3/4)^e[/i]

Hence, the probability of getting exactly n charges is q(n) = ∑ _{a,b,c,d,e ← ℕ0, a+b+c+d+e+5 = n} p(a,b,c,d,e). The cumulative probability of getting at least m rolls is ∑ _{n ← [1,m]} q(n). And if we limit ourselves to being interested only in the cumulative probabilities for m ≤ 100, we can always discard partial results for x+y+2 > 100, since ∀ _{x,y,z ← [0,99]} (x+y+2 ≥ 100) ⇒ (x+y+z+3 > 100).

I think the argument I saw going on before was because you can’t sum medians meaningfully. That is, you can’t take the median value for 1d20 (13) and combine it with the median value for 1d12 (7) in order to get the median value for 1d20+1d12 (26).

In this case, the individual probabilities of each number of charges are being computed exactly, and then the median is found by looking at the cumulative probability distribution.

Anyway, it was a fun little analysis.

P.S. I wasn’t going to post this, because it’s pretty obvious. But then I decided that I wanted to play around with vB and unicode and see how well notation comes out. I wonder how much of this is only readable to me because I have extra math fonts installed.

 

[Spatzimaus]

You HAD to bring the Greek letters into it, didn't you? (And yes, it's readable.) Heh, just kidding.

maggot: The 5d10 system WOULD actually be easier to work with, in most ways. It's easy to price. It's easy to keep track of. It's got a mean of 50, and the median would still be pretty high. But, to me, there's a certain style to the ever-decreasing dice. You'd really FEEL like the wand is getting weaker and weaker, beyond just the tick mark on the character sheet. Once you're down to using a d4, it'd really be an "any minute now..." sort of thing.

But it doesn't really matter, we could come up with a bunch of progressions that'd work. For instance, what if you use 8 d6s instead (which'd REALLY have a low chance of <20 charges) Or, 4 d12s?

 

[Agback]

we randomly determine whether a wand's charges have expired (using a "charge die"). A fully charged wand uses a d20, and on a roll of a 1, it degrades to a d12, a d10, and finally a d8. (As a side question, can anyone with any level of sophistication in math tell me how close I will get to 50 using this method).

Congratulations! The expected number of spells a wand will fire under this scheme is exactly 50. (It is a sum of four geometric distributions with p=1/20, 1/12, 1/10, 1/8, mean for the geometric distribution is 1/p, and the mean of the sum is the sum of the means.)

The standard deviation is a bit of a brute, though: 26.2. Over 30% of your wands will produce either fewer than 24 spells or more than 76. (Variance of the geometric distribution is (1-p)/p^2, the variance of the sum is the sum of the variances, and standard deviation is the square root of variance.)

 

[Agback]

Be wary of 'averages' here.

Lets make a deal. Everyday you give me a dollar, and I will roll a D(million) die. On any roll except a 1, I keep the dollar. On a roll of a 1, I will give you 1,000,000 dollars.
On average, we will both break even.

Who wants to take me up on this?

Do you really have the million? Are you prepared to put it in escrow? Because you are offering better odds than any lottery, and millions of people buy tickets in million-dollar lotteries every week.

 

[Agback]

d6 = price of three scrolls of the spell. (Yes, at this point I'd start treating it as a semi-scroll)
d4 = price of two scrolls of the spell

Those last two are good buys!

 

[Agback]

How about five dice and all of them d10's?

The mean is still 50.

Standard deviation is reduced to 21.2, which is a better statistic.

Better yet, you could make it 10 d5. (Std. dev. =14.14)

Or for a really accurate system, which still requires only value to be recorded, and save the bother of all that die-rolling, why not 50 d1? (Std. dev. = 0)

 

[Agback]

I think the argument I saw going on before was because you can’t sum medians meaningfully. That is, you can’t take the median value for 1d20 (13) and combine it with the median value for 1d12 (7) in order to get the median value for 1d20+1d12 (26).

That's right. Because of the Central Limit Theorem, as you add in more random variables, the distribution of the sum of their values approaches Gaussian Normal, which is a symmetrical distribution with median equal to its mean. The sum of five geometric distributions isn't particularly close to symmetrical, but it is closer than any of its components.

 

[babomb]

Ok. I get how you are calculating those numbers. It's not something my University math professors would approve of; I bet that the theory of medians has all sorts of subtleties that a linear interpolation just doesn't capture. But I understand where you are coming from.

Maybe. Arguably, there's no meaning to saying the median is a result that is impossible to occur. I looked on the 'Net only briefly, and didn't see a better approach, but I don't doubt that there is one. Still, I believe the results are a reasonable approximation. A real mathematician might get mad at me, but sometimes close enough is close enough.

You are right. I showed why it does not follow simply adding them, and babomb has been able to use the brute force method to get a more accurate median.

Brute force isn't pretty, but it always works, at least.

I think the progessive wand degridation is quite cool, but I would suggest five rounds of d10, which would give a similar mean, but a bit less of a chance of very few or very many charges.

A 5d10 wand would have a median of 46.3745. The most probable outcomes are 40 and 41, occuring 2.05887% of the time each. There is over a 20% chance for the wand to be in the range [37,46], so it is less of a gamble.

Have we all agreed that D20-D12-D8-D6-D4 would produce near the desired results and provide an elevated roleplaying experience at the table?
I lost track of that in the discussion of probabilities and such

Yes, I think so. It has a slightly higher median than the d20-d12-d10-d8 variety (44.7765) and a slightly smaller standard deviation, but still pretty high. You'll get pretty much the same results with either.

While the PCs can be sure there's a reasonable amount of charges most of the time (over 91% have at least 23 uses), there's enough variability to create some anxiety/excitement, and that builds as the wand dies down (pun intended). The most probable result is 36, with 1.96235%. And, given a sufficient number of wands, the mean charges per wand will be about 50. So, yes, I believe such wands would have the desired effect.