Can we make "charge die" core, please?

[WhatGravitas]

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.

That's the charm of the rule! It makes magic random, without the usual "Oh, I killed myself!"-backlash effect. While I like the rule very much, I'd rather see the near-death of charged items in 4E. Keep some, where the charges are a feature of it, so a "charge die" would just be another kind of item - but get rid of the spell batteries. But I guess the "Wizard Implements" suggest exactly that, so we can put our charge die wands back, whenever we want (which would make me very happy).

Cheers, LT.

 

[kenmarable]

No matter the odds, somewhere there will be the unlucky player with a 4 charge wand.

 

[Lackhand]

I suspect pretty heavily that a 4e wand and a 3e wand are divergent concepts. I don't think that they're spell batteries at all; the evidence indicates to me that they're more like polearms, modifying the specifics (reach, critical multiplier) or special maneuvers (bonus to disarm, drop to avoid trip) of attacks which still originate with the caster.

In fact, other than potions, I'm not sure what would stay in that would be charged. But I'm sure some Wondrous Items will still be charged, because any random rules mechanic can get put there.

 

[Nifft]

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.

Here's a trivial Python program to do the simulation:

[sblock]

#!/usr/bin/env python

import random

def trial():
    i = 0

    while 1:
	i += 1
	if random.randint(1,20) == 1: break

    while 1:
	i += 1
	if random.randint(1,12) == 1: break

    while 1:
	i += 1
	if random.randint(1,8) == 1: break

    while 1:
	i += 1
	if random.randint(1,6) == 1: break

    while 1:
	i += 1
	if random.randint(1,4) == 1: break
    
    return i

A = 0
trials = 10000
for x in xrange(trials):
   A += trial()
print A / float(trials)

[/sblock]

The average value is indeed 50. So my math (in my previous post above) must be off.

Cheers, -- N

 

[BartD]

The average value is indeed 50. So my math (in my previous post above) must be off.

The calcluations are probably fine but when you add up medians you don't get the mean (expected) number of charges.

Anyway, the math you need is actually much simpler: On a d6, the expected number of rolls to roll a 6 is 6, just like the expected number of rolls to roll a 20 on a d20 is 20. It's like taking 20.
So the expected number of rolls to roll a one on d20, d12, d8, d6, d4 is then 20 + 12 + 8 + 6 + 4 = 50.
(Not the best explanation but I hope helpful)

By the way, I *like* this variant

 

[comrade raoul]

Pretty sure that was me, actually! Though the original post might've died with a server at some point or another, and I could be misremembering.

It's really cool to hear someone used rules I came up with.

Perhaps a better progression--and I suspect a statistically equivalent one--is the following: d20-->d12-->d8-->d6-->d4. I like ending with a d4, where every use of the wand stands a really large risk of being the last one.