Dice Probability

[Gale]

I’ve realized recently that I don’t understand probability as well as I had originally thought. An example I thought of today has been bugging me and I can’t find an explanation for it online I can fully grasp so I figured I would ask the forum about it.

Say you had a d20, and you rolled it exactly twenty times, what are the odds that you roll a 20 at least once?

Normally, I’d consider this to be a fairly simple question and I would say the answer 100%, but this obviously can’t be true; but I don’t know what the right answer would be here or even how to begin calculating it. Can someone explain this to me in a simple manner?

[BRC]

A quick googlign finds me the formula I was looking for
1 - (1 - P)^N
P=probability of success
N=Number of Trials

5% chance of rolling a 20 on a D20, so

1-(1 - .05)^20

1-(.95)^20 (.95 being the chance of the thing NOT happening in any given trial)


plugging that into Google gives me roughly .64


so, if you roll a d20 20 times, you have about a 64% chance of getting any given number at least once.

[Jay R]

It's easier if you turn it around. What's the probability that you don't roll a 20 in 20 rolls?

Each roll has a .95 probability to not be a twenty. Since the rolls are independent, you multiply .95 by itself 20 times.

(0.95)20 = 0.358485922

Either you get at least one twenty or you don't get a 20.

So the probability that you do get a 20 is 1 - 0.358485922 = 0.641514078

[Flickerdart]

Say you had a d20, and you rolled it exactly twenty times, what are the odds that you roll a 20 at least once?
BRC posted the math, but let me walk you through it.

The odds that you do not roll a 20 whenever you roll a d20 are 95% (0.95).

Thus, the odds that you do not roll a 20 when you roll twice are 95%*95% or 0.952 (the odds that you do not roll a 20 the first time, and the odds that you do not roll a 20 the second time).

The odds that you do not roll 20 when you roll thrice are 95%*95%*95% or 0.953, and so on until you get to the odds that you do not roll a 20 after rolling 20 times, 0.9520. This is equal to slightly less than 0.36 (36%).

Since you actually want to roll a 20 at least once, you need the opposite probability. The likelihood of rolling no 20s and the likelihood of rolling at least one 20 should add up to 100% since they encompass every possible result; any trial where you rolled at least one 20 is a trial in which you did not roll no 20s. So you subtract the 36% likelihood of no 20s from 100%, and get 64% - the likelihood that you will roll at least one 20.

[CharonsHelper]

While all of the % answers above are true - assuming that you have 1/20 chance of rolling a 20 on a d20 - it's just that. An assumption.

If you are rolling a physical die - it WILL NOT be perfectly balanced. The closest that physical dice come is the casino d6 - and those they replace from time to time because the wear of rolling them wears them down slightly.

A plastic mold d20? Not even close to as well balanced. Those casino d6s are machined - not created in a mold - and they're pretty expensive. A machined d20 would be far more expensive if someone ever did build a machine to make them.

Technically even computer based dice-rollers aren't PERFECTLY even between the 20 options - but the imbalance is statistically insignificant.

[Strigon]

While all of the % answers above are true - assuming that you have 1/20 chance of rolling a 20 on a d20 - it's just that. An assumption.

If you are rolling a physical die - it WILL NOT be perfectly balanced. The closest that physical dice come is the casino d6 - and those they replace from time to time because the wear of rolling them wears them down slightly.

A plastic mold d20? Not even close to as well balanced. Those casino d6s are machined - not created in a mold - and they're pretty expensive. A machined d20 would be far more expensive if someone ever did build a machine to make them.

Technically even computer based dice-rollers aren't PERFECTLY even between the 20 options - but the imbalance is statistically insignificant.

Technically true, but irrelevant; if you don't know how the die is imbalanced, there's an equally high chance it could be biased towards twenty as any other number. Thus, if you take a random die and roll it, you have a 1/20 chance of getting a 20. This chance only changes if you know what the imperfections are.

[Gale]

Alright, I understand now. I'm kind of surprised though, I always figured the odds would be much higher.

[GrayDeath]

Probably because you have "Aimrollers" in your groups.
I had one once too, his probability of rolling 20s was about 1 in 4.
Until we made him use an electric roller that is. ^^

[Lvl 2 Expert]

The average number of times you'll roll 20 in 20 rolls is still 1. One third of the time you don't roll any twenties at all, but other times you'll roll 2, 3 or even 20 twenties. Those balance each other out.

[Storm_Of_Snow]

While all of the % answers above are true - assuming that you have 1/20 chance of rolling a 20 on a d20 - it's just that. An assumption.

If you are rolling a physical die - it WILL NOT be perfectly balanced. The closest that physical dice come is the casino d6 - and those they replace from time to time because the wear of rolling them wears them down slightly.

A plastic mold d20? Not even close to as well balanced. Those casino d6s are machined - not created in a mold - and they're pretty expensive. A machined d20 would be far more expensive if someone ever did build a machine to make them.

Technically even computer based dice-rollers aren't PERFECTLY even between the 20 options - but the imbalance is statistically insignificant.
It'll also depend on how you hold it in your hand, the nature of the surface it's rolled on, how hard it's rolled, whether it hits anything on the way, how it's released from the hand, whether you shake it in your hand (and how you shake it) or just pick it up and roll it, the height between your hand and the surface...

[Cluedrew]

The way probabilities work is the question "getting at least 1 20 in 20 rolls" comes out to "what is the chance of rolling 1 or 2 or 3 or ... 20 20s in 20 rolls". This is because you have to roll at least 0 twenties and at most 20 twenties. Now to simple version of this question what is the chance of NOT rolling 0 20s in 20 rolls. Jay R covered the rest of that train of thought, in fact you could consider this as prologue to his answer.

[Raimun]

The average number of times you'll roll 20 in 20 rolls is still 1. One third of the time you don't roll any twenties at all, but other times you'll roll 2, 3 or even 20 twenties. Those balance each other out.

That's my experience. You do that usually once for every set of 20 rolls.

Of course, it could be more or it could be less.

Also, sometimes the dice don't follow any rhyme or reason. Play Savage Worlds regularly and eventually, you could swear that Spiral Energy exists and rules of probabilities, well... don't. I mean, what are the odds of rolling a total result of 30+ with a six sided die, twice in a row?

[Lvl 2 Expert]

That's my experience. You do that usually once for every set of 20 rolls.

Of course, it could be more or it could be less.

Also, sometimes the dice don't follow any rhyme or reason. Play Savage Worlds regularly and eventually, you could swear that Spiral Energy exists and rules of probabilities, well... don't. I mean, what are the odds of rolling a total result of 30+ with a six sided die, twice in a row?

Without knowing the system those odds sound pretty low. I've never even rolled 7+ with a six sided die. :smallbiggrin:

[Flickerdart]

Without knowing the system those odds sound pretty low. I've never even rolled 7+ with a six sided die. :smallbiggrin:
I assume the dice are exploding - so you roll a 6, roll again for another 6, roll again for another 6, and then roll two more 6es. The chances of rolling a 6 once are 0.167, and 0.1676 gives us 0.00013. So in 10,000 rolls you will explode past 30 1.3 times.

[Knaight]

Expanding on the answer a bit: What we're looking at here is a binomial distribution, where you either get a success (rolling a 20) or a failure (rolling anything else). Lets call those P and Q, where Q=(1-P). We also have a number corresponding to the number of trials (rolls), which by convention is called n.

The way calculating binomial probabilities works is basically that you first calculate the probability of getting the combination of successes and failures as if there is one way to do it, then multiply by the number of distinct combinations that produce the same combination. For a simplified example, consider flipping coins. Both success and failure have a 0.5 probability, so for any combination of heads and tails, the probability before considering duplicate results is 0.5^n. However, duplicates make things more likely. With a three coin example, there's a 1/8 chance of all heads or all tails, and a 3/8 chance of two heads and a tail or two tails and a head. The reason is that for the first you just have HHH and TTT; for the second you have the combination of HTT, THT, TTH and THH, HTH, HHT.

Coming back to dice, lets call the number of successes you're looking for k. With the already answered example question k is 0. Ignoring duplicates, you have the probability of PkQ(n-k) . The reasons behind that should be pretty intuitive. You then need to compensate for the duplicates. The formula for that is n!/((n-k)!*k). That is just multiplied by the result you want. If you need multiple added up, you can either manually add them all up, or subtract the ones you don't have from 1, which must be the sum of all the probabilities. With this question, the latter is much easier.

The explanation above is a bit short, and might be confusing*. Fortunately, I did a quick check, and Khan Academy has a series of videos (https://www.khanacademy.org/math/probability/random-variables-topic/binomial_distribution) for the binomial distribution. I haven't actually watched them, but they probably handle this nicely.

*It isn't if I explained it well, but my optimism on that point is limited.

[Jarawara]

I assume the dice are exploding - so you roll a 6, roll again for another 6, roll again for another 6, and then roll two more 6es. The chances of rolling a 6 once are 0.167, and 0.1676 gives us 0.00013. So in 10,000 rolls you will explode past 30 1.3 times.


If you have exploding dice, the concern is not the result of die toss, but rather how far away you can toss them! I mean seriously, you could lose a finger depending upon the blast radius.

As to getting a result of 30+, I suppose it's possible the fragments of dice could fall in a way that the "3" face falls next to a lower half of the "6" face, making it look like a small zero, like a "o". And if a fragment of the "4" face, appearing now to be a "+", could fall next to it.

But I don't know how to calculate the odds of that happening.
At least not by hand, especially if I had just lost a finger.

[RedMage125]

I just want to point out that the numbers posted by BRC are absolutely correct....

IF

...you are using something scientifically proven to give you EXACTLY even probabilities of each side being rolled. Most virtual dice apps are designed to do exactly this. As are some dice manufacturer's products (like GameScience dice).

HOWEVER, most dice are not that accurate. Chessex, and other major producers of polyhedral dice are certainly not. Chessex, especially, is guilty of making dice with the primary focus of LOOKING good. Their dice are mass-produced in molds without any means of forcing air bubbles out, so right there you have an uneven weight distribution inside the die. Some of them are painted in a mass of dice being rolled through swirls of liquid paint. Now, the uneven distribution of paint might not SEEM like a significant amount of weight, but when the object in question weighs a tenth of an ounce or less, even bit can make a difference. Finally, one of the most telling factors is that many dice manufacturers put their dice in rock tumblers as part of the process. This rounds off the edges, the purpose of which is to make the dice roll farther, which is much more dramatic. The side effect of which is that not all edges are rounded evenly, so some edges of the die inhibit momentum more than other edges.

The cumulative effect of all of this is dice that are more likely to produce some number more than others. This might be evenly balanced (i.e. a die that rolls 1-3 and 18-20 a lot, but rarely the numbers in between), or it may just be skewed towards one or the other end of the spectrum. You know how many gamers have superstitions about "lucky" d20s? They may well be correct. I have a few d20s that roll an alarmingly high amount of 17+. So much so that-as a matter of professional ethics-I never use them when I DM, only when I play.

So, while I do not contest the math that people much better at statistics than myself have posted, hat math is only correct if you have a die that CORRECTLY has an even chance of giving you each number on the face. You can buy "precision" dice from some manufacturers. These dice are usually transparent or translucent, so you can see the lack of air bubbles inside them. They also usually have very sharp edges, much like the 6-sided dice you will find at a casino (as opposed to the ones that you see in a Milton Bradley board game). These sharp edges arrest momentum much faster, so the dice do not roll as far, and of course, they are not as pretty. They also often still come with a few burrs of plastic left from the mold tree, which you must file down yourself.

But bottom line, if you want to know the probability of numbers for YOUR dice, the only scientific way to determine this is to roll your d20 100 times (or more, if you want to be SUPER ACCURATE), and record the number of times each number shows up. Ideally, a perfect die shows each number 5 times, but there's always an element of chance and other factors (sweat on your hands, imperfections in the surface you roll on, etc). I'd say if no single number shows up more than 7 or 8 times, you have a fairly even die. Or you might have a die that rolls 20 more than 10 times out of the 100 and NEVER gets an 8 or 9. This is likely due to the factors I have been discussing.

Me? I'll keep my "lucky" dice. I have 2 bags of dice, one filled with lots of beautiful sets and a few oddball outliers (one of my luckiest dice is plain white with red numbers). The other bag is my "loaner" dice, full of dice I got in mass quantities (like buying them by the pound). These I lend to players who forgot theirs, or are new and do not have any yet.

[Lvl 2 Expert]

But bottom line, if you want to know the probability of numbers for YOUR dice, the only scientific way to determine this is to roll your d20 100 times (or more, if you want to be SUPER ACCURATE)

As a side comment, 100 rolls for a 20 sider is a very small number if you want to do statistics with it. (It's a lot of work, but the accuracy of the test will be too low to tell you much of anything.) I've forgotten how to calculate this and am too lazy to look it up, but if you model the binomial distribution as a normal distribution you can calculate how much of a deviation in a test would prove with 95% (or any other number you like) certainty that the zero hypothesis "this die rolls 20 ones every 20 rolls" is not true. I think that number will be pretty far from the average 5 times with just 100 rolls. I wouldn't be surprised if I rolled only 2 twenties in such a test for instance, or 10 of them. However, if you roll a thousand times you're probably not going to roll only 20 twenties with a fair die, or a hundred of them, meaning your test gets more accurate for finding imperfect dice.

As a side note to that side comment, apparently (according to an article I also can't find right now) if you test dice in these manners, most cheap dice roll more 1's than 6's. This is kind of surprising, as one might expect that the hollowed out pips make the sides with large numbers lighter. But apparently the air that gets trapped in the pips slows down the die as it rolls over those sides or something, increasing the odds of them stopping there. Dice with numbers on them or pips that have been filled with one material or another should not suffer the same deviation.

And as a note to a note to a note: The thing that brings everything full circle is of course that if you prove for each number on your dice that they are fair with a 95% certainty you're still on average going to find one bad number per die, if you're testing perfectly fair dice. This means that you'll find one or more errors in about two third of your dice, because we already did that calculation. :smallcool:

[Flickerdart]

I just want to point out that the numbers posted by BRC are absolutely correct....

IF

...you are using something scientifically proven to give you EXACTLY even probabilities of each side being rolled. Most virtual dice apps are designed to do exactly this.
Surely this is not true, since computers use pseudorandom number generation by necessity. There are true random number generators like random.org that use hardware to sample atmospheric noise or whatever, but most virtual dice apps just go Math.random() or whatever the equivalent is in your favourite language.

[Tanarii]

As a side comment, 100 rolls for a 20 sider is a very small number if you want to do statistics with it. (It's a lot of work, but the accuracy of the test will be too low to tell you much of anything.)Statistical noise is about SQRT(X)/X. So for 100 tests, it's around or about +/- 10%. For 1000 tests it'd be around +/- 3%. TBH though that's for a truly random subset of a much larger population, so it may not apply.

[Khedrac]

Something to watch out for is that the apparent probability of thinks like rolling 30+ using exploding dice (or reroll and add on a max result) is not the same as the actual probability....

Assuming honest dice:
Consider the likelihood of rolling two natural 100s in a row using D% (because the maths are easy).
Fairly obviously this is 0·01 × 0·01 = 0·0001 = 1 in 10000.

However, most people will only think about this as a possible occurrence after the first 100 has been rolled.
So, what's the chance for rolling 100 in a single roll just after 100 has been rolled? = 0·01 or 1% - one hundred times as much.
This is because the two rolls are independent - they do not affect each other and by saying that 100 has just been rolled you have drastically reduced the number of possible outcomes.

So, although the chance of two consecutive 100s is one one-hundreth of a percent, it will occur every hundred rolls of 100 :smalleek:

Looking at exploding D6s - one in 6 initial 6s will roll a second 6 and continue exploding - a lot more than people expect because they think about the overall chance (1 in 36).

So to get 30+, this requires five consecutive 6s (the 6th roll is then irrelevant)...
The probablity of 5 consecutive 6s is 0·0129%
But this means that 0·0772% of all initial rolls of a 6 will go on to be over 30.
Or 0·463% of all double 6 rolls.
2·778% of all triple 6 rolls.
Etc.

(I have hit something close to the reverse effect in play. One player thought that using the percentages and a D% was a more accurate way of determining a random number from 1 to 7 than rolling a D8 and re-rolling 8s. This was because he assumed one would not keep re-rolling 8s until a non-8 resulted - we never did find out what he assumed the 8 would count as if not re-rolled.)

[Lvl 2 Expert]

Statistical noise is about SQRT(X)/X. So for 100 tests, it's around or about +/- 10%. For 1000 tests it'd be around +/- 3%. TBH though that's for a truly random subset of a much larger population, so it may not apply.

Only 10% So if you'd throw 20 4 or 6 times instead of 5 it'd already be suspect? That sounds very counter intuitive. (I'd also get other numbers if I went after the chance to throw "not 20", so that can't be right.)

So, are we talking 10% of the total number of rolls here? it's still not suspect if you throw 20 14 times out of hundred?

Let's just roll that out with anydice:
7, 2, 17, 11, 5, 20, 18, 10, 19, 6,
7, 6, 20, 16, 1, 9, 20, 2, 12, 4,
9, 3, 5, 9, 1, 10, 8, 17, 15, 15,
8, 12, 2, 11, 13, 16, 6, 14, 2, 9,
18, 3, 9, 19, 5, 8, 15, 11, 8, 17,
6, 1, 2, 12, 9, 16, 4, 16, 12, 18,
14, 14, 14, 14, 16, 6, 1, 20, 4, 2,
19, 2, 15, 5, 4, 12, 6, 18, 15, 2,
13, 2, 10, 20, 6, 5, 13, 14, 4, 4,
5, 9, 19, 3, 5, 13, 2, 7, 19, 4

Using Ctrl+f I find 5 20's, 5 19's, 4 18's, 3 17's, 5 16's, 5 15's, 6 14's, 4 13's, 5 12's, 3 11's, 3 10's, 7 9's, 4 8's, 3 7's, 7 6's, 7 5's, 7 4's, 3 3's, 10 2's, 4 1's.

Now, these are only twenty tests, and thus not very accurate, but if a 10% deviation would fall outside of a 95% certainty interval this die was seriously crooked. Based on these numbers I'd say a number wouldn't be suspect until it's at least 3 points higher or lower than 5, but 10 points seems a stretch as well...

Not to mention that the normal distribution assumes the odds flare off in the same way on both sides, which doesn't really work if the minimum number is 1, the average is 5 and the maximum is 100... :smallfrown:

[veti]

Now, these are only twenty tests, and thus not very accurate, but if a 10% deviation would fall outside of a 95% certainty interval this die was seriously crooked. Based on these numbers I'd say a number wouldn't be suspect until it's at least 3 points higher or lower than 5, but 10 points seems a stretch as well...

That's a subtle statistical fallacy, summed up here (http://xkcd.com/882/). It's a very common mistake even by people who are paid and expected (https://en.wikipedia.org/wiki/Lucia_de_Berk#Statistical_arguments) to know about this stuff.

Perform a statistical test once, to a 95% confidence level, and the chance of a false positive is only 5%. Perform the same test 20 times, and the chance of at least one false positive is - about 64%...

[Walnut]

That's a subtle statistical fallacy, summed up here. It's a very common mistake even by people who are paid an to know about this stuff.

Perform a statistical test once, to a 95% confidence level, and the chance of a false positive is only 5%. Perform the same test 20 times, and the chance of at least one false positive is - about 64%...
In this case it's a little more complicated than that, since the twenty aren't independent of each other. Though calculating the exact chance isn't going to be fun.

[Lvl 2 Expert]

Perform a statistical test once, to a 95% confidence level, and the chance of a false positive is only 5%. Perform the same test 20 times, and the chance of at least one false positive is - about 64%...

Yes, I know. I stated that a few posts up, it's the same calculation as the one that started this thread.


In this case it's a little more complicated than that, since the twenty aren't independent of each other. Though calculating the exact chance isn't going to be fun.

Right, another complication.

This was not a very clever way to avoid looking up the actual calculation, was it? :smallbiggrin:

[Raimun]

The funny thing about Savage Worlds and its exploding dice is that some people (including me) sometimes keep some (or even all) of the Skill-dice pretty low (D4 or D6) just because they experience that a lower skill rating can quite often be drastically more effective than a high Skill-die (D10 or D12).

Now, I remember I once read calculations that proved otherwise (high dice are about as balanced) and as far as I could tell, the math was solid (even taking the Wild Die to account) and that was how it should be. Still, my experience says otherwise and I know for a fact that relying on low exploding dice is usually drastically more effective: you will roll higher. As an added bonus, they also save your Skill points and make you overall more versatile, since you have a ton of low die skills.

[Lvl 2 Expert]

Crap, crap crap crap.

I typed out a post, but I got logged out and lost it. The gist of it was this:

D4:
Chance to roll 3 or better = 1/2, chance to roll 7 or better = 1/8, chance to roll 11 or better = 1/32, chance to roll 15 or better = 1/128. chance to roll 19 or better = 1/512

D10:
Chance to roll 3 or better = 4/5, chance to roll 7 or better = 2/5, chance to roll 11 or better = 1/10, chance to roll 15 or better = 3/50. chance to roll 19 or better = 1/50

In the real world weird things can happen, but part of the explanation may be how our memory works. A D10 exploding ones is not memorable, a D4 exploding thrice is. Thus we remember the times it happens better. If you write down all rolls made over several sessions the D10's should roll higher on average, and will probably produce the highest results overall. If not you may have a lucky D4...

[Raimun]

Crap, crap crap crap.

I typed out a post, but I got logged out and lost it. The gist of it was this:

D4:
Chance to roll 3 or better = 1/2, chance to roll 7 or better = 1/8, chance to roll 11 or better = 1/32, chance to roll 15 or better = 1/128. chance to roll 19 or better = 1/512

D10:
Chance to roll 3 or better = 4/5, chance to roll 7 or better = 2/5, chance to roll 11 or better = 1/10, chance to roll 15 or better = 3/50. chance to roll 19 or better = 1/50

In the real world weird things can happen, but part of the explanation may be how our memory works. A D10 exploding ones is not memorable, a D4 exploding thrice is. Thus we remember the times it happens better. If you write down all rolls made over several sessions the D10's should roll higher on average, and will probably produce the highest results overall. If not you may have a lucky D4...

Generally, D4s and D6s are lucky in Savage Worlds, no matter who is throwing them.

While your math is correct in a certain point of view, it's also true that no die roll in real world has nothing to do with what you rolled previously. Every time you roll a D12 in Savage Worlds, there is a 8,333% chance that your die explodes. Every time you roll a D4, there is 25% chance. While I know that math proves me (eventually, in a long run and with empirical tests) wrong, there is still a lot to learn from practical experience.

Of course, all this is most likely skewed because D4s and D6s get thrown a lot more, since they are more easily acquired. It's quite possible to have something like over 15 different Attributes/Skills/Damage-dice that use D4 or D6 but getting 3-4 D10s and/or D12s gets a bit tricky and unversatile. Wild Die and Bennies affect the outcome as well, if they are appliable to the particular roll.

[Lvl 2 Expert]

While your math is correct in a certain point of view, it's also true that no die roll in real world has nothing to do with what you rolled previously. Every time you roll a D12 in Savage Worlds, there is a 8,333% chance that your die explodes. Every time you roll a D4, there is 25% chance.

That's true in statistics as well...

When you've already rolled three fours the chance of the next roll being four is still 1/4. But the chance that the next four rolls will all be fours, no matter what you rolled previously, are not 1/4 (because that would mean there's a 1/4 chance of rolling infinite fours), they're 1/4*1/4*1/4*1/4, and that's 1/256. If you'd use a single sixteen sided die your chances of rolling the same numerical value would be 1/16.

You're warning me for the famous roulette wheel fallacy (no idea if anyone calls it that, but the example is always a roulette wheel). When red has come up 25 times the odds of red on the next run are still (a little less than) 1/2, not one in a bazillion. But you're trying your best to fall for the reverse idea. The odds that the next 26 runs will all end up red really are much lower than the odds of them being say 13 times red and 13 times black, and that's because there are a lot more different orders to get those 13 reds and 13 blacks in. You can demonstrate that to yourself by imagining the first 13 rolls. Any series that comes up, even one of the worst case scenarios, leaves at least one string of 13 more rolls that will lead to the result being 13 reds and 13 blacks. However, there is only a single string of 13 rolls (all red) that leaves only one single string of 13 more rolls (all red as well) open to reach the goal of 26 reds. If the odds of every exact string of rolls are the same (and they are, 1/2^26, ignoring the zeroes for simplicity), the amount of ways something has to randomly happen tell you what the odds are.

Back to dice: The value of your roll goes up through addition, every success is four more points, while the odds go down by multiplication, the next success always only has a 1/4 chance of happening. The smaller die are never going to keep up to the larger ones, multiplication will take care if that. The odds would become a lot "fairer" if the value went up by multiplication too. If you roll four and four your value is four times four, that's sixteen. And the chance of that happening is 1/4*1/4, which is 1/16, the same odds you got with the 16 sider.

[Raimun]

That's true in statistics as well...

When you've already rolled three fours the chance of the next roll being four is still 1/4. But the chance that the next four rolls will all be fours, no matter what you rolled previously, are not 1/4 (because that would mean there's a 1/4 chance of rolling infinite fours), they're 1/4*1/4*1/4*1/4, and that's 1/256. If you'd use a single sixteen sided die your chances of rolling the same numerical value would be 1/16.

You're warning me for the famous roulette wheel fallacy (no idea if anyone calls it that, but the example is always a roulette wheel). When red has come up 25 times the odds of red on the next run are still (a little less than) 1/2, not one in a bazillion. But you're trying your best to fall for the reverse idea. The odds that the next 26 runs will all end up red really are much lower than the odds of them being say 13 times red and 13 times black, and that's because there are a lot more different orders to get those 13 reds and 13 blacks in. You can demonstrate that to yourself by imagining the first 13 rolls. Any series that comes up, even one of the worst case scenarios, leaves at least one string of 13 more rolls that will lead to the result being 13 reds and 13 blacks. However, there is only a single string of 13 rolls (all red) that leaves only one single string of 13 more rolls (all red as well) open to reach the goal of 26 reds. If the odds of every exact string of rolls are the same (and they are, 1/2^26, ignoring the zeroes for simplicity), the amount of ways something has to randomly happen tell you what the odds are.

Back to dice: The value of your roll goes up through addition, every success is four more points, while the odds go down by multiplication, the next success always only has a 1/4 chance of happening. The smaller die are never going to keep up to the larger ones, multiplication will take care if that. The odds would become a lot "fairer" if the value went up by multiplication too. If you roll four and four your value is four times four, that's sixteen. And the chance of that happening is 1/4*1/4, which is 1/16, the same odds you got with the 16 sider.

Heh, you clearly know more about statistics than me. I'm not doubting any of this in the long run, even if my experience playing Savage Worlds does not correlate with the math. For example:


I assume the dice are exploding - so you roll a 6, roll again for another 6, roll again for another 6, and then roll two more 6es. The chances of rolling a 6 once are 0.167, and 0.1676 gives us 0.00013. So in 10,000 rolls you will explode past 30 1.3 times.

Result of 30+ about once every 10, 000 rolls? That's not my experience. I haven't kept track but I'm willing to bet that in three on going Savage Worlds-campaigns I've rolled 30+ with a D6 (not counting D4s) something like well over 10 (perhaps even close to 20?) times. What's notable is that by quick estimation, the number of D6s I've rolled in three campaigns is not even near to half of 10,000.

Of course, "1.3 times out of 10,000" is not the real probability. Calculating overall statistics in Savage Worlds might not even be possible. Sometimes, circumstances, cinematic rules and even player agency affects them.

First the Wild Die. Every time a PC (or important NPC) makes a Trait-roll (ie. Attribute or Skill), he also rolls the Wild Die (D6) as well. Wild Die can also explode but you only re-roll (and add to the total) those dies that actually expolode. So, everytime you roll a Skill of D4, you also roll a Wild Die of D6. Let's say D4 gives you '3' and D6 '6'. You re-roll the '6' (until you cease to roll '6') but don't re-roll '3'. This time, you naturally use Wild Die's result and completely ignore '3'.

Then there are Bennies. Every Benny is a re-roll (among other things) and you can choose when and how many times to use one. At that point, you re-roll all the dice used for the Trait-roll (ie. (Attribute or Skill)+Wild Die). You can even re-roll a re-roll, as long as you have Bennies to spare (even if it might not be wise :smalltongue: ). You use the highest of these results, not the last result. You generally begin a session with 3 of them (could be more, could be less) and you can earn more by roleplaying your character and doing cool stuff.

Finally, there are some fixed and circumstance-based modifiers (could be positive or negative) but these are generally pretty low numbers. If you're an excellent driver (or have an Improved Trademark Weapon) you would get +2 to your roll every time you make a Skill-roll. This is added after you have rolled and added up the dice. It's pretty rare to get +6 or +8, though sneak attack (using Stealth-skill) gives you +4 to attacking Skill-roll and Damage. GM might give other modifiers depending on favorable or unfavorable circumstances.

Surely, these factors help a lot when you try to roll 30+ with a D6 but that doesn't change the fact that you still have to roll a '6' six (or five, w/outstanding modifiers) times in a row, with a single go.

This kind of unpredictable chaos is one of the reasons I like Savage Worlds so much. :smallsmile:

[Flickerdart]

Surely, these factors help a lot when you try to roll 30+ with a D6 but that doesn't change the fact that you still have to roll a '6' six (or five, w/outstanding modifiers) times in a row, with a single go.
Well...they kind of do change that fact. Because you're no longer rolling six in a row, just six within a bunch of dice. Of course the probability is going to be orders of magnitude higher!

[Aliquid]

Alright, I understand now. I'm kind of surprised though, I always figured the odds would be much higher.I find it helps to visualize these things by bringing it down to a more manageable scale.

Picturing what would happen if you rolled a d20 20 times... it is hard to imagine all the possible outcomes.

Change that to "If you rolled a d2 2 times what are the odds of rolling a 2 at least once?". That we can picture in our heads.

Possible outcomes for rolling a d2 2 times:
a) First roll: 1, Second roll: 1
b) First roll: 1, Second roll: 2
c) First roll: 2, Second roll: 1
d) First roll: 2, Second roll: 2

You can easily see that 3 of those 4 possible outcomes have at least one 2. So you have a 75% chance to roll a 2 at least once. (or 75% chance to get a 'heads' at least once if you flip a coin twice)

Now increase the scenario to "if you roll a d3 3 times, what are the odds of rolling a 3 at least once?"... the outcomes increase quite a bit.
1 1 1
1 1 2
1 1 3
1 2 1
1 2 2
1 2 3
1 3 1
1 3 2
1 3 3
2 1 1
2 1 2
2 1 3
2 2 1
2 2 2
2 2 3
2 3 1
2 3 2
2 3 3
3 1 1
3 1 2
3 1 3
3 2 1
3 2 2
3 2 3
3 3 1
3 3 2
3 3 3

If you look in there, 19 of the 27 outcomes have at least one 3. Which makes the odds are 70.3%

So you can see the trend. The bigger the die, the lower the probability of rolling a x at least once out of x tries.

[Raimun]

Well...they kind of do change that fact. Because you're no longer rolling six in a row, just six within a bunch of dice. Of course the probability is going to be orders of magnitude higher!

You still need to throw six sixes in a row. You just get multiple tries (at the most 3-5*) if you use Bennies.

*Using that many for one roll is usually wasteful and an excellent way to get your character killed later on when there's no Bennies left for Soaking damage.