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

[babomb]

This just didn't sound right to me.... and it has caused me to bang my head against an excel spreadsheet, and some notepaper, to prove it to myself.

You cannot add medians together to get an 'overall' median. It does not work that way. The median of rolling a 1 on a D10 is 6.591 rolls, and rolling a 1 on a D12 is 7.968. But to roll a 1 on a D12, then roll a 1 on a D10 will have a median *higher* than 15.

You are correct. Blast.

 

[babomb]

The problem with my previous approach was that it failed to take into account the combinatorics. For example, there is only one way to get a wand with 4 uses, but 4 ways to get a wand with 5 uses and 10 ways to get one with 6, and so on. The number of possiblities increases rapidly as you go upward, partly offsetting the decreased probability of each given permutation.

Unfortunately, since different dice are involved, there doesn't seem to be a "nice" way to calculate this. So, I wrote a program that goes through each possiblity and calculates the probability. That is, for example, the program calculates the probability for each of the 4 ways to have a wand with 5 uses and adds them together. I have also made the algorithm general, so I can plug in any numbers and kinds of any dice I want and compute the result.

I can't say for certain that it is entirely correct. However, I have verified that the computer counts the same number of possibilities predicted by the theory. In addition, in the degenerate case of exactly one die, it finds the same medians as I posted previously. So I think it's right. The algorithm is much too complicated to post here, but I can upload the source somewhere, if you want.

Apparently, the wands in question have a median of 44.7449 uses. Adding just the d6 brings the median to 50.8943. Adding the d4 as well gives a median of 54.9645.

By the way, the most probable result on the d20-d12-d10-d8 is 36 uses (1.85055%).

 

[Borlon]

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.

 

[Stalker0]

What would be really nice is to see some kind of graph of uses and probability. Basically, some people are saying the means are around 20, some say 50. But regardless, it would be nice to see a spectrum and say in general, what range of charges I'm I going to get out of a wand?

A few 60 charge wands are nice, but if I'm going to get a lot more 25 or 30 charge wands, its good to know.

 

[Halcyon]

The problem with making a complete table is that for so many dice the conditional probabilities are kind of a pain to sort out. For example, you could get 50 charges by having 25 rolls on a d20 until a 1, 11 on a d12, 9 on a d10, and 5 on a d8. The probability of this occurnce would be given by p(25 rolls until a 1 on a d20)*p(11 rolls until a 1 on a d12)*p(9 rolls until a 1 on a d10)*p(5 rolls until a 1 on a d8). However, this is only one possible way to get 50 charges. There are a theoretically huge number of ways that you could get a number of rolls on each of the four dice that add up to 50. This is only for 50 charges. Then the same could be done for 49, 48, etc etc. I could make a series of tables that plot these probabilities, and when combined calculate the total probability of getting a certain number of charges. However, just plotting them roughly indicates that using these four dice in combination in the way suggested a majority of the time will yield less than 50 total charges. However, the mechanic sounds a lot more fun and probably will promote players using charged items more...so why not

 

[Spatzimaus]

The problem with my previous approach was that it failed to take into account the combinatorics.

Apparently, the wands in question have a median of 44.7449 uses. Adding just the d6 brings the median to 50.8943. Adding the d4 as well gives a median of 54.9645..

You're right, we missed the combination factor. Although for d20-d12-d10-d8-d6-d4 I'm getting a median of 58.5, so we might have slightly different algorithms. But the point is, your median moves much closer to the mean.

 

[radferth]

stop the algorithms

This problem does not call for iterations. It is a discrete probability question with a discrete answer. Borlon is right for the probablity that it will take x number of rolls for a 1 to come up (for d20 5% for one roll, 5(100-5)% for two rolls, etc.). If one takes this all the way out, you end up with the median number of rolls needed to get a 1 and just a bit over half the number of sides on the die. This means that for the d20-d12-d10-d8 method, the average would be a bit over 25 (probably 30-35, I'll do the real math at home tonight). However, their would be a decent chance that it could last much longer than average, especially if that 1/20 takes a long time to come up. 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. The original way causes much of the wand's logevity to depend on when that d20 comes up one. Once it does, you probably have only about 15 charges left. Of course, if you want a campaign where wands just work and work for a while, but then degrade quickly once they start to go, use the original degradation progression.

BTW, the title of this thread inspired the following magic item:

Staff of Popiel

Purify Food and Drink (1 charge) [Its the perfect kitchen tool]
Grease (1 charge) [It even waxes the floor]
Blade Barrier (3 charges) [But remains sharp enough to make mounds and mounds of julien fries]

 

[ARandomGod]

You're right, we missed the combination factor. Although for d20-d12-d10-d8-d6-d4 I'm getting a median of 58.5, so we might have slightly different algorithms. But the point is, your median moves much closer to the mean.

Good! I like it, and I don't mind "giving" out a few more (possible) charges, considering that they almost certainly will have some very bad wand experiences.

As said above, calling the wands the "house", the house does indeed have an unlimited supply (there are generally as many wands as you can imagine in any given store). The PC's have a limited supply. If the odds are exactly even, the PC's would eventually lose big. Moving the odds a little in the PC's favor will make the whole thing taste so much better.

Plus this way you get to use ALL of the dice. No die needs to feel left out or underappreciated.

 

[Coredump]

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.

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.

What would be really nice is to see some kind of graph of uses and probability. Basically, some people are saying the means are around 20, some say 50. But regardless, it would be nice to see a spectrum and say in general, what range of charges I'm I going to get out of a wand?

The mean *is* going to be 50, that is a done deal. (Took me by surprise...but kind of cool.) The discussion has now moved to the median, which will help answer the other part of your question.

Babomb says the (new) median is about 44 (This is a bit higher than I would have guessed, but seems reasonable, and he did do the work.) And the mean is 50.
So, in general, you will get half of the wands having 44 or fewer charges, and half having 45 or more.
The vast majority of these will end up between 35 and 80. But to give you more detailed info, will take a lot more work. Also, it is kind of irrelevant. The variance is (likely) large enough that you cannot get a much better prediction for only a few wands.

Good! I like it, and I don't mind "giving" out a few more (possible) charges, considering that they almost certainly will have some very bad wand experiences.

I think you have this a bit backwards. I would counsel against raising the median above 50. In fact, the original 20-12-10-8 seems about perfect. It is hard to have a 'very bad' wand experience, but almost all of them will be above 50, and some way above.

The PC's have a limited supply. If the odds are exactly even, the PC's would eventually lose big. Moving the odds a little in the PC's favor will make the whole thing taste so much better.

Not quite. The odds *are* exactly even with the 20-12-10-8 method; that means that the PC's will eventually come out....exactly even. By adding a D6 and D4, that means that the PCs will eventually come out...20% ahead. (60 charges per wand.)
20-12-10-8 means that most will be in the 40-50 range, some a little lower, some a lot higher, eventually balancing out to 50.
20-12-10-8-6-4 means that most will be in the 50-60 range, some a little lower, some a lot higher, eventually balancing out to 60.

The original method makes it a bit of a gamble, but still a very fair gamble. The second method makes it a no-brainer, and the PC's will almost always come out ahead, and on average quite a bit ahead.

 

[Coredump]

I am not sure if you have been reading the entire thread. Some of this has been covered. Also, we need to be careful with using average, since we are using mean and median for this discusion. (and a couple of times even mention mode.)

If one takes this all the way out, you end up with the median number of rolls needed to get a 1 and just a bit over half the number of sides on the die.

Well, that depends on how you define 'bit'. If you check the last page, a d20 has a median about 13.5ish.

This means that for the d20-d12-d10-d8 method, the average would be a bit over 25 (probably 30-35, I'll do the real math at home tonight).

Now you lose me. The mean will be 50, and the median will be quite abit over 35, babomb has it around 44. Not sure where you are getting 25, or for which version of average.

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.

It would give the exact same mean, but I am not convinced it would give a higher median. And not sure if I would want a higher median.

The original way causes much of the wand's logevity to depend on when that d20 comes up one. Once it does, you probably have only about 15 charges left.

On average, the wand will have 30 charges left. It is more likely that you will get around 25 charges, but sometimes a lot more.

Of course, if you want a campaign where wands just work and work for a while, but then degrade quickly once they start to go, use the original degradation progression.

Well, I think you have the 'quickly' part a bit off, but I do think it 'speading up' is part of the fun. You have to be careful when you are on the 'last die'

Staff of Popiel

KICK-ASS!