Can we make "charge die" core, please?


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Alright, let me see if I can have some fun with this idea. First off, we really aren't using charges anymore, it's more like power levels that randomly drop off. So lets call it that, and label the power levels as such:

d20: Full power
d12: High power
d10: Low power
d8: Minimal power

Rolling a 1 on these die rolls results in a loss of power (but the spell still goes off normally). Rolling a one on minimal power consumes the magic entirely.

Now then, this works for wands, that are based on one casting = one charge. But what about staves that have multiple spells with different charge costs? Just take the any use that uses more than one charge, and reduce the die type by one category for each charge higher.

Going down the list, die types used are: d20, d12, d10, d8, d6, d4, d3, d2

So using a fully powered staff for a 2 charge ability would need a roll of 1d12, and the overall power level would drop on the result of one. For a 3 charge ability, that same staff would use 1d10. A minimally powered staff would use a 1d6 for a two charge ability, or 1d4 for a 3 charge ability. Most staves only use 1-3 charges, but a staff of resurrection can use 5, and at minimal power thats a 1d2, a 50% chance of depleting the staff. If there's anything that uses more charges than that, it could go right to a 100% power loss chance.


Yeah, I think this would make a fine replacement for the charge mechanic in 4e. Of course, staves and wands will have different uses, but I guess you could charge anything.
 

If you're using a costs-X-charges power from a staff currently at dN, why not just roll XdN? Reduce by a level per 1 rolled.
 

Lackhand said:
I'm pretty sure that my thinking *wasn't* incorrect, though.

You will roll a 1 in 14 trials 50% of the time -- (20^14 - 19^14)/(20^14) ~= 0.51233, meaning that roughly 50% of wands will fail after 14 trials (downgrading to the next die size), and roughly 50% will keep going (more trials needed!).
I actually did some calculations on the back of a napkin a few months ago regarding this very question, and came up with this: the chance of rolling any particular number (in this case, 1) on an n-sided die over n trials approaches ~63% as n approaches infinity. Having only basic finite math and calculus, I can't tell you why 63% in particular, but that's the number I got. I didn't bother to actually calculate the limit, to determine whether it's actually asymptotic or not, or figure out the actual value of the limit. ~63% is what came back for a d1000000, and all other arbitrarily large values I tried, so that's good enough for me.

Anyway, if a 1 comes up in 20 trials about 63% of the time, then rolling a 1 in 14 trials 51% of the time seems like a pretty reasonable estimate.

Edit:

I ran the numbers again in Excel, using the formula that the chance of rolling 1 on ndn is equal to 1-((n-1)/1))^n. I got the following:
Code:
die          chance
4            0.684
6            0.665
8            0.656
10          0.651
12          0.648
20          0.642
30          0.638
100        0.634
1000       0.632
10000     0.632
100000   0.632
etc.

edit again: I also used my formula to double check your calculations, and they're all correct.
 
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well, yeah, my math checks out. Sheesh! :p

[offtopic]
I suspect the 63 percent thing is a standard dev on the bellcurve-of-love you've generated with NdN rolls, but am neither certain of this fact nor qualified to make that statement, not having really thought about it at all.
[/off topic]

So, in short, if you use the die size as the charge equivalent, 2/3 of wands will fail by the time you get That Many Rolls.

I, personally, wouldn't use die size as the charge equivalent, though, because in fact, "most" (meaning, more than half, meaning more than 50%) wands will fail before this point by a significant margin.

Figure out what X makes ([die size]^X - [die size -1]^X)/([die size]^X) just greater than .5, and that's the charge equivalency for that die size, as that's the minimum value such that "most" wands will have failed by that roll.

a d2 crosses over in 1 trial (well, obviously : p)
a d4 crosses over between 2 and 3 trials.
a d6 crosses over between 3 and 4 trials.
a d8 crosses over between 5 and 6 trials.
a d10 crosses over between 6 and 7 trials.
a d12 crosses over between 7 and 8 trials.
a d20 crosses over between 13 and 14 trials.
a d30 crosses over between 20 and 21 trials.
a d100 crosses over between 68 and 69 trials.

Thus, use the larger value for each of the above as a die value, I'd suggest.
Therefore, a 50 charge wand (today) should start with a d30 and end with a d8, for 54 uses from "most" wands.
10 charge objects can be approximated with a d10 (13 expected uses);
25 charge objects with a d12 (21 expected uses).

I second the motion that something taking more charges simply require more rolls on whatever die the charged object is already on.
 

Here's an easier way of thinking, to help us figure it out: What are the numbers that lead to your wand NOT burning out? That's all the numbers in the range 2-20, or a 0.95 probability. The chance of still having your wand after ONE use is 95%. :)

Now we can multiply out by the number of uses:
2 tries = .95^2 = 0.9025 = 90.25%
3 tries = .95^3 = 0.8574 = 85.74%
4 tries = ... = 0.8145
5 tries = 0.7738
6 tries = 0.7351
7 tries = 0.6983
...
12 tries = 0.5404
13 tries = 0.5133
14 tries = 0.4877

So your d20 die is expected to last ~14 charges. That's the half life of this phase of the wand's deterioration.

Next we move on to the d12. Here we are happy with 2-12, which are 0.9166 of the available choices. We have only a 0.4985 chance of lasting through 8 uses.

d10 is easy: 90% chance of success each try. We last 7 before hitting 0.4783.

d8 lasts for 6 uses before sinking to 0.4488.

d6 breaks .5 after 4 uses (4 tries = 0.4823).

d4 is expected to last a mere 3 uses (3 tries = 0.421875).

- - -

So, we have 14 + 8 + 7 + 6 + 4 + 3 = roughly 42 expected charges. This is on the high side though, since I'm merely taking the highest integer below the 50% mark.

An actual statistician could no doubt tell you lots more. :)

Cheers, -- N
 


I beleive the original thread got lost in the Crash..

however, here is the text of the rule I use:

[sblock]Instead of having a flat 50 charges, wands are charged by Die..

When you use a wand, roll the die type. On a roll of ‘one’ the die type decrements down the following scale:
20-12-8-6-4

Expended wands cannot be recharged, but they can be re-enchanted

Cost in the Market:

All five charges still in: 100%
d20 gone: 60%
d12 gone: 36%
d8 gone: 20%
d6 gone: 8%
d4 gone: 8%
Expended:

Re-enchanting a wand cost the % of the step you are going to, so d8 wand re-enchanting one step costs 36% of the normal cost.[/sblock]

I didn't save all the stats and analysis we did. I had created an Access program that ran through the iterations the hard way instead of just basing it on statistics. The result was that the mean charge was just over 50, with the oddballs down at 7ish and up to 130ish. There was one instance of a wand > 200.... but I am talking about massive iterative runs before that happened. My 'standard' test was 100 batches of 100 wands, run 20 times :)


And yes, multiple charges use multiple die rolls.
 

Well, I quoted the original rule. The way I actually use it, the charge die goes down to d6 and d4. I didn't mention it because I didn't want to be seen as too soft.
In a recent encounter, one of the enemies was a dual-wand wielder with a total of 3 wands. His wand of Levitation was a d4 charge die and he ruined it in the first round of combat. The players were pissed not to get it.
 

I bet. Sucks to be them! On the bright side, there's no such thing as "using up luck" -- a wand can fail at any time, and that d4 wand could, just possibly, have lasted 20 charges. Or more.

Unlikely, sure. But possible, and just think of the lucky player :)
 

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