Instead of charges....

[Doom]

An idea came up as my group was retro-gaming AD&D (just finished Keep, moving on to Bone Hill, fwiw).

I haven't been using staffs and wands with charges, as I'm so burned out with the record keeping from 4e that I just wanted to minimize it.

Instead, I just give the item a % chance of disintegrating with each use. So, the wand of fireballs with 10 charges before, now has a 10% chance of being done every time it's used.

It seems like it's working quite well, and I'm rather curious if it would do much to fix the "CLW wand" issues of 3.5. Probably not, but I'm definitely finding the players more willing to use items now that the charges aren't set.

[fectin]

Unless I mathed wrong, that gives you an expected number of additional uses between six and seven. A failure chance of ~7% gets you ten expected additional uses.

Not that it really changes your main point.

[Doom]

Nah, the expected number of uses is the reciprocal of the chance of failure, so failure chance 1/2, expect 2 uses, for 1/3, expect 3 uses, etc.

[Caedrus]

A friend of mine actually suggested a system whereby not only wands, but magic in general works by something like this principle (somewhat different because he suggested making magic fairly dangerous with various side effects and such). You can't know exactly when you will be spent, thus encouraging the player to be frugal with their casting regardless of the time of day.

Personally, I am not a huge fan of this solution just because I as a player want more things directly under my control. I like tactical maneuvering more than rolls of the dice, etc etc.

[Parthenon]

Nah, the expected number of uses is the reciprocal of the chance of failure, so failure chance 1/2, expect 2 uses, for 1/3, expect 3 uses, etc.

What the fuck Doom? How can you be so wrong? Its like you've forgotten basic probability and wilfully ignored all the discussion about the fact that if the chances of someone dying in a battle is even 5% then the chances of any PCs surviving a campaign is very low.

Looking at your example of 10% chance of failing:

	Use	Cumulative chance of failing
	1	10%
	2	19%
	3	27.1%
	4	34.4%
	5	41.0%
	6	46.9%
	7	52.2% (more than half the wands will fail at this point)
	8	57.0%
	9	61.3%
	10	65.2%

So, its pretty obvious that over half the wands will fail in 7 casts. Not 10. You have failed at probability.

Now, if you wanted the chance of failure to be about 50% at 10 charges, then your wanted probability of failure is actually between 6.5% and 7%. Not 10%.

Edit: Sorry fectin, I just noticed that I seem to have repeated you exactly where I thought I was repeating only the general gist and didn't notice you saying 7%. Oh well, Doom seems to have ignored you completely. Maybe the table will make it clear.

Edit 2: If you really want to do this, then you are better off using 5% (since you can roll a d20, and 14 on average), giving the first charge for free, and saying that wands have 15 charges on average. Or, using 1% for a 70 charge wand.

[hogarth]

I hate gambling; the amount of irritation I get from losing (e.g. having my wand die the first time I use it) greatly exceeds the amount of pleasure I get from winning (e.g. having my wand last for 20 times or more).

[Kaelik]

Doom. If the failure chance is one half, and you expect two uses, you are fucking retarded.

Take a Coin. A Coin flips heads 50% of the time.

What is the average number of times you get heads in row? Give you a hint, it's not 2.

50% of all wands with 2 charges never even go off the first time.

[Caedrus]

To explain the math... It's really quite simple.

If you have a cumulative 10% chance of failure...

The chance that you will fail the first time is 10%. Then in the other 90% of times you get a second charge. The chance of failing that second charge is 10% of that (so another 9%), and then there's 81% left, so you tack on 10% of that. (8.1%) and so forth. So the progression looks like the one that was plotted out above: 10%, 19%, 27.1%, and so on and so forth.

[Grek]

It seems like it's working quite well, and I'm rather curious if it would do much to fix the "CLW wand" issues of 3.5. Probably not, but I'm definitely finding the players more willing to use items now that the charges aren't set.

Assuming the CLW wands have a 2% chance to fail on each use, giving them an expected lifetime of ~35 uses and are priced as being 35/50ths of a 50 charge CLW wand would be, then it simply means that you need to carry a single backup wand in order to reduce your chances of loosing all of your healing before getting to use it down to 1 in 500 odds. A second backup wand, for 3 in total, reduces that to 1 in 125000.

So, you just carry (N/5)+2 wands, where N is the amount of healing you expect you'll need. Comes out to 525 per wand, or 3150 for a sixpack that you can expect to last you for your next 1470 hit points, give or take. Half the price or twice the health if you have a party member making them.

[fectin]

My bad. Doom is right: the mean number of uses (and expected number of additional uses) is 10. The median number is just under seven. My fault for saying it badly.
That does mean there's a really strong skew to the sad side of the curve (just like every other exponential decay).

So is that a chance of failure on every use or after every use?

[Doom]

Doom. If the failure chance is one half, and you expect two uses, you are fucking retarded.

Take a Coin. A Coin flips heads 50% of the time.

What is the average number of times you get heads in row? Give you a hint, it's not 2.

50% of all wands with 2 charges never even go off the first time.

Normally, I just ignore Kaelik, but godDAMN is he stupid.

Anyway, yeah, I'm talking fixed chance, not cumulative, after each successful use. It shouldn't make a difference ultimately, but sure seems like it does. Or maybe it's just the "it's new so it's cool" factor.

[fectin]

Memoryless distributions are weird, and the exponential is more so than usual. Overall, the balance point is the same, but this is significantly more likely to hurt your players than to help them, and that's going to be more true with everyday utility like CLW. For one off utility like water walk, they'll just carry scrolls (because they need it to work right, and the marginal benefit of having extra uses afterwards is low).

Not saying any of that is good or bad.

[Kaelik]

Doom. If the failure chance is one half, and you expect two uses, you are fucking retarded.

Take a Coin. A Coin flips heads 50% of the time.

What is the average number of times you get heads in row? Give you a hint, it's not 2.

50% of all wands with 2 charges never even go off the first time.

Normally, I just ignore Kaelik, but godDAMN is he stupid.

Do you have any actual evidence or arguments?

Or just stupidity?

You claimed that the average item that explodes 50% of the times you use it will give you two uses.

This is false in literally every single way, whether you use mean, median, or mode. This could not possibly be more false if you hired Glenn Beck to give it a spin.

And when you are called on what is literally the most retarded thing you could possibly say... You respond by not even defending yourself with anything close to an argument.

[Caedrus]

Overall, the balance point is the same

It's really not. There is a huge difference between having a resource you can rely on and one that's a gamble.

[fectin]

Kaelik, you forgot the weighting in your summation. There is a really strong skew, and you're talking about the median, not the mean.

Consider the set [1,1,1,1,1,1,1,1,1,1,10]. The average result (as described by Doom) is 2. That's also the expected result for infinite random selections from that set ("expected result" is a specific term of art and doesn't mean what a normal person would think it means).

You are pointing out that it's overwhelmingly likely that a result less than the expected result will come up. That's true. I also happen to agree with you that setting the number so 10 is the median is more fair to players.

The facts you two are citing do not disagree.

Details here:
http://en.m.wikipedia.org/wiki/Exponential_distribution

[Caedrus]

("expected result" is a specific term of art and doesn't mean what a normal person would think it means).

Would you mind clarifying what it actually means, then?

[UmaroVI]

I think Kaelik is assuming the wand blows up and does NOT work on the "final" use that destroys it. IE, 50% chance failure gives you a 50% chance of NO uses. Everyone else is I think assuming that it breaks AFTER working, so a 50% chance that it works once and then immediately disintegrates, 50% chance that it works once and then is fine and can be reused.

For the curious: this is a geometric distribution. A wand with chance of breaking AFTER use (ie, it works then breaks, not you try to use it then it breaks instead of doing anything) has a mean of 1/(p) uses, and a wand with the chance of breaking BEFORE use has a mean of 1/(p) -1 uses, as mentioned above.

Medians are a bit harder to find in general. The median number of uses (assuming breaking after use) is -ln2/ln(1-p) (or rather, the medians are all the numbers between the floor and ceiling of -ln2/ln(1-p)). If you want the median number of uses to be x, you'd want p = 1 - 2^(-1/x). So for example, for median 50 uses (equivalent of a fully charged wand), you'd want a 1.3% failure chance.

Having items that might or might not hold out is less good for you, because you cannot afford to have your only CLW wand break. I would based on this suggest you go for median rather than mean uses, and round percentages down. This means the wands will on average give you a lot more use, but you accept the risk of them exploding.

[fectin]

Geometric isnt quite right either, because practically, uses caps out (i.e., doesn't go to infinity). So that skews everything down a couple percent more. It's already a bit off anyway, so it's much easier to pretend it's continuous instead of discreet. Finding the median is much simpler for exponential, which is the continuous version of the geometric: (1/2)*ln(2).
If I were looking for a system though, I'd take chance of breaking = 1/(1.5*charges). It overestimates it by about 4%, but is really easy.

[Doom]

Wow, I didn't mean to spark so much discussion over a mostly irrelevant attempt on my part to indirectly address the too-easy manufacture of magic items in 3.5.

Anyway, I think a proof for expected number of uses for general probability of failure after use is a bit much for here, but let's go with proving it for failure probability 1/2, since it's particularly easy, just for the simple pleasure of making Kaelik look like an ignorant loudmouth idiot in complete detail. I begin with what might be a well known result, but I include it for completeness, and because I can recycle the method of proof for the expected value we seek:

Lemma:

1/2 + 1/4 + 1/8 + 1/16 + ... = 1.

Proof: Let 1/2 + 1/4 + 1/8 + 1/16 + ... = X

Multiply both sides of the equation by 2. Then we get:

1 + 1/2 + 1/4 + 1/8 + ... = 2X

1 + (1/2 +1/4 +1/8 + 1/16 + ...) = 2X

1 + X = 2X

1 = X //


Now, let's suppose a wand has a 50% chance of failure after each use. The probability of getting 1 use is 1/2. The probability of getting 2 uses is 1/4 (1/2, not failing on first use, times 1/2, failing on the second use), and so on.

The expected number of uses:

1 (1/2) + 2 (1/4) + 3 (1/8) + 4 (1/16) + 5(1/32) + ...

= 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + ... (I'm deliberately not simplifying fractions)

You can check on a calculator that that comes very close to 2, but as I can find no proof online, allow me to fill in the details.

Theorem: The expected number of uses of a wand with failure probability 1/2 as given above is 2.

Let 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + ... = X (call this equation **), our expected value.

Let's multiply both sides by 1/2.

We get:

1/4 + 2/8 + 3/16 + 4/32 + ... = (1/2)X

Let's subtract this infinite sum on the left, and (1/2)X on the right, from equation ** to get:

1/2 + 2/4 + 3/8 + 4/16 + 5/32 + ... - (1/4 + 2/8 + 3/16 + 4/32 + .... ) = X - (1/2)X

On the left hand side, leave the 1/2 alone, and do the other subtractions term by term, i.e., 2/4 - 1/4, 3/8-2/8, 4/16 - 3/16, etc. On the right hand side, X - (1/2)X = (1/2)X, of course.

This leaves us with:

1/2 + (2/4 -1/4) + (3/8 - 2/8) + (4/16 - 3/16) + ... = (1/2)X

1/2 + 1/4 + 1/8 + 1/16 + ... = (1/2)X

By the Lemma, the sum on the left is 1.

So we get:

1 = (1/2)X

So, X, our expected value, is 2. //

For a general probability of say, 1/k, the proof is easier if we appeal to techniques in stochastic processes, albeit still beyond my ability to notate properly in this online format, but I'll spare you that in any event.

[Kaelik]

Great, you can do math, you still have no brain.

If you pull out a wand and attempt to use it, and it explodes 50% of the time, then you start with (0).5, then 1/4 (1), then 1/8 (2) ect.

Limit is not 2. You are still retarded.

[Doom]

Wow, I now know why it's so understandable that threads can start off "Kaelik please do not post here". Truly stunning how your confusion can be so complete and, most importantly, unalterable.

Even if your initial confusion were somehow cleared (despite numerous people trying to explain to you), considering things your special way would mean little, simply subtracting one, not much of a difference for items with many charges.

But, you've demonstrated completely your inability to learn, so it's a moot point.

"Kaelik, please do not post in this thread."

[fectin]

("expected result" is a specific term of art and doesn't mean what a normal person would think it means).

Would you mind clarifying what it actually means, then?

Wikipedia is smarter than I am:
http://en.wikipedia.org/wiki/Expected_value

Basically, it's the geometric mean. There's some weird stuff with non-convergence, but it's all edge cases you don't care about, and even if you do I can't explain it. The big thing to notice is that it doesn't mean "most probable outcome," or anything similar.

It matters a lot for cost/benefit comparisons. Like with roulette, your expected return for any bet is 36/38 of your wager. I.e., if you made a 38 dollar wager infinite times, your average return is 36 dollars. That says nothing about what will happen next time, just what happens over the (infinitely) long run.

[Caedrus]

("expected result" is a specific term of art and doesn't mean what a normal person would think it means).

Would you mind clarifying what it actually means, then?

Wikipedia is smarter than I am:
http://en.wikipedia.org/wiki/Expected_value

Well, see, that's why I couldn't find anything when I searched for "Expected Result" as a specific term of art.

So... yeah.

[fectin]

It's the "expected" part. Value, result, cost, etc. all get used depending on the field (economics is the worst), and all have the same weird meaning. But I should have linked it the first time or made sure it was googleable.

[Ice9]

If you pull out a wand and attempt to use it, and it explodes 50% of the time, then you start with (0).5, then 1/4 (1), then 1/8 (2) ect.

The wand disintegrates after working. Doom is correct on the mean result - the fact that a majority of people will get less than 10 uses is balanced (mathematically) by the few lucky people that get a much larger number of uses.