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

[Spatzimaus]

So i am reluctantly forced to consider such hypotheses as a systematic mis-perception of risk.

I'm not disagreeing with you on that. Far too many people gamble more than they can afford to, perceiving it as a minor expense when in reality it adds up. But, that's not true of everyone (I'd argue it's not even true of the majority), and that was the flaw in the original statement; it's not that lotteries are inherently a bad choice, as the "tax on the stupid" sort of comments imply, it's that in many situations it's not a good choice.

Let's take an example.
I'm a graduate student. I earn just enough to live decently on, supplemented by a little money I saved away from a job I had between undergrad and grad school. It'd be pretty stupid for me to play the lottery right now, all things considered. I just don't have money to spare, so I'd have to dip into the money I allocate for "essentials".
Within the next year, I'll get my PhD. My salary will at least double at that point (more likely triple); it's not going to make me "rich", but it's going to be a dramatic change. While I'll be able to buy nicer stuff, I'm not going to immediately switch to a far more expensive lifestyle, so I'll have a certain amount of money that can be spent on other things. Using a little of the money to play the lottery at that point wouldn't really be a bad thing, not that I will.

And again, this ignores the non-financial aspects. Some people ENJOY gambling. My grandmother, before she died, played maybe $10 or $20 per week, just for fun. She didn't need the money or anything, it was just something she and her friends did. Then, look at all the people who go to Vegas, because there's a whole gambling "experience" there; in college we'd head off to Vegas every once in a while (I was in L.A.) and gamble a hundred dollars or so. More often than not we'd lose overall, but it was still fun.
If you're going to ignore the enjoyment aspect of it, then I'd like to point out many of the other things in life that cost you money in exchange for enjoyment. Girlfriends, for instance; what are your chances of making a fortune there? Now, the main reason I won't play the lottery is because I don't enjoy that form of gambling, but not everyone's the same.

If you're intending to make money over time, then sure, lotteries aren't a good choice. Join a mutual fund, buy a CD or some bonds, tweak your 401(k), whatever earns you a profit higher than the inflation rate. But viewed as a hobby, it's not that bad; the cost per week is relatively low (for most people, that is), and there's always the chance you'll score big and be set for life, no matter how slim.

 

[nycdan]

Math!

Okay, let me settle the math issue here for you guys. You've all been great help to me with rule interpretations so I'm glad to find a way to give back.

To determine the AVERAGE number of uses of a wand as you described is easily done in Excel.

First take the d20 series. You have a 95% of NOT rolling a 1 (19/20). Keep multiplying (19/20) times itself until your number drops below .5. This is the roll that, on average, you will get a 1. In the case of a d20, you get 14 rolls (the real average is between 13 and 14, but I'm rounding for ease of interpretation).

Next comes the d12. You have an 11/12 chance to NOT roll a 1. Multiply this number by itself until you get below .5 This adds 8 rolls.

The d10 adds 7 more rolls and the d8 adds 6 more rolls.

Adding these up, you get a total of 35 rolls. The real average is slightly less, probably between 33 and 34 but this is close enough. It's overall, a little short of 50 uses on average.

If you want to get to 50, you can do it by adding up the rolls for each die until you reach 50. For example, if your progression went d20, d20, d20, d12, you would get 14 + 14 + 14 + 8 which equals 50. You can create an average of almost any number by summing up the numbers provided in some combination.

Hope this helps. Good luck!

-Dan

 

[scottin]

FYI, I just used Excel to simulate this. Random(20) + Random (12) + Random(10) + Random(8) for 100 wands. I ran thru the simulation 10 times. This is the chart that I got for the min and max number for charges.

Run Min Max Avg
1 9 46 27.79
2 10 43 26.28
3 12 43 27.55
4 9 42 28.5
5 11 40 27.73
6 12 44 27.45
7 9 41 28.06
8 12 42 26.04
9 11 44 27.03
10 9 42 26.38



Seems to me that the players get gipped, unless they find ALL of the wands, in which case they do all right (avg 27.281 vs avg 25.5 ((1+50)/2) on found wands)

Happy gaming

 

[Primitive Screwhead]

Excell wont give you an easy simulation.. check the first page for the results of a function I built in Visual Basic.

But this did remind me..

I have decided to use this variant in game, which leads to two questions:

Cost of a wand at each step?

Cost to restore a wand one or more steps?
{this assumes that recharging a wand is both possible and economically worth it... altho for my ebberon campaign I could imagine old battlefields littered with 'disposable' wands..some still with a D4 worth of charges }

The one area I have always shied away from is item creation... I look forward to seeing what kind of consensus can be reached

 

[Thaniel]

I'm just gonna say that I just tried this myself 20 times with real dice. None of them got anywhere close to 50.

I had anything between 12 and 36 charges. What a rip. I'll stick with my bookkeeping wand charges, thank you very much. Is it that hard to make a little tick mark everytime you use a wand?

 

[Spatzimaus]

Okay, let me settle the math issue here for you guys. You've all been great help to me with rule interpretations so I'm glad to find a way to give back.

nycdan, as we explained before, simply adding the 50% expectation value (i.e., the median) for each individual die DOES NOT WORK. It's a variation on the Central Limit Theorem; the more dice you add to the chain, the closer the median approaches the mean. And the mean number of rolls for an individual die is simply the number of sides on the die; the mean on a d20 is 20, on a d12 is 12, and so on.
The reason for this is the distributions aren't symmetric; while the median falls at 14ish on the d20, and 8ish on the d12, the distributions skew to the high side; that is, if one of the two dice goes high, it'll go high by a larger margin than a die that goes low could possibly make up for. As a result, the median of the sum is higher than the sum of the medians, both of which are less than the sum of the means.

And scottin, the equation isn't Random(20) + Random(12) + Random (10) + Random(8). That's just the wrong math. You're not equally likely to run out of d20 charges in 1 as 20, and it's possible to go higher than 20.

It's nice that you're both trying to help with the math on this, and more feedback is always good, but you're making math mistakes that we discussed in this thread two weeks ago. The math behind the medians we gave has been proven both explicitly (working all possible combinations) and iteratively (running millions of test cases).

And Thaniel: That's very strange, because you should have had at LEAST one wand go for 100 or more charges, assuming your dice aren't biased in any way. I suppose you could have just been unlucky on the rolling, but to have NONE of your wands last longer than 36 implies to me that your d20 is a bit iffy, since it's really the most important die in this process. Ever tested it? (This isn't an idle point. One of my d20s rolls a LOT of natural 20s, well above 5%. Naturally, as of 3E it's now my favorite.)

Primitive Screwhead: Which die progression are you using?
The simplest way is to price it by the mean. That is, if you're using the 20-12-8-6-4 progression (which I feel is the best, and has a mean of 50), then it'd go like this:
All five charges still in: 100%
d20 gone: 60%
d12 gone: 36%
d8 gone: 20%
d6 gone: 8%
d4 gone: BOOM!
And if you included a way to repair, just move back up a step. You could price by median, in which case we'd need to rederive the medians for the 4-, 3-, and 2-step progressions (the final 1-step is easy). But it's really not worth it.

Now, IMC we have "expended" wands and staff still retain 50% of their original value (the expensive materials, which can be salvaged), which changes this as you'd expect.

 

[Primitive Screwhead]

Spatzimaus, Yup.. thats what I was looking for

Thanks for the help with the math.

Just for fun I altered my VBA script for the 20-12-8-6-4..
Each line is a run of 100 wands, so the 20 lines cover 2,000 wands....

Wand 100, Average charges = 41.18 , best charge = 101
Wand 100, Average charges = 36.7 , best charge = 97
Wand 100, Average charges = 40.77 , best charge = 161
Wand 100, Average charges = 50.85 , best charge = 119
Wand 100, Average charges = 48.58 , best charge = 108
Wand 100, Average charges = 47.68 , best charge = 136
Wand 100, Average charges = 65.26 , best charge = 163
Wand 100, Average charges = 46.66 , best charge = 105
Wand 100, Average charges = 63.28 , best charge = 118
Wand 100, Average charges = 51.33 , best charge = 96
Wand 100, Average charges = 56.99 , best charge = 128
Wand 100, Average charges = 36.54 , best charge = 97
Wand 100, Average charges = 63.39 , best charge = 104
Wand 100, Average charges = 43.54 , best charge = 84
Wand 100, Average charges = 51.2 , best charge = 124
Wand 100, Average charges = 39.37 , best charge = 126
Wand 100, Average charges = 56.48 , best charge = 117
Wand 100, Average charges = 41.14 , best charge = 100
Wand 100, Average charges = 45.28 , best charge = 101
Wand 100, Average charges = 44.32 , best charge = 79

So, overall higher potential.. {not to mention I found a logic error in my previous run goes to show what happens when coding too fast}

 

[Spatzimaus]

So, overall higher potential..

Right. The mean's 50, which you basically confirmed, but there's something like a 6% chance of going over 100 charges, so 15 out of your 20 batches of 100 had at least one wand over 100 (and three of the other five were high-90s). Two out of the 2000 even made it to 160!

The thing I really like about this system is that I can FINALLY sell partially-charged wands without feeling like I'm unbalancing anything. In the core rules, selling a wand that's had 10 or 20 of its charges used really only matters if it's a spell they'd use constantly. Most spells simply don't get used 30 times. So, to the players, buying a half-charged wand was just as good as buying a fully-charged one, and buying a wand with a handful of charges remaining was cheaper than buying the equivalent number of potions or scrolls. This was especially true of the Psion who had Master Dorje; he always wanted dorjes with one charge remaining, so that he could use them to feed his ability.

But now, if you sell the players a wand that has its d20 charge gone, it might EFFECTIVELY be the same as a 30-charge wand due to the mean, but the odds of it being a "weak" wand (running out after a handful of uses) have gone way up, and the chances of it going for 80+ charges have gone way down. It's the reverse of what we talked about before; as the number of dice involved reduces, the median gets further and further below the mean. Take the extreme case, where the wand only has its d4 left; the mean's 4 (giving its price as 4/50ths of the original), but the median's only 2-3, and the chances of only having one or two charges left is nearly 50%.

 

[Coredump]

I'm just gonna say that I just tried this myself 20 times with real dice. None of them got anywhere close to 50.

I had anything between 12 and 36 charges. What a rip. I'll stick with my bookkeeping wand charges, thank you very much. Is it that hard to make a little tick mark everytime you use a wand?

Just to be clear. Did you roll each of them once and add them? Or did you keep rolling the D20 over and over, until you got a 1. Then start rolling the D12 over and over....

I think people are confused about the mechanic. The way it works is you do *not* get D20 charges.

The way it works is each time you use the wand, you roll a D20, and as long as you roll a 2-20, nothing happens. Use it again, roll it again,and again, and again, etc. One of these times, it will roll a '1'. Then the next time you use it, you roll a D12, as long as you roll a 2-11, nothing happens. When you finally roll a '1', you need to switch to the D8, then D6, then D4.

Now, what is not obvious, is the average.
Take a D12, keep rolling until you get a '1'. Count how many times that was.
Now do it again, and again, and again......
Most of the time, it will take about 7 to 10 rolls, sometimes only 1, or 5, or 11. But sometimes, it will make it to 15, or 21, or 38. If you do it enough, and add the numbers up, you will find the average is..... 12.
This holds for all the dice, the *average* will be a 20 from the D20, a 12 from the D12, 8, 6, 4... for a total of 50.

Now, like the D12, most of the wands will have charges around 40-50, some will have only 20, or 32, or 39. But some will have 57, or 73, or 162. On the *average*, they will have 50 charges each.

Hope that helps. None of this was obvious to any of us, it was a combined effort from a number of geeks working out the fine points. We may still have gotten some things wrong, but we caught most of our mistakes... I think.

 

[nycdan]

Spatz,

Thank you. Your explaination makes sense and I didn't consider the sum of medians issue. I'm embarassed because I should have thought through that more but that's what I get for jumping on it while at work.

-Dan