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

[Denaes]

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.

Oh, Cam just wasn't entirely clear in his/her statement.

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

Looks like Cam is saying that it's more likely to get a 1 than any other specific number on the first roll. What Cam meant was that the chances of rolling a 1 on specific successive die rolls is less than on the first. I'm not saying I can explain it better...

Yeah, that makes sense so long as you're thinking about the right kind of specific numbers

 

[Spatzimaus]

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

He was vague in how he stated it. What he was trying to say, as Borlon gave the math for, was "...as any other specific NUMBER OF ROLLS". That is, getting it on the first roll is more likely than getting it on the 5th roll, because you're less likely to ever make it TO the 5th roll (if you got a 1 on rolls 1-4, it stops there). Once you get TO that roll, it's still the same 5% chance you've always had, but you have to include the possibility that the situation will never come up.

 

[Denaes]

He was vague in how he stated it. What he was trying to say, as Borlon gave the math for, was "...as any other specific NUMBER OF ROLLS". That is, getting it on the first roll is more likely than getting it on the 5th roll, because you're less likely to ever make it TO the 5th roll (if you got a 1 on rolls 1-4, it stops there). Once you get TO that roll, it's still the same 5% chance you've always had, but you have to include the possibility that the situation will never come up.

yes I know. Once Borlon specifically mentioned what "specific numbers" were being talked about, I understood.

 

[Coredump]

Yep, I wasn't clear enough. Thanks for covering for me guys.

 

[ARandomGod]

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.)

Now, I took statistics once a long time ago. So I do know what you're talking about... and I recogize that (assuming you're correct in your math, there's no way I can duplicate it and I'm not going to pretend to be able to say 'that looks right') a 33 charge wand is a better place to set one of these wands at.

Now, as per above, can you

1) Come up with a dice degredation set that is closer to 50? (ie, will adding a d6 and a d4 do that? How close will that come??)

2) Give us (me) a breakdown of what each die degredation would yeild up as the median from that point on.

Meaning I can see that using the above it's 33 charges for a "full" one. What about one that's at d12?

And more important than using the one above, assuming adding a d6 and a d4 would bring it closer, how much closer does that bring it, and what would each step yeild as a mean from that point on? (So what would a d12 wand buy and sell for? What about a d6 wand? Etc.)

(And if you get and do this, thanks! I really, really wanted it.)

 

[Borlon]

Do you have a link or something that explains where those numbers come from? 2.444 for d6, and so on? Or maybe it is easy enough to just explain?

 

[Spatzimaus]

1) Come up with a dice degredation set that is closer to 50? (ie, will adding a d6 and a d4 do that? How close will that come??)

Per his numbers, which I have no reason to doubt, adding a d6 and a d4 increases the median to 39.5. But remember, the MEAN would fall at 60, so that 39.5 would be a set of a lot of 20s and 30s, mixed with a small number of 100+ charge versions.

If you really want the MEDIAN to fall around 50, I'd suggest adding a d30 on the top end, if you have one. Its median would be somewhere around 20. It's too bad there isn't a d16, because the 20-16-12-10-8-6-4 would work even better. I suppose you could have a d16 by doing 2d4 and only failing if both roll 1's, but that would just be a bit awkward.

2) Give us (me) a breakdown of what each die degredation would yeild up as the median from that point on.

He gave you that already; since these are independent events, the median number to go from d12 to the next die will always be 7.97 rolls, regardless of what else is in the progression, so just add them up. It takes a median of 13.5 to go from d20 to d12, 7.97 to go d12 to d10, and so on. It'd take a median of 20 rolls to go from d12 to d10 to d8 to dead; if you're currently at the d12 stage, you should expect the wand to last you another 20 activations. (Adding the d6 and d4 extends it by another 6.2.)

There are a few other ways you could do it. Instead of decreasing die sizes, how about increasing thresholds? Like this:
Each wand starts with 10 "charges". When you use a new wand, you roll a d20. If you roll a 1, it uses a charge, and the threshold increases by 1. So once you've used a charge, the next charge expends on a roll of 2 or higher, then 3, and so on; as you use up the wand, it becomes easier and easier to fail the next check (but you stay with the d20). Once you fail on a threshold-10 roll (costing your tenth charge), the wand is expended. This'd have a slightly longer lifespan than what you had previously (mean 58, median ~40), but it's far less susceptible to a string of bad rolls.
If you use your d20-d12-d10-d8-dead system, it's possible that you'd only get 4 charges out of a brand-new wand. Even adding the 6 and 4 would still allow a wand to fail after 6 uses, even if the odds are horrendous. But the system I mention guarantees a minimum of 10 uses.

This'd also allow you to customize things. A well-made wand might use a d20 and have 10 charges, while a cheaper one might only have 5 charges, or use a d12.

And Borlon: I'm not him, but I got the same numbers using the following logic in a computer. (Use whatever language you want)

chance = 1.0 / (die size)
A[0] = 0.0
for N = 1,100
A[N] = A[N-1] + (chance)*(1.0-chance)^(N-1)
endfor

(1.0-chance)^(N-1) is the probability of making it to roll number N, so A[N] is the probability that you would have finished on or before roll N. Then, just find where the array A crosses 0.5. That's your median. Technically, since these are discrete events, the median should be a whole number, but you can interpolate between the two nearest values to get a decimal.

 

[Coredump]

a 33 charge wand is a better place to set one of these wands at.

No, it is still a 50 charge wand... over the long haul.

1) Come up with a dice degredation set that is closer to 50? (ie, will adding a d6 and a d4 do that? How close will that come??)

2) Give us (me) a breakdown of what each die degredation would yeild up as the median from that point on.

Meaning I can see that using the above it's 33 charges for a "full" one. What about one that's at d12?

Let me see if I can explain this a bit better.

If you get 10 CLW wands now, each one will provide 50 charges, for a total of 500 charges.

With this new system 10 wands will provide (for example) 19, 24, 28, 30, 33, 41, 49, 59, 78, 139 charges; for a total of 500 charges.

Now, you can see that most wands provide less than the 50 charges of 'normal', but a couple make up the difference. There are a couple of issues with this example however.
1) If you only have 2-3 wands, it is very unlikely it will balance out to 50 each. Though there is a chance of hitting it big.
2) I am not sure how many wands it would take to be fairly sure it would balance out. (5, 10, 25...??)
3) The above example was based on a median of 34, that is too low for the real median. So the actual 'half-way' point will be higher. (A median of X means that 50% of the wands will run out before X charges, and 50% will run out after X charges. But regardless of median, they will always balance out at 50 charges per wand...eventually.)

 

[Coredump]

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.)

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.

Unfortunately, my math skills are too rusty to provide a totally rigorous proof, I can show I deduced this.

The trick it to determine how many of the combined rolls will be above the proposed median. According to the assumption above, a bit less than 50% of the rolls should be above 15.

Above and below the median is a 50-50 chance. So to roll above the median on a D12 (I call that 12H) is going to happen half the time, and below the median is half the time (12L). Since we are doing these rolls in order, we get 4, equal, possible types of outcomes.
12H-10H (25%), 12H-10L (25%), 12L-10H (25%), 12L-10L (25%).

Of those, the first and last are easy.
Looking at the 12H-10H combinations, if both rolls are above the median, than the total will be above the median, then 100% of those combinations will be above the median. That means that 100% of that 25% segment will be above the median (Total above=25%)

Looking at the 12L-10L combinations, if both rolls are below the median, than the total will be below the median, then 0% of those combinations will be above the median. That means that 0% of that 25% segment will be above the median (Total above=25%+0%=25%)

That leaves us with going high-low, and low-high.

Looking at the 12H-10L combinations, we can start by ignoring the 10L roll for now. The 12H rolls will give results of 8 and up. Any roll of 14 or higher will put the total at or above 15, and thus above the median. (This is about to get a bit confusing..) About 66% of the *12H* rolls will be at 14 or above, thus MORE THAN 66% of the 12H-10L combniations will be above the median. That means that more than 66% of that 25% segment will be above the median; 66% of 25% gives us 16.5% (Total above=25%+16.5%=41.5%)
[NOTE: this is quite an underestimation. For illustration: This only counts if the 12H roll is 14 or higher, yet if it is 13, the 10L roll *has* to be a one, or the total will be above the median. There are lots of 'above the median' combinations that we are not counting.]**


Looking at the 12L-10H combinations, we can start by ignoring the 12L roll for now. The 10H rolls will give results of 7 and up. Any roll of 14 or higher will put the total at or above 15, and thus above the median. (This is about to get a bit confusing..) About 50% of the *10H* rolls will be at 14 or above, thus MORE THAN 50% of the 12H-10L combniations will be above the median. That means that more than 50% of that 25% segment will be above the median; 50% of 25% gives us 12.5% (Total above=41.5%+12.5%=54%)
[NOTE: Same as above]

So we can that even while underestimating the results, we are getting more than 54% of the combination totals will be at 15 or above. Thus the median cannot be 14.5 as projected. We cannot just add medians together to get an overall median. And this is just two dice, the more dice we add together, the higher the resultant median will creep.

**Some may ask why I didn't figure the rest out and come up with a true median. Two reasons, This started as an excercise to see if adding medians is okay, and it is not. Second, do figure the others out will take a lot more tedious maths....

Hope this made sense, and I realize it is probably more info than anyone really cared about, but hey, it was fun.

 

[babomb]

Do you have a link or something that explains where those numbers come from? 2.444 for d6, and so on? Or maybe it is easy enough to just explain?

I can give you the C++ code I used to figure it out. Will that help?

#include <cmath>
#include <cstdlib>
#include <iostream>

using namespace std;

const int DIE_TYPE = 10;
const double PROB_NOT_ONE = (double)(DIE_TYPE-1)/DIE_TYPE;
const double PROB_ONE = 1/(double)DIE_TYPE;
const double CUTOFF_PROB = 0.50;

int main()
{
	double cum_prob = 0;
	double cum_prob_not_one = 1;
	double prob = 0;
	double last_cum_prob = 0;
	int i = 0;
	double median = 0;
	
	cout << "d" << DIE_TYPE << endl;
	
	for(; cum_prob < CUTOFF_PROB; ++i)
	{

		prob = cum_prob_not_one * PROB_ONE;
		last_cum_prob = cum_prob;
		cum_prob += prob;
		
		cout << i + 1 << ": " << cum_prob << endl;
		
		cum_prob_not_one *= PROB_NOT_ONE;
	}
	
	median = (0.50 - last_cum_prob)/(cum_prob - last_cum_prob) + i - 1;
	cout << "median: " << median <<endl;
	
	system("PAUSE");
	return 0;
}

The output it gives me is

d10
1: 0.1
2: 0.19
3: 0.271
4: 0.3439
5: 0.40951
6: 0.468559
7: 0.521703
median: 6.59162
Press any key to continue . . .

If you don't understand C++, here's the math class version:
First, find the cumulative probability that a one is rolled. You can stop when you get a cumulative probability of greater than 50%.
The cumulative probability for the nth roll is as follows:
c = p(1), n = 1
= p(not 1)^(n-1) * p(1) + c(n-1), n > 1

I'll use m for the lowest roll with a cumulative property <=.5. For the d10, m=6.
Now find the following:
A = c(m); For a d10, that's 0.468559 (the 6th roll).
B = c(m+1); For a d10, that's 0.521703 (the 7th roll).
deltac = .5 - A; For the d10, that's .031441
I = the distance between intervals. This is 1 for any die, but if we could find the cumulative probability at half-rolls, we could use 1/2.
R = B-A; For a d10, 0.053144.
deltax = deltac * I/R; For a d10, that's .59162.
median = m + deltax; in this case, 6.59162

If geometry helps, you plot the points (m, c(m)) and (m+1, c(m+1)). Then draw a line between them. Find the slope of that line (R/I). Since you know a point on the line (namely, (m, c(m)), you can write the equation of the line as
c - c(m) = (R/I) * (median-m)
.5 - A = R/I * (median - m); c = .5, A = c(m) (as above)
deltac = R/I * (median - m); deltac = .5 - A (as above)
deltac * I/R = median - m
deltac * I/R + m = median
deltax + m = median; deltax = deltac * I/R
QED.