[X] My Assistant

[Lionhearted]

The hell does the exclamation mark mean?

[Epicedion]

Factorial.

3! = 3*2*1
7! = 7*6*5*4*3*2*1

And so on.

EDIT: And 0! = 1, strangely enough.

[Draco18s]

EDIT: And 0! = 1, strangely enough.

Not really. A factorial is a set of numbers which are multiplied. Specifically n! = Π(k=1 -> n).

Π = product the same way Σ = sum.

So 0! follows the rules of the Empty Product.

[Falconer]

Your math is wrong...... I can instantly prove you wrong as well.
n=1 in your formula. Produces a value of 8. Meaning 1 d6 can roll to 8 different results.

The actual number of combinations without respect to order of 16 d6 dice is in nCr notation... 21 C 5 == 20349.

But this number is meaningless to us as this only tells us how many total distinct combination bins there are to count results into. It does not tell us how many results fall into each bin or their probability density. So if you roll toss out 10,000 results at random... you fill less than half the 'bins' even if no bin gets more than 1 result (more likely some bins get multiple balls, and others get none like some monstrous 'plinko' machine).


And monte carlo won't tell you the odds of rolling 16 1's.... it can't give you even a close guess on it until you've rolled a good order of magnitude over that 3 trillion figure I just gave you. At which point it may have dropped 1 ball into the 16'1s bin, or 20... depending on how fickle lady luck is feeling.

For the record the odds of rolling 16 1's on 16 dice are exactly
16: 1 in (6^16)!~= 2.82111... trillion
15: 80 in
14: 3000 in
13: 70000 in...
....
0 1's: 1.52588E11 out of this. Roughly 5.4% of the results have no 1's at all.

At the end of the day each of the sub categories summed together must result in precisely 6^16 power to cover 100% of the probability distribution.


That's why I said the monte carlo numbers aren't reliable. You're not doing it enough to invoke the law of large
numbers. You're not even doing it enough to cover all your 'without respect to order' bins.

[Sengir]

Not really. A factorial is a set of numbers which are multiplied. Specifically n! = Π(k=1 -> n).

Π = product the same way Σ = sum.

So 0! follows the rules of the Empty Product.

Or more descriptively, n! is the number of different sequences which can be produced from a set containing n different elements. How many (distinguishable) ways are there to arrange nothing? One obviously

[Epicedion]

Your math is wrong...... I can instantly prove you wrong as well.
n=1 in your formula. Produces a value of 8. Meaning 1 d6 can roll to 8 different results.

Hmm.. you're right, but what is it missing? Ah well, I was going from memory, which is painfully unreliable. So on 1 die you've got 7 ticks and 1 star:

| | | | | | | *

Ah, now I see, the 8 was coming from allowing the star outside the ticks, you want to go by the whitespace (bins). This would rather be written as...

(5 + n)! / (5! n!)

which for 16 is 20349, and for 1 is 6.

But this number is meaningless to us as this only tells us how many total distinct combination bins there are to count results into. It does not tell us how many results fall into each bin or their probability density. So if you roll toss out 10,000 results at random... you fill less than half the 'bins' even if no bin gets more than 1 result (more likely some bins get multiple balls, and others get none like some monstrous 'plinko' machine).


And monte carlo won't tell you the odds of rolling 16 1's.... it can't give you even a close guess on it until you've rolled a good order of magnitude over that 3 trillion figure I just gave you. At which point it may have dropped 1 ball into the 16'1s bin, or 20... depending on how fickle lady luck is feeling.

It's not meaningless, but there are a few trillion fewer bins than you made it seem. And Monte Carlo will get you the odds, but you'll have to increase the sample size by a lot -- Monte Carlo is useful when computational cycles are cheaper than actual figuring. In this instance you still get the practical answer: the odds of rolling 1s on all 16 dice is so low that it may as well be counted as zero. We're not launching a rocket to Alpha Centauri, after all.

[Epicedion]

Not really. A factorial is a set of numbers which are multiplied. Specifically n! = Π(k=1 -> n).

Π = product the same way Σ = sum.

So 0! follows the rules of the Empty Product.

I don't actually think it's strange, but it certainly doesn't follow by the short (read: awful) way I presented it.

[Falconer]

Since you seem lost the general form for rolling any arbitrary number of dX's is.
n= number of sides, r=number of dice
(n+r-1)!/(r!(n-1)!)


But no you completely missed the point. People have posted that the 'glitch' rate going up to 16 dice is about 1% based on flawed monte carlo assumptions. Because they didn't do the basics to determine the 'confidence'... generally you need to have far more than the expected rate of occurance for the law of large numbers to start working. Whoever did the monte carlo did not calculate the confidence interval and realize that his figures were wholly innaccurate.

https://docs.google.com/viewer?a=v&q=ca...MTo4KL0U7-IR5Dg

Good example of craps... to get to 0.1% confidence interval on a nearly 50:50 event... nearly 1million!!! results need to be run. On an event with a far lower expectation rate (below 1%) far more than even that need to be run. Note the line graph on page 2 of the same program run 100x in a row and notice how often a single run 'spikes' outside the expected error showing as high as 70% win odds or as low as 30%.


How do I know this because I know the *EXACT* non-random guestimation is ~0.21% (6.06E9/2.82E12). Because I can enumerate all these without enumerating all them, simply by enumerating 16 coinflips probability distribution and then assigning weight to each side of the coin (a coin 5x more likely to flip heads than tails) to calculate precise odds.

Even at 11 dice... the odds of 6 or more 1's is a paltry 0.46%


So anyone claiming 1% on 16 dice is off by a factor of *5*. That would get them an 'F' in most prob and stat courses. Monte carlo would generally not be used for this at all because an exact mathematical derivation is easily obtained.

[Epicedion]

Since you seem lost

Mmmno.

But no you completely missed the point. People have posted that the 'glitch' rate going up to 16 dice is about 1% based on flawed monte carlo assumptions.

Mmmalsono. Since the percentage drops below 1% somewhere around the 10 dice mark, as I've stated previously, it's safe to assume that at the 16 dice mark it's far less. It's great that you calculated the actual percentages, but there's no reason to get increasingly anal about it.

So anyone claiming 1% on 16 dice is off by a factor of *5*. That would get them an 'F' in most prob and stat courses. Monte carlo would generally not be used for this at all because an exact mathematical derivation is easily obtained.

Easy, but a little tedious. Since this isn't a prob and stat course, you probably shouldn't fault a programmer for coming up with a programming solution, especially one you can write in a minute and run a few hundred thousand iterations of in the time it takes to make a cup of coffee.

[sk8bcn]

Easy, but a little tedious. Since this isn't a prob and stat course, you probably shouldn't fault a programmer for coming up with a programming solution, especially one you can write in a minute and run a few hundred thousand iterations of in the time it takes to make a cup of coffee.

Exactely.

I mean, I'm ok to spend 5 mins for a result I find accurate enough.

Now I've could have calculate the exact numbers but it would have taken me time and really, really, it doesn't bring a lot to the thread.

By the way, this is misleading:

Good example of craps... to get to 0.1% confidence interval on a nearly 50:50 event... nearly 1million!!! results need to be run. On an event with a far lower expectation rate (below 1%) far more than even that need to be run. Note the line graph on page 2 of the same program run 100x in a row and notice how often a single run 'spikes' outside the expected error showing as high as 70% win odds or as low as 30%.

Where the hell would I need a 0,1% Interval?


You (Falconer) come here with a very good mathemacial point of view (nice) but a very narrowed mental state (EDIT: badly written, too mathematical straight, too cartesian, too scholar. I don't know how to explain rightly what I mean. In french: "trop droit"). To me, if my 16,3% glitch chance through 10 000 trials is enough to be at 1% interval confidence at a probability à 80% (standartly, most tests use 95% but I don't need that accurracy just to discuss of an RPG glitch-system proposal). -ps: still not wanting to go in exact calculations-.


Anyway, just calculate them. I didn't feel like doing it. Just do it.

To me it was enough to find out that glitches just falls off with high pools and under which dice ranges they're still likely to occur.



ps:
(for non math users, in the case of craps: you flip a coin and you tell yourself: I'll look how many times I'll win. How many times should I flip the coin to have my calculated value at 0,1% difference from the real win probability (with a likeliness of 95%). That's 1 million.

But there's two factors you can play with: the interval-the confidence probability.

I made a test on computer:

CODE

Wins    520    5056    49920    249940    500380
Trials    1000    10000    100000    500000    1000000
%    52,00%    50,56%    49,92%    49,99%    50,04%

My point is:
5056/10000=50,56% is enough for me to get the feeling of a rule.

But I welcome the calculation of the real values.

[Sengir]

Good example of craps... to get to 0.1% confidence interval on a nearly 50:50 event... nearly 1million!!! results need to be run.

Which is a) overkill for the purposes of a discussion like this and b) generating one million random numbers plus checking how many of them are successes still happens while fetching coffee, even without CUDA tricks.

[_Pax._]

Sk8 and Sengir make good points. For the purposes of idly discussing mechanic, "back of the napkin" accuracy is really all we need in order to get a broad feel for the mechanic in question. Most people playing the game around their dining room table aren't going to be more precise than that, in deciding whether or not they like a system, after all.

There comes a time where too much accuracy (or rather, the effort needed to achieve it) is actually more detrimental than not enough of it.

[Sengir]

As an example, my simulation of hacking probabilities used 10k samples for each dice pool. The resulting graphs show some deviations, but they are easily recognized by just looking at them and do not diminish the information value of the graphs one bit. Burning more CPU cycles on smoother curves would effectively just be eye-candy.

Also, the simulation code was written on the back of a letter before going to bed, whereas I spend a good week trying to figure out a closed formula for it.