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Message: <blockquote data-quote="Spatzimaus" data-source="post: 2633401" data-attributes="member: 3051">Per his numbers, which I have no reason to doubt, adding a d6 and a d4 increases the median to 39.5.  But remember, the MEAN would fall at 60, so that 39.5 would be a set of a lot of 20s and 30s, mixed with a small number of 100+ charge versions.If you really want the MEDIAN to fall around 50, I'd suggest adding a d30 on the top end, if you have one.  Its median would be somewhere around 20.  It's too bad there isn't a d16, because the 20-16-12-10-8-6-4 would work even better.  I suppose you could have a d16 by doing 2d4 and only failing if both roll 1's, but that would just be a bit awkward.He gave you that already; since these are independent events, the median number to go from d12 to the next die will always be 7.97 rolls, regardless of what else is in the progression, so just add them up.  It takes a median of 13.5 to go from d20 to d12, 7.97 to go d12 to d10, and so on.  It'd take a median of 20 rolls to go from d12 to d10 to d8 to dead; if you're currently at the d12 stage, you should expect the wand to last you another 20 activations.  (Adding the d6 and d4 extends it by another 6.2.)There are a few other ways you could do it.  Instead of decreasing die sizes, how about increasing thresholds?  Like this:Each wand starts with 10 "charges".  When you use a new wand, you roll a d20.  If you roll a 1, it uses a charge, and the threshold increases by 1.  So once you've used a charge, the next charge expends on a roll of 2 or higher, then 3, and so on; as you use up the wand, it becomes easier and easier to fail the next check (but you stay with the d20).  Once you fail on a threshold-10 roll (costing your tenth charge), the wand is expended.  This'd have a slightly longer lifespan than what you had previously (mean 58, median ~40), but it's far less susceptible to a string of bad rolls.If you use your d20-d12-d10-d8-dead system, it's possible that you'd only get 4 charges out of a brand-new wand.  Even adding the 6 and 4 would still allow a wand to fail after 6 uses, even if the odds are horrendous.  But the system I mention guarantees a minimum of 10 uses.This'd also allow you to customize things.  A well-made wand might use a d20 and have 10 charges, while a cheaper one might only have 5 charges, or use a d12.And Borlon: I'm not him, but I got the same numbers using the following logic in a computer.  (Use whatever language you want)  chance = 1.0 / (die size)  A[0] = 0.0  for N = 1,100    A[N] = A[N-1] + (chance)*(1.0-chance)^(N-1)  endfor(1.0-chance)^(N-1) is the probability of making it to roll number N, so A[N] is the probability that you would have finished on or before roll N.  Then, just find where the array A crosses 0.5.  That's your median.  Technically, since these are discrete events, the median should be a whole number, but you can interpolate between the two nearest values to get a decimal.</blockquote>

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