The probability and predictability of dice.

[Dust Raven]

Re: The probability and predictability of dice.

 

Actually, the RKA 3d6+1 has a mean Stun result of 30.667. An RKA doesn't produce the same range and mean as an EB of the same cost, in part because of the -1 on the stun multiple, and in part because it's a multiple. Multiplying a dice probability graph by a number doesn't produce a graph identical to a larger group of dice.

 

I will be posting a graph of the potential Stun damage of a 3D6+1 RKA soon.

That was the point of my example n fact. The possible results of either roll don't graph the same, with the KA having are more evenly distributed or flat spread.

[SteveZilla]

Re: The probability and predictability of dice.

 

That was the point of my example n fact. The possible results of either roll don't graph the same' date=' with the KA having are more evenly distributed or flat spread.[/quote']

 

But because of the differeing mechanics for getting the result, it's comparing oranges to apples. A closer comparison (one that doesn't use a multiple) would be comparing an Xd6 to an (X-Y)d6 + Y*3.5 (like 12d6 vs. 4d6+28).

 

And Prestigitator was correct in that Xd6 will, on average, get more damage through any defense greater than X, than an (X-Y)d6 + Y*3.5.

 

Here's the graph I promised for the 3d6+1 RKA Stun (html file & it's folder zipped together).

[SteveZilla]

Re: The probability and predictability of dice.

 

Mighty cool and repped when I can.

 

Any chance I can get a copy of that page/program to save to my computer? I currently don't have internet access at home and that could come in handy when I least expect it to.

 

Thanks. Here is the group of plain html pages. The zip file contains all 30, and a folder with the stylesheet file and the red "bar" image. The 30 files have links to all 30 at both the top and the bottom.

 

Enjoy!

[Dust Raven]

Re: The probability and predictability of dice.

 

Thanks. Here is the group of plain html pages. The zip file contains all 30, and a folder with the stylesheet file and the red "bar" image. The 30 files have links to all 30 at both the top and the bottom.

 

Enjoy!

 

I was about to say a huge THANK YOU but when I opened the zip it was empty except for a single file called "dice_graphs" with no extension. Don't worry, it's not your fault. For some reason I have this problem occasionally dowloading files attached to posts on these boards. Could you email it to me? dustraven(at)tlo3t.net

[Dust Raven]

Re: The probability and predictability of dice.

 

But because of the differeing mechanics for getting the result' date=' it's comparing oranges to apples. A closer comparison (one that doesn't use a multiple) would be comparing an Xd6 to an (X-Y)d6 + Y*3.5 (like 12d6 vs. 4d6+28).[/quote']

Agreed, though the X vs X-Y example is really no better. Apples and Lemons if you will. The only comparisons you can make to determine the difference in predictability between a greater and lesser number of dice is to just use greater and lesser numbers of dice. Adding (or multiplying or subtracting or any other modification) something to one group distorts the results. Though if you do the same to both groups it balanced again.

[PhilFleischmann]

Re: The probability and predictability of dice.

 

This is not an example of the Law of Large Numbers.

I didn't say that it was. You gave me something to review, so I gave you something to review.

 

This is not an example of less dice vs more dice. Is that clear? You wanted clarity.

It is clear to me that you consider the fixed number 7 to be "two dice". and thus, to you, 10d6 and 2d6+28 are both examples of "ten dice." Is it clear to you that when I say "dice," I mean dice that are actually rolled? Clarity is a two-way street.

 

I take this an an insult and an end to our discussion.

And nothing you've said was an insult to me, right?

[PhilFleischmann]

Re: The probability and predictability of dice.

 

I presumed he was trying to compare two relatively similar dice rolls, when in fact, he isn't. So I really don't see how any sort of relevant conclusion can be made concerning predictability.

 

10d6 vs 2d6+28

 

These two dice rolls don't have enough in common to come to any conclusion about thier predictablity.

Dude! What did you think we were talking about this whole time?

[SteveZilla]

Re: The probability and predictability of dice.

 

I was about to say a huge THANK YOU but when I opened the zip it was empty except for a single file called "dice_graphs" with no extension. Don't worry' date=' it's not your fault. For some reason I have this problem occasionally dowloading files attached to posts on these boards. Could you email it to me? dustraven(at)tlo3t.net[/quote']

 

Emailed. It worked fine for me when I downloaded it. Strange.

[SteveZilla]

Re: The probability and predictability of dice.

 

Dust Raven, PhilFleischmann, can we please keep this discussion cordial? I know that it's aggravating when we think we are correct and the other person is not being swayed into agreement. Phrasing one's posts neutrally goes a long way to keeping it civil.

 

Agreed' date=' though the X vs X-Y example is really no better. Apples and Lemons if you will. The only comparisons you can make to determine the difference in predictability between a greater and lesser number of dice is to just use greater and lesser numbers of dice. Adding (or multiplying or subtracting or any other modification) something to one group distorts the results. Though if you do the same to both groups it balanced again.[/quote']

 

I have to disagree with you that adding a number to the dice distorts the results. Multiplying the bell curve by another die (or dice) definately does distort the result. The Stun graph for a 3d6+1 RKA bears practically no resembelance to a regular bell curve. It has gaps, and the bars vary in a very non-uniform way. Adding a number to the dice doesn't change the shape of the curve, only the end result values.

 

I can understand that as the fixed amount added to the dice gets larger, the percentage of variance gets smaller. After all, the random element is making up a smaller portion of the total as the fixed element grows. But I fail to see how that fact applies in a game, or affects the chance of rolling any particular result.

 

I.e., The chance to roll 18 on 4d6 is the same as the chance to roll 46 on 4d6+28. The chance to roll 20 or better on 4d6 is the same as the chance to roll 48 or better on 4d6+28. While adding a number to the dice obviously affects the final total, the chance of getting that total depends only upon the random element.

[PhilFleischmann]

Re: The probability and predictability of dice.

 

One more way of explaining what I mean by "predictabiliy":

 

If you try to predict the result of 2d6, you'll want to pick 7, as it gives you the best chance of being right in your prediction. You'll have a 16.667% (1 in 6) chance of being right.

 

If you try to predict the result of 2d6+28, you'll want to pick 35. You'll have a 16.667% (1 in 6) chance of being right. By this measure, 2d6 and 2d6+28 are equally predictable.

 

If you try to predict the result of 10d6, you'll want to pick 35. You'll have a 7.269% (1 in 13.76) chance of being right. By this measure, 10d6 is less predictable than 2d6 or 2d6+28.

 

Let's say you don't mind being off in your prediction by a little, let's say by one. Then you have a 44.444% (4 in 9) chance of being within 1 of your prediction for the 2d6 or 2d6+28. You'll have a 21.575% (less than 2 in 9) chance of being within one of your prediction for the 10d6. Again this measure gives 10d6 as being less predictable than 2d6 or 2d6+28.

[SteveZilla]

Re: The probability and predictability of dice.

 

One more way of explaining what I mean by "predictabiliy":

 

If you try to predict the result of 2d6, you'll want to pick 7, as it gives you the best chance of being right in your prediction. You'll have a 16.667% (1 in 6) chance of being right.

 

If you try to predict the result of 2d6+28, you'll want to pick 35. You'll have a 16.667% (1 in 6) chance of being right. By this measure, 2d6 and 2d6+28 are equally predictable.

 

If you try to predict the result of 10d6, you'll want to pick 35. You'll have a 7.269% (1 in 13.76) chance of being right. By this measure, 10d6 is less predictable than 2d6 or 2d6+28.

 

Let's say you don't mind being off in your prediction by a little, let's say by one. Then you have a 44.444% (4 in 9) chance of being within 1 of your prediction for the 2d6 or 2d6+28. You'll have a 21.575% (less than 2 in 9) chance of being within one of your prediction for the 10d6. Again this measure gives 10d6 as being less predictable than 2d6 or 2d6+28.

 

I was going to point this out, but got beaten to it.

 

Then, after thinking about it some more, I think I might see a flaw with defining predictability this way. By only picking the mean, this method seems to ignore the fact that all the rest of the results have an even less chance of coming out. It would think that 2d6 is equally predictable as 1d6.

 

We could take the average of all the percentages, but there is still a problem. Take the following probability results for 5d4-2 and 3d6:

 

	5d4-2		3d6
Result	%chance		%chance
3	0.09765625	0.462962963
4	0.48828125	1.388888889
5	1.46484375	2.777777778
6	3.41796875	4.62962963
7	6.34765625	6.944444444
8	9.86328125	9.722222222
9	13.18359375	11.57407407
10	15.13671875	12.5
11	15.13671875	12.5
12	13.18359375	11.57407407
13	9.86328125	9.722222222
14	6.34765625	6.944444444
15	3.41796875	4.62962963
16	1.46484375	2.777777778
17	0.48828125	1.388888889
18	0.09765625	0.462962963

 

Since both produce a total of 16 results, both will have the same average %chance. Yet clearly they are not equal. It's possible that they may have different means/ The 5d4-2 has a higher "hill", and lower "ends". The change from being higher to lower happens around 7.5 and 13.5.

 

However, it's easy to realize that the more possible results there are, the lower the average will be. The same 100% total is simply being divided more ways with greater numbers of results. The same happens with a single die with greater numbers of faces. The reason that when comparing the graphs peak-to-peak, some of the results seem to be *more* likely

 

Dust Raven has (IIRC) pointed out that for growing numbers of dice, in terms of numbers of results, the "hill" in the middle doesn't grow as fast as the "plains" on either side of it. The question I have with this method is: What determines the dividing point? Without a precise definition (there may be -- I just don't know if it exists), it becomes subjective. When looking at a graph, most will often agree where the "hill" stops for low numbers of dice, but what about huge numbers of dice where the graph is very close to totally flat?

[PhilFleischmann]

Re: The probability and predictability of dice.

 

Then' date=' after thinking about it some more, I think I might see a flaw with defining predictability this way. By only picking the mean, this method seems to ignore the fact that all the rest of the results have an even less chance of coming out. It would think that 2d6 is equally predictable as 1d6.[/quote']

My post was not a precise definition of predictability (nor was it intended to be), merely an illustration of it.

 

We could take the average of all the percentages, but there is still a problem. Take the following probability results for 5d4-2 and 3d6:

The average of the percentages doesn't tell you anything except the number of possible results in the range. Both 5d4 and 3d6 have an average percentage of 6.25% (1/16). That's why we use things like Standard Deviation to give us more info. The SD of 5d4 is 2.5. The SD of 3d6 is about 2.9. So 5d4 deviates less from the mean, hence it it more predictable.

 

The question I have with this method is: What determines the dividing point?

The Standard Deviation is useful for this purpose.

 

And one more example based on my previous post:

 

Let's say you don't mind being off in your prediction by as much as five. Then you have a 100% chance (complete certainty, complete predictability) of being within 5 of your prediction for the 2d6 or 2d6+28. You'll have a 70.74% (about 7 in 10) chance of being within 5 of your prediction for the 10d6. Again this measure gives 10d6 as being less predictable than 2d6 or 2d6+28.

[SteveZilla]

Re: The probability and predictability of dice.

 

My post was not a precise definition of predictability (nor was it intended to be)' date=' merely an illustration of it.[/quote']

 

I misunderstood. Sorry.

 

The average of the percentages doesn't tell you anything except the number of possible results in the range. Both 5d4 and 3d6 have an average percentage of 6.25% (1/16). That's why we use things like Standard Deviation to give us more info. The SD of 5d4 is 2.5. The SD of 3d6 is about 2.9. So 5d4 deviates less from the mean, hence it it more predictable.

 

The Standard Deviation is useful for this purpose.

 

So, in essense (and in general), the smaller the standard deviation, the more predictable it is?

 

And could you post a formula that can be used for varying numbers, and varying sizes (# of sides) of dice?

 

Let's say you don't mind being off in your prediction by as much as five. Then you have a 100% chance (complete certainty' date=' complete predictability) of being within 5 of your prediction for the 2d6 or 2d6+28. You'll have a 70.74% (about 7 in 10) chance of being within 5 of your prediction for the 10d6. Again this measure gives 10d6 as being less predictable than 2d6 or 2d6+28.[/quote']

 

That seems to make sense.

 

Though I am wondering just how applicable Standard Deviations are to Dice. It seems to me that either a result falls inside a SD or outside of it, which means that, for Dice, SD has to be an integer (or rounded to one).

 

When we round the SD, we get something like this:

 

Nd6	SD		Rounded SD
1	1.707825128	2
2	2.415229458	2
3	2.958039892	3
4	3.415650255	3
5	3.818813079	4
6	4.183300133	4
7	4.518480571	5
8	4.830458915	5
9	5.123475383	5
10	5.400617249	5
11	5.664215156	6
12	5.916079783	6
13	6.157651067	6
14	6.390096504	6
15	6.614378278	7
16	6.831300511	7
17	7.041543391	7
18	7.245688373	7
19	7.444237145	7
20	7.637626158	8
21	7.826237921	8
22	8.010409894	8
23	8.190441584	8
24	8.366600265	8
25	8.539125638	9
26	8.708233652	9
27	8.874119675	9
28	9.036961141	9
29	9.196919774	9
30	9.354143467	9

 

While this shows (I think) that SD doesn't apply well to dice because of the rounding issue, it also shows that SD increases at a rate that slows with the number of dice.

[SteveZilla]

Re: The probability and predictability of dice.

 

The statistic that is used to measure this is the coefficient of variation' date=' which divides the standard deviation by the mean to give an idea of how large the deviation is.[/quote']

 

I thought the size of the deviation itself would tell how large it was?

 

The CoV of 4d6 is .24 while 10d6 is .15. The smaller number shows that 10d6 is grouped tighter around the mean than 4d6 is.

 

It is grouped tighter in terms of all possible results. But that "tighter group" still gets larger with greater numbers of dice. For example, the "tight group" for 10d6 goes from 21 to 49, a spread of 29 results; for 30d6 it goes form 87 to 123, a spread of 37.

 

2d6 has a mean of 7' date=' a standard deviation of 2.4 and a coefficient of variation of .34. 2d6+28 has a mean of 35 a standard deviation of 2.4 and a coefficient of variation of .07. Xd6+Y is more predictable than Xd6 because it transfers an absolutely smaller Standard Deviation to a larger mean, making the deviation less significant (a smaller percentage of the result). Xd6+Y is also more predictable than the number of dice normally thrown at that damage class for the same reason, a smaller than normal standard deviation for that mean.[/quote']

 

IMO 2d6+28 has a lower CoV than 2d6 because the relative variation is smaller. I.e., the variation makes up a smaller portion of the total. But the absolute variation is still the same -- 2d6 -- either way. And to me, it's solely the absolute variation (the dice alone) that governs predictability (the odds of getting any particular result). Adding a number to that result doesn't change the odds of that happening.

 

Here are some more problems I have with equating the CoV to predictability. The Coefficient of Variation for Xd6 - 3.5X is impossible to calculate, as that would be a division by zero. Xd6 + Infinity produces a CoV of zero, which means it's absolutely predictable (in theory)? And Xd6 - 3.51X has negative predictability? "Predictability" should not depend upon what the final numbers are -- only the chance of arriving at those numbers. Or, another way of putting it, the location of the bell curve on the number line should have no bearing upon its predictability -- only its shape should.

 

It seems akin to shooting a shotgun with buckshot at a target. Normally, getting further from the target would mean that less of the shot hits the circles near the center, thus greater scatter. Greater scatter would tend to indicate less accuracy/predictability. Yet the angle of the shot spread (each pellet's path) has not changed, and IMO is the true measure of the gun's accuracy/predictability. Looking at just the results on the target leaves out accounting for varying distances. This would lead one to think the same gun has wildly varying accuracy/predictability when that isn’t the case. Distance to the target is a factor that is "external" to the gun, and thus shouldn't be considered when figuring the accuracy/predictability of the gun.

 

I know it isn't a perfect analogy, in that adding distance would be akin to reducing the +Y component from a very large value (they're inversely proportional). And leaving out distance to target is akin to including the fixed component (+Y) in predictability calculations. But otherwise I think it holds up, and illustrates why I think the way I do.

 

The reason is that 5 is a smaller percentage of 35 than it is of 7. This is the point where you have to start adding application. If a result of 7 would be sufficient for what ever I want (maybe an attack against 5 defense) 5 is going to make a big difference. If 35 is sufficient' date=' 5 makes less of a difference.[/quote']

 

I'm not sure what you mean by "This is the point where you have to start adding application". But otherwise, I'd have to say that the above is not necessarily the case in Hero System. Five over what ever you want is still 5 over. It's 5 more of whatever the character is doing. Anything external to the dice is irrelevant to the discussion of the predictability of the dice.

[PhilFleischmann]

Re: The probability and predictability of dice.

 

So' date=' in essense (and in general), the smaller the standard deviation, the more predictable it is?[/quote']

Yes. Although Coefficient of Variation, as Conduit points out, can also be used as a measure of predictability. There is no specific, quantifiable statistical definition of "predictability." The appropriate metric to use depends on the context.

 

And could you post a formula that can be used for varying numbers, and varying sizes (# of sides) of dice?

The one I posted on the previous thread:

 

SQRT[(M^2-1)*N/12]

 

Where M is the number of sides (assuming the dice are numbered in a "normal" way, 1 - M, or 0 - M-1), and N is the number of dice, works as long as all the dice are the same type. For mixing different types together, it's a little trickier. AFAICT, the formula for the sigma of mixed dice would be:

 

SQRT{[(a^2-1)*A+(b^2-1)*B+(c^2-1)*C ... ]/12}

 

where a, b, c, ... are the number of sides on the different dice and

A, B, C, ... are the number of dice of each type.

 

Thus the sigma for d4+d6+d8, which produces a slightly less steep bell curve than 3d6, ought to give a slightly higher Standard Deviation. Let's see:

 

SQRT[(15+35+63)/12] = SQRT(113/12) = SQRT(9.4166667) = a little more than 3 (<3.07)

 

While the sigma for 3d6 is a little less than 3 (>2.94).

 

Though I am wondering just how applicable Standard Deviations are to Dice. It seems to me that either a result falls inside a SD or outside of it, which means that, for Dice, SD has to be an integer (or rounded to one).

If we're dealing with a smooth, continuous phenomenon that statistically produces a normal distribution curve, then 68.2% of the results are within one sigma of the mean. If we're dealing with a discrete, integral phenomenon, such as dice, the 68.2% figure will be off somewhat. The decimal portion of the Standard Deviation will show us just how far off we are. If the sigma is 3.9, then significantly less than 68.2% of the results will be within one sigma, since any result of 4 more than the mean won't be within the SD. But if the SD is 4.1, we'll be much closer to the 68.2% within one sigma.

[PhilFleischmann]

Re: The probability and predictability of dice.

 

I thought the size of the deviation itself would tell how large it was?

You are correct. The Standard Deviation is the size of the standard deviation. That's why it's called that. The Coefficient of Variation is the deviation *relative* to the mean. This can be a useful number to know in some circumstances. AFAICT, it's not that useful in regard to dice games, however.

 

If you want to know how tall i am, it doesn't help much for me to say that I'm 4% shorter than the average adult caucasian male - a *relative* answer. You would then need to know the height of the AACM, and do some math. It would be much more useful to get an *absolute* answer: I'm 5'8". (Whether this is actually 4% shorter than average, I don't know or care.)

 

Here are some more problems I have with equating the CoV to predictability. The Coefficient of Variation for Xd6 - 3.5X is impossible to calculate, as that would be a division by zero. Xd6 + Infinity produces a CoV of zero, which means it's absolutely predictable (in theory)? And Xd6 - 3.51X has negative predictability? "Predictability" should not depend upon what the final numbers are -- only the chance of arriving at those numbers. Or, another way of putting it, the location of the bell curve on the number line should have no bearing upon its predictability -- only its shape should.

Precisely! It may be why you don't hear about CoV as often as you hear about SD. I've seen many spreadsheet programs and calculators that have functions for calculating the Standard Deviation, but I don't recall seeing the term "Coefficient of Variation" that many places outside of this thread. If the CoV is undefined because the mean is 0, does that mean that you can't say anything about the phenomenon's predictability? If everyone starts wearing platform shoes, does that mean their effective height (including the shoes) is more predictable because the mean is higher?

 

If you're discussing temperature of a region, the CoV will vary wildly depending on whether you use Farhenheit, Celsius, or Kelvin. But the Standard Deviation, in absolute terms, will stay the same.