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

[Henry]

Since this one is firmly in House Rules territory, off it goes!

 

[Coredump]

I think part of the point of averages is that "on average" you'll get X results. It's not more or less likely to go over. You're just as likely going to roll 5 numbers above 10 in a row as 5 numbers below 10 in a row.

Just to be clear. This statement is incorrect. 'Average' does *not* mean an even distribution. It is very easy to have an average of (lets say) 20, but have lots of numbers below 20, and a few numbers much higher than 20. Because 10,10,10,50 averages to 20.
In this case, rolling a D20, you are as likely to get 13 or fewer charges before it degrades, as to get 14 or more charges. And you are twice as likely to get 20 or fewer charges than to get more than 20 charges.
We are not talking about rolling a die, we are talking about rolling it until a '1' crops up, and that is *not* an even distribution.
(Note for geeks, yes, I am assuming average=mean here.)

So whatever results you think are more likely towards a lower number on single die rolls are purely imagined. It's an even spread throughout the die.

This is true for a single roll(s). But we are talking about counting rolls until getting a one, and that is not an even spread.

Chances are that after 20 times on a d20 you'll roll a 1, but it could happen sooner or later. Could be on the first roll or on the 50th roll. Yes, chances are against both results, ie - they're not average results.

Actually, the *most* likely result is to get a 1 on the first roll, as opposed to any other specific number.

 

[ARandomGod]

Be wary of 'averages' here.

...


It is true, on average, it will take 20 rolls to roll a 20. But that is because you get a lot of the time it only takes 12 or 13 times, but it will rarely take 35 times.

So, if you have 3-4 wands, chances are you will *not* get 50 charges on any of them. If you go through 30-40 wands, chances are you will get over 50 on some of them. But may still average at less than 50 charges.

Still a fun mechanic....

"do ya feel lucky punk...?"

The answer is no, I do not feel lucky. Which is why I think that if I were to introduce this mechanic (which I agree is fun) I'd give them a d6 and a d4 too.

As for what to charge...

Well, I haven't decided.
Probably :
d20 = Full price (as if 50 charges)
d12 = 70%
d10 = 40%
d8 = 20%
d6 = price of three scrolls of the spell. (Yes, at this point I'd start treating it as a semi-scroll)
d4 = price of two scrolls of the spell

And I'd only very, very rarely allow PC's to buy a wand at less than a d10 in stores. Finding 'em is a different matter altogether!

 

[Felnar]

Lets make a deal. Everyday you give me a dollar, and I will roll a D(million) die. On any roll except a 1, I keep the dollar. On a roll of a 1, I will give you 1,000,000 dollars.
On average, we will both break even.

Who wants to take me up on this?
---------------------
Or the flip side....

I will give you $1000, and you roll a D10. On any roll besides a 1, you keep the money. On a 1, you give me $10,000. On average, we will both break even.

Who wants to play?
--------------------

this beats Vegas any day

 

[Borlon]

[thread hijack]

Another fun trick with averages is to introduce an "infinity sword." It does d10 damage, but if you roll a 0 you reroll, multiplying the result by 10. If you roll another 0 you reroll yet again, but multiply the result by 100. And so on. After you have calculated the dice damage, you add other modifiers; strength, sneak attack, flaming, whatever.

If I have done the math correctly its mean damage is infinite. But I eyeball it as being worth about a +3 enhancement. (Depending on the base weapon; a d10 dagger is worth more than a d10 greatsword; but this adds about +3 above that. I think.)

[/thread hijack]

 

[babomb]

Be wary of 'averages' here.

Lets make a deal. Everyday you give me a dollar, and I will roll a D(million) die. On any roll except a 1, I keep the dollar. On a roll of a 1, I will give you 1,000,000 dollars.
On average, we will both break even.

Who wants to take me up on this?

Indeed. There's a phenomenon called "Gambler's Ruin". What it means is that even though you come out even on the "average", if you go to Coredump's casino and play this game long enough, you'll almost definitely go bankrupt. That's assuming that Coredump, being the casino owner, starts with a lot more money than you do. And it's even worse in a real casino, where the odds are against you.

The point is, "average" can be very misleading. In this case, even though the mean number of charges is 50, most such wands will have significantly fewer than 50 charges (and if you buy enough wands, you may get one that lasts for 1000 charges). It is NOT a normal distribution.

The median number of rolls before getting a 1 on the d20 is 13.5198. The median before getting a 1 one the d12 is 7.96757. The median on a d10 is 6.59162. The median on a d8 is 5.20134. So the median for the whole thing is 33.2803. That is, over 50% of the wands will have fewer than 34 charges. (By the way, the median for a d6 is 3.816, and the median for a d4 is 2.444.)

 

[glass]

I am not that hot on means and medians.. averages are easy

Mean and median are both measures of average (as is mode).


glass.

 

[glass]

I gotcha. I thought that was a chart for rolling a 1 on Nth sided die. It's the percentage of rolling a 1 on Nth roll.

By the Nth roll. The chance of rolling a 1 on the Nth roll is always the same (1 over the number of side).


glass.

 

[Denaes]

Just to be clear. This statement is incorrect. 'Average' does *not* mean an even distribution. It is very easy to have an average of (lets say) 20, but have lots of numbers below 20, and a few numbers much higher than 20. Because 10,10,10,50 averages to 20.
In this case, rolling a D20, you are as likely to get 13 or fewer charges before it degrades, as to get 14 or more charges. And you are twice as likely to get 20 or fewer charges than to get more than 20 charges.
We are not talking about rolling a die, we are talking about rolling it until a '1' crops up, and that is *not* an even distribution.
(Note for geeks, yes, I am assuming average=mean here.)

This is true for a single roll(s). But we are talking about counting rolls until getting a one, and that is not an even spread.

I kneel before your superior math skills ::kneels::

Actually, the *most* likely result is to get a 1 on the first roll, as opposed to any other specific number.

This I don't get. It should be the same percent chance to get any single number on a single roll.

The only time I've heard differently is for computer die rollers because of how they roll and round numbers.

How is a 1 more likely to appear than a 5, 10, 13 or 4?

 

[Borlon]

There is a 0.05 chance of getting a 1 on your first roll. The chance of getting a 1 on your second roll is (0.95 * 0.05), a little bit less. The chance of getting a 1 on your third roll is (0.95 * 0.95 * 0.05) and so on. And so on. The chance that it will be your 225th roll that comes up 1 is very, very small, about 2 million to 1. Much less than 0.05.