Can we make "charge die" core, please?

[howandwhy99]

Wow. This is one of the coolest house rules I've heard in a while. For 3e it makes a lot of sense.

The Flavor in the world would have to follow suit, of course. Some easily sensed means of whether the wand was d20, d12, d10, or d8 would be needed - if the player rolled the die. Otherwise it sounds awesome.

 

[Moon-Lancer]

so, this is a different d20 roll then the umd, and its only purpose is to not roll a one? it sounds kindof neet. I may have to try it sometime. Do you have charts that show equivalent die size to charge amount? Lets say you are running a written module and a bbeg has a wand with x number of charges. what do i start the wand out with?

anyone? is their an easy convert for x number of charges to die size.

 

[Abstraction]

Add up the top number on each die. A fully charged wand has 20+12+10+8 = 50 charges. A D12 wand is equivalent to 12+10+8 = 30. D10 = 18. D8 = 8.

 

[wgreen]

anyone? is their an easy convert for x number of charges to die size.

# of charges     die size
31-50              d20
19-30              d12
9-18               d10
1-8                 d8

That seems about right.

(Or, what Abstraction said above.)

-Will

 

[Lackhand]

Using really fast and loose, only slightly motivated by an actual backing from statistics, I'd say each die size is worth half its number of faces, plus one, uses.

Thus, over 50% of all three-roll runs of a d4 have a 1 in them, while under 50% don't (and this ratio is reversed for two-roll runs).

For the larger die-sizes, the probability of rolling a 1 hovers just under 50% in this average-roll-round-up scheme.

So, I'd say that the table is (before any downgrade: this is just roll-to-fail-forever)
d4 :3
d6 :4
d8 :5
d10 :6
d12 :7
d20 :11
d30 :16
d100 :51

(the bigger the die, the wonkier my eyeballing was)

If you use the downgrade solution, just add the charge-estimates for the bottom of the scale up to the die size you're using together. Some wands will fail well before this. Some will fail well after.

in the terminate-at-d8 world, a 50-charge wand is approximated by a d30 (16+11+7+6+5 = 45; have it drop to a d6 for 49, but my math is approximate enough that the 45 is probably about right as, I've erred towards lower expectations of 1s).

ninja edit!
Heh: Or what they said. The problem is that since the wands are being given repeated trials, the number of rolls before you reasonably expect to see a one _isn't_ the die size, but a function of the die size.

Thus, the chance of rolling a 1 on a d20 is 1/20.
In two rolls, you expect to see a one on either die on 39 of the 400 trials (which is more like 1/10!)
In three rolls, you expect to see a one on any of the dice on 1141/8000 of the trials, which is pretty similar to 1/7.

Basically, it's where you want to draw the lines. One seventh of all d20 wands will downgrade after three uses. Half of them will downgrade after 14 uses. My rule of thumb was very rough.

 

[Abstraction]

Your thinking is incorrect. The chance of rolling a 1 on a d20 is 1 in 20, not 1 in 10. Half the time you will roll a one before you roll twenty times, but half the time you will roll more than twenty times before you get a 1.

 

[Lackhand]

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!).

 

[Lackhand]

This is easiest to show on a d4:
Rolling just once, we get (1), (2), (3), and (4); 1/4 of our wands fail.

Rolling twice, we get (1 followed by 1, 2, 3, 4), (2 followed by 1, 2, 3, 4), (3 followed by 1, 2, 3, 4) and (4 followed by 1, 2, 3, 4).
In this, one sample from each of the series that started with 2, 3, and 4 fails, and anything that rolled a 1 on the first die fails. That's 4+3 = 7 of our 16 possible wands that fail.

Thrice, the possibilities are:
111 121 131 141 211 221 231 241 (and because 2 != 1, just run that last set
112 122 132 142 212 222 232 242 of sixteen twice more for me, swapping
113 123 133 143 213 223 233 243 3 and 4 for 2 in the hundreds place)
114 124 134 144 214 224 234 244

In this, 1/4 fail because they start with 1, 1/4 of each of the remaining sets of 16 fail because they have a 1 in the hundreds place, and 1/4 of each of the 3 remaining sets (within those 16) fail for the ones place.
Thus, 16+3*4+3*3*1 = 16+12+9 = 37 of 64 possible wands have failed, or over half.

Why do wands that have failed keep rolling? Well, they don't, actually. Once they've gone kaput, they've used up their charged.
But they still dominate the probability space: If a wand has a 1/4 chance of failing on the first roll, fully 1/4 wands will fail on that first roll.
Of those that remain, 1/4 will fail on the second, and so that's 1/4+3/16 that fail in two rolls.

And so on. The math works out the same.

How do I make a table?

And, on topic: I use something like this rule myself, but would hate for it to be core!

 

[Moon-Lancer]

thanks Abstraction and wgreen

Im going to see if the players like this and try it out.

 

[Abstraction]

I wish I could remember who had the idea originally, I got it off the ENWorld rules forum about a year ago. Thanks, whoever.