Dice math.

[Chaldfont]

Roll when they find the wand and keep track of the number

Your method is not good, since the wand has 1:50 chance of having 1 charge, but succesively smaller chances of having more. Anyway, without an in-depth analysis I'm drawn to say that there's no simple method that would give you the uniform distribution you're probably after.

We have an easy answer to this: If someone buys or creates a new wand we don't make them roll for the first few charges.

I'm really hoping someone can tell me the stats behind this. I am no statistician and I don't want to completely hose my players.

 

[orsal]

If you want to replicate the official probability distribution (any number from 1 to 50 has a 2% chance of being the number of charges), then you can't do it with a fixed probability. By the official distribution, the probability is 1/50 that the first charge is the last, but if it isn't, the probability that the second charge will be the last will go up to 1/49. This will continue: if you've already used 48 charges, and know that the wand isn't depleted, there are either 1 or 2 left, so the probability that then next charge is the last is as high as 1/2.

The official probability distribution is called a uniform distribution, because every possible number of charges is equally likely. What you propose is called a geometric distribution. One of the properties of a geometric distribution is that there is no limit to how high the random variable can get. For example, if you use 2% as your probability (that each given charge is the last), then 13% of wands will have over 100 charges, and about 1.8% will have over 200.

On the other hand, if you do want a geometric distribution (so that arbitrarily high numbers are possible, although less and less likely the higher you go), then decide what the average number of charges per wand should be. If you want them to average n charges, the probability you use should be 1/n. So your suggested method would produce 50-charge wands on average, whereas the rulebooks have that number as the maximum.

 

[Cheiromancer]

Here's a question for you guys:

Instead of keeping track of wand charges, I would rather roll dice every time the wand is used. If some number is rolled, that was the last charge on the wand.

I have been making my players roll d100 and saying that a result of 01-02 means that the charges have been depleted--but this mechanic is based on just gut feeling. What is the most statistically accurate way to do this?

I like this idea. I'd yoink it, but I had thought it up previously.

For randomly acquired wands try using d20 or d10; on average the wand will have 20 or 10 charges left. Good if there is a chaos-motif in your magic system.

One other item of mine from around the same vintage is the "infinity sword." It does 1d10 damage, but if you roll a "0" on the d10 you roll again and multiply the result by 10. If there is another "0" you roll again and multiply by 100, and so on.

The arithmetic mean damage is infinite, but in practice it is about a +2 or +3 damage.

 

[jmucchiello]

My probability knowledge is a bit rusty but basically what you've done is make it all but impossible to get 50 charges out of a wand.

The chance you can get one charge out of the wand is 98%, the chance of getting two charges (i.e. both the first and second charge roll outside the 01-02 range) is .98 * .98 = .9604, the third roll requires that both the first two rolls and its roll succeed or .98 ^ 3 = .941. Each additional roll is another multiplication by .98.

IOW, ask the question: Will I get 25 charges out of this wand? For that to be true, each charge before that charge has a .98 chance of success. Thus I have a .98 ^ 25 chance of getting 25 charges, or about a 60% chance. That means 4 out of 10 wands will fail by their 25th charge.

Really, it is much better to just keep track of the charges.

 

[orsal]

I realized after rereading some of the posts on the wand question it's not clear: is this system supposed to be for new wands or for found wands? I had been thinking of found wands when I posted earlier.

If you intend that *new* wands have the geometric distribution, then OK -- provided you want a system with no theoretical maximum to the number of charges. In which case you do have the right probability to get an *average* (mean) of 50 charges per wand.

The beauty of the geometric distribution is that if you find a wand with at least one charge left, it's exactly the same distribution, exactly the same average number of charges.

 

[CRGreathouse]

One other item of mine from around the same vintage is the "infinity sword." It does 1d10 damage, but if you roll a "0" on the d10 you roll again and multiply the result by 10. If there is another "0" you roll again and multiply by 100, and so on.

The arithmetic mean damage is infinite, but in practice it is about a +2 or +3 damage.

Yes, the mean damage is infinite, but that's only because you multiply by such a large number. If you did this with a d12, or multiplied by 9 instead of 10, the mean damage would be finite.

For example, using a d12 (10 = x10, 1-9 and 11-12 normal) the mean damage would be (exactly) 34.

Edit: Also, if you add damage instead of multiplying, you get finite damage -- much smaller finite damage, I might add. 1d10 where 10 means "add 10 and reroll" averages 6 1/9 = 6.1111....

 

[ichabod]

Here's a question for you guys:

Instead of keeping track of wand charges, I would rather roll dice every time the wand is used. If some number is rolled, that was the last charge on the wand.

I have been making my players roll d100 and saying that a result of 01-02 means that the charges have been depleted--but this mechanic is based on just gut feeling. What is the most statistically accurate way to do this?

This is a geometric distribution, which is a special case of the negative binomial distribution. The probability that X (the number of charges you get from the wand) is equal to x is q^(x-1)*p, where p is the probability that the wand is used up, and q = 1-p. According to my reference material*, the mean is q/p. So in your case the mean is 0.98/0.02 = 49. My reference gives the second moment as q/p^2, so the variance would be 49 also, giving you a standard deviation of about 7. By rule of thumb that means 2/3 of the wands will end up with between 42 and 56 charges, and only 1 in 20 will have less than 35 or more than 63.

On the other hand, this is a pretty skewed distribution, so that isn't such a good rule of thumb. It turns out 1 in 20 wands will have 3 or less charges. 1 in 10 will have 6 or less. 1 in 4 will have 15 or less, 1 in 4 will have 69 or more, 1 in 10 will have 114 or more, and 1 in 20 will have 145 or more.

That makes this probably not a good house rule.

*Univariate Discrete Distributions, 2nd ed., Johnson, Kotz, and Kemp.