Mathematical Help Required - "4d6, drop lowest" weighted average?

[Larkas]

As everyone knows, the standard way to generate ability scores in 3.X is to roll 4d6 six times, always discarding the lowest roll in a each group of die.

According to the Player's Handbook, the average PC is above average, with mean scores between 12 and 13.

That this generation method provides above average results is quite obvious - at least, above the regular arithmetic mean (which is the same as with 3d6, that is, 10,5).

I've been trying to find the "average that matters" - that is, the weighted arithmetic mean. If I can plot a graphic showing the probability of each result, all the better.

However, the way to do so is not obvious at all. I have to find a way to filter each of the 1,296 results of 4d6 (6^4), taking away the lowest die and summing up the other three for the result. And I have no idea how to do it!

Would a kind, mathematical soul please help me out?

[Segev]

I'll try to hunt it down again myself, but can't recall the actual math off the top of my head. However, if you look up Legend of the Five Rings dice mechanics and math behind them, you may find a clue. They use a "roll and keep" system, and I recall seeing some extensive probabilities done for it.

What the 4d6-drop-low mechanic is is really 4k3, but with d6s instead of d10s. (Read "4k3" as "four-keep-three.")

[Larsen]

i did it on an excel file some time ago, and here is the result :

score-----occurrence-----chance(%)
3 ---------1 ----------- 0,077160494
4 ---------4 ----------- 0,308641975
5 ---------10 ----------- 0,771604938
6 ---------21 ----------- 1,62037037
7 ---------38 ----------- 2,932098765
8 ---------62 ----------- 4,783950617
9 ---------91 ----------- 7,021604938
10---------122----------- 9,413580247
11---------148----------- 11,41975309
12---------167----------- 12,88580247
13---------172----------- 13,27160494
14---------160----------- 12,34567901
15---------131----------- 10,10802469
16---------94 ----------- 7,25308642
17---------54 ----------- 4,166666667
18---------21 ----------- 1,62037037

[GilesTheCleric]

There's also a site (http://catlikecoding.com/blog/post:4d6_drop_lowest) with the graphs made if you prefer the visual.

[prufock]

Anydice.com. The actual average is 12.24 with an SD of 2.85.

What exactly are you planning to use as your weighting factor, though?

[bekeleven]

in 6 rolls you'll get 16 14 13 12 10 9-8.

Here, I built a sim for it. (https://docs.google.com/spreadsheet/ccc?key=0AvDPPQoXw7vOdC1WcVRXQjVWZi1mdWVWTFloS1hGO Hc&usp=sharing) The 6th roll is almost exactly 8.5. If you keep the sims at 10k it may take 30 seconds to run.

[Larkas]

Wow, thanks for all the great and absurdly speedy responses, everyone!


What exactly are you planning to use as your weighting factor, though?

I was going to account for the frequency of appearance of any given total, but with all that has been posted above, I believe it won't be necessary!

Thanks again, guys!

[Eldan]

Here. (https://docs.google.com/spreadsheet/ccc?key=0ApCuoSxQMcvddEtoTWtwNWZLZXItZWVSd0N0VFhPQ 0E&usp=sharing) I did this ages ago. All possible 4d6b3 rolls. Average is 12.24

It took a while to do, too. In case it wasn't clear, the colour of a cell shows which number was dropped. In all gree ncells, the lowest roll is 1, all blue cells, the lowest roll is 2, etc.

[Chronos]

Incidentally, what you're looking for is the just-plain-average. As in, if you rolled up a bunch of ability scores, added them together, and divided by the total number of ability scores, you'd get 12.24 .

While we're at it, the Elite Array and 25 point buy are both designed to be roughly equivalent to the typical results of 4d6 drop lowest.

[Larkas]

Here. (https://docs.google.com/spreadsheet/ccc?key=0ApCuoSxQMcvddEtoTWtwNWZLZXItZWVSd0N0VFhPQ 0E&usp=sharing) I did this ages ago. All possible 4d6b3 rolls. Average is 12.24

It took a while to do, too. In case it wasn't clear, the colour of a cell shows which number was dropped. In all gree ncells, the lowest roll is 1, all blue cells, the lowest roll is 2, etc.

Now that's practically a work of art! o.O Thanks for showing it, Eldan!

[bekeleven]

While we're at it, the Elite Array and 25 point buy are both designed to be roughly equivalent to the typical results of 4d6 drop lowest.
They're really not.

See the link that I posted earlier. "Typical results" rounds closer to a 27 or 28-point buy. WotC rounded 15.65 down to 15 and it screwed everything up.

[Larkas]

Incidentally, what you're looking for is the just-plain-average. As in, if you rolled up a bunch of ability scores, added them together, and divided by the total number of ability scores, you'd get 12.24 .

But that IS a weighted average. :smallbiggrin: If you're adding up all the possible results and dividing by the total number of results, you are automatically giving more weight to the results that show up more frequently. It just so happens that when using 3d6 the weighted average (sum of all 216 results divided by 216) is equal to the plain average ( (3+18)/2 or, if you prefer, (3+4+5+...+16+17+18)/16). That is not the same in 4d6 drop lowest. :smallwink:

[Chronos]

It just so happens that when using 3d6 the weighted average (sum of all 216 results divided by 216) is equal to the plain average ( (3+18)/2 or, if you prefer, (3+4+5+...+16+17+18)/16).Neither (min+max)/2 nor (3+4+5+...+16+17+18)/16 is an average in any meaningful sense for 3d6. There might be some term for either of those concepts, but they're so seldom useful.

[tzar1990]

Neither (min+max)/2 nor (3+4+5+...+16+17+18)/16 is an average in any meaningful sense for 3d6. There might be some term for either of those concepts, but they're so seldom useful.

Actually, they are!

The key thing to realize is that a 3 and an 18 are equally likely, a 4 and a 17 are equally likely, and so on. As such, it actually balances out.

Note that this doesn't normally work - it only happens in this case because the curve is symmetrical.

[Fax Celestis]

http://img.photobucket.com/albums/v216/FaxCelestis/bellCurve3d6.png
http://img9.imageshack.us/img9/5264/4d6droplowest.png
Relevant.

Edit: spoiled for screenstretch.

[Larkas]

Note that this doesn't normally work - it only happens in this case because the curve is symmetrical.

Exactly this! Because the bell curve is symmetrical, the plain average coincides with the highest point in the curve. A weighted average, as explained, will always coincide with this point, regardless of (a)symmetry, but in the case of 3d6, the weighted average is the same as the plain average.


http://img.photobucket.com/albums/v216/FaxCelestis/bellCurve3d6.png
http://img9.imageshack.us/img9/5264/4d6droplowest.png
Relevant.

Edit: spoiled for screenstretch.

Thanks, Fax! :smallsmile:

[Starmage21]

Actually, they are!

The key thing to realize is that a 3 and an 18 are equally likely, a 4 and a 17 are equally likely, and so on. As such, it actually balances out.

Note that this doesn't normally work - it only happens in this case because the curve is symmetrical.

On 4d6 drop lowest, 3 is less likely than 18, because you can only get a 3 with 1 combination of the dice (four 1s), but an 18 can be scored on a larger number of dice thanks to keeping 3 dice which are 6s and the fourth die can be anything.

[Larkas]

On 4d6 drop lowest, 3 is less likely than 18, because you can only get a 3 with 1 combination of the dice (four 1s), but an 18 can be scored on a larger number of dice thanks to keeping 3 dice which are 6s and the fourth die can be anything.

We were talking about 3d6 here.

[Ekul]

My group usually promotes MAD builds, so instead of the conventional method, we do 4d6, reroll all ones, drop lowest.
The expectation of the original is 12.24. Our version has an expectation 13.43.

The probability of getting an 18 rises to 2.7%, odds of a 17 rises to 6.72%, odds of a 16 rises to 11.2%. For 15, 14.5% and for 14, 16%.

All in all, the odds of getting a +2 or higher rises from 35.5% to 51.2%

Pretty nifty.

[Larkas]

While we're at it, the Elite Array and 25 point buy are both designed to be roughly equivalent to the typical results of 4d6 drop lowest.

Analyzing the data, I must say that they missed the target by a bit, then. As you can read here (http://catlikecoding.com/blog/post:4d6_drop_lowest) and confirm in AnyDice (http://anydice.com/) with the formula below, a more spot-on Elite Array would be 16, 14, 13, 12, 10, 9 and a more precise point buy would be 28 points.

ABILITIES: 6 d [highest 3 of 4d6]
loop P over {1..6} {
output P @ ABILITIES named "Ability [P]"
}

Myself, I would be a little less conservative with Ability 5 and up it to 11. As shown in the data below (if you're more of a graph guy, you can also perceive that there), even though the average is 10.41 (and thus suggests rounding down), the probability of getting an 11 is actually higher than getting a 10. Hence, a more appropriate Elite Array would be 16, 14, 13, 12, 11, 9, with a point buy of 29.

"Ability 1",15.661886272460427,1.427252687505418,3,18
#,%
3,2.11042533009e-17
4,3.29732853573e-13
5,2.40060881297e-10
6,4.56989751645e-8
7,0.00000341951599652
8,0.000130072106015
9,0.00275398332265
10,0.0352473739053
11,0.27992456955
12,1.49069021444
13,5.39581352104
14,13.3982261101
15,22.6399788298
16,26.6873037109
17,20.7331397967
18,9.33678835276

"Ability 2",14.174227877278518,1.4373365097302584,3,18
#,%
3,1.64001152401e-13
4,5.11020584269e-10
5,1.22905367555e-7
6,0.00000956978963161
7,0.000337133312915
8,0.00662069212451
9,0.0775083676596
10,0.574525690987
11,2.72701980263
12,8.75221441626
13,18.8517274413
14,26.8467826565
15,24.3130026304
16,13.5530770794
17,3.92004460622
18,0.37712978997

"Ability 3",12.955107933993867,1.4639666317837485,3,18
#,%
3,5.3104940703e-10
4,3.29736928888e-7
5,0.0000260912590551
6,0.000827407525721
7,0.0136682413814
8,0.138170044816
9,0.892334020316
10,3.82572865325
11,10.8457573886
12,21.0009225125
13,26.9916793171
14,22.1652497967
15,10.8611870388
16,2.92500721717
17,0.331239267542
18,0.0082026726789

"Ability 4",11.760929505983505,1.529983986177594,3,18
#,%
3,9.17193183557e-7
4,0.000112937346976
5,0.00290704516701
6,0.0372259355189
7,0.28638601114
8,1.48331938715
9,5.26637032767
10,13.0323703033
11,22.0385695402
12,25.7630512027
13,19.8032810281
14,9.4208566823
15,2.51739717853
16,0.332848175775
17,0.0152025841094
18,0.000100743828584

"Ability 5",10.411356502012051,1.6580459100525937,3,18
#,%
3,0.000891225290976
4,0.0212066045571
5,0.172719397297
6,0.879496020131
7,3.11785373614
8,8.21937571256
9,15.9669468663
10,22.7409354462
11,22.8963023332
12,16.1617919989
13,7.44223057809
14,2.05839586757
15,0.301689158645
16,0.0197944027742
17,0.000369991132654
18,6.61178080321e-7

"Ability 6",8.504084500863689,1.9490207684732168,3,18
#,%
3,0.462070819948
4,1.8305319797
5,4.45397697276
6,8.80466324356
7,14.1743440511
8,18.8560877949
9,19.9237160644
10,16.2726740138
11,9.73094488423
12,4.14614447007
13,1.14489774399
14,0.184562960884
15,0.0148933119665
16,0.000487932586174
17,0.00000375429364592
18,1.81002994222e-9


My group usually promotes MAD builds, so instead of the conventional method, we do 4d6, reroll all ones, drop lowest.
The expectation of the original is 12.24. Our version has an expectation 13.43.

The probability of getting an 18 rises to 2.7%, odds of a 17 rises to 6.72%, odds of a 16 rises to 11.2%. For 15, 14.5% and for 14, 16%.

All in all, the odds of getting a +2 or higher rises from 35.5% to 51.2%

Pretty nifty.

Hey, now that's interesting! I've been toying with 5d4+2, drop lowest myself. Check out the formula below in AnyDice!

ABILITIES: 6 d [highest 4 of 5d4]
loop P over {1..6} {
output P @ ABILITIES+2 named "Ability [P]"
}

[Barstro]

the colour of a cell shows which number was dropped. In all gree ncells, the lowest roll is 1, all blue cells, the lowest roll is 2, etc.

Thank you for explaining. I could not figure the color scheme out.

Another nice visual, if you were bored enough to do it, would be a color for each result number (all 15s are blue, etc) and comparing the same to a chart of 3d6.

[Chronos]

Quoth tzar1990:

Actually, they are!
Both of those "methods" happens to give the correct answer in this particular situation, but then, so would the "method" of "take 173.7-163.2". In order for a method to be right, it has to work in all relevant situations, not just a few particular ones.


Quoth Larkas:

A weighted average, as explained, will always coincide with this point, regardless of (a)symmetry, but in the case of 3d6, the weighted average is the same as the plain average.
No, the weighted average could be anything, depending on how you weight it. The plain average will, however, always coincide with the peak for a monomodal symmetric distribution.

Actually, strictly speaking, all averages are weighted averages, but we default to using 1 for all of the weights. Thus, for instance, the average of 3d6 is (1*3 + 1*4 + 1*4 + 1*4 + 1*5 + 1*5 + 1*5 + 1*5 + 1*5 + 1*5 + ... + 1*16 + 1*16 + 1*16 + 1*16 + 1*16 + 1*16 + 1*17 + 1*17 + 1*17 + 1*18) / 216. If we wanted some other weighted average, we could replace all of those 1s by other numbers.

[Bacchus]

I wonder what gives a better average.
4D6 drop lowest
or
3D6 with 1 being a trick die with 2 6's and no 1's

Obviously rolling 3 trick die will result in the highest average.

Is there a way to check this on any of those websites?
Or the stats equation. i can't remember at the moment.

[Larkas]

No, the weighted average could be anything, depending on how you weight it. The plain average will, however, always coincide with the peak for a monomodal symmetric distribution.

Actually, strictly speaking, all averages are weighted averages, but we default to using 1 for all of the weights. Thus, for instance, the average of 3d6 is (1*3 + 1*4 + 1*4 + 1*4 + 1*5 + 1*5 + 1*5 + 1*5 + 1*5 + 1*5 + ... + 1*16 + 1*16 + 1*16 + 1*16 + 1*16 + 1*16 + 1*17 + 1*17 + 1*17 + 1*18) / 216. If we wanted some other weighted average, we could replace all of those 1s by other numbers.

Okay, sorry, I've been very unclear about the concepts I'm using here, and maybe ended up confusing people in the process. Let me try to explain.

I wanted the weighted average of the values between 3 and 18 for 4d6, drop the lowest. To find this, I'd have to know how many times, in an universe with all the possible rolls, each of the values show up. This number of times would be the weight of each value. The plain average between the values would be 10.5, the weighted average, 12.24. This happens to be the plain average of all the possible rolls too, since there you take into account each roll (and not the value thereof), and each roll has a weight of one (because, even if other rolls with the same value show up, that -specific- roll is singular). This is in fact a much easier way to find what I was looking for. Talk about taking the long road.

When we started talking about the average of 3d6, all the values converge, since the bell curve is symmetric. If we make a weighted average for the values, each value and its "mirror" will have the exact same weight, thus cancelling out any influece of weight. A plain average will thus reach the same results. And a plain average of all rolls will also reach the same results, for the exact same reason.

Eh... I don't know if that clarified anything, but one can hope? :smallfrown:

[erikun]

I wonder what gives a better average.
4D6 drop lowest
or
3D6 with 1 being a trick die with 2 6's and no 1's

Obviously rolling 3 trick die will result in the highest average.

Is there a way to check this on any of those websites?
Or the stats equation. i can't remember at the moment.
Here you go:
http://anydice.com/program/2950

The first table is just the general 3d6. (output 3d6)
The second table is 4d6, taking best three. (output [highest 3 of 4d6])

The third table is what you suggested, using 2d6 plus one trick die labeled "A", with the sides 2, 3, 4, 5, 6, and 6. (output 2d6+1dA)
The fourth table is with all three dice as trick dice. (output 3dA)

[Chronos]

I wanted the weighted average of the values between 3 and 18 for 4d6, drop the lowest. To find this, I'd have to know how many times, in an universe with all the possible rolls, each of the values show up. This number of times would be the weight of each value. The plain average between the values would be 10.5, the weighted average, 12.24. This happens to be the plain average of all the possible rolls too, since there you take into account each roll (and not the value thereof), and each roll has a weight of one (because, even if other rolls with the same value show up, that -specific- roll is singular). This is in fact a much easier way to find what I was looking for. Talk about taking the long road.
No, what you want is the unweighted average. You don't want to weight by how common each result is, because that's already taken into account by doing the average to begin with. I mean, you could introduce an additional weighting factor that happened to be equal to how common each number was, since you can after all use any weighting you like, but that wouldn't really give you anything meaningful.

I'm pretty sure I understand what it is you actually want; it's just that you (and several other posters in the thread) are using muddled terminology for what that is.

[Bacchus]

Here you go:
http://anydice.com/program/2950

The first table is just the general 3d6. (output 3d6)
The second table is 4d6, taking best three. (output [highest 3 of 4d6])

The third table is what you suggested, using 2d6 plus one trick die labeled "A", with the sides 2, 3, 4, 5, 6, and 6. (output 2d6+1dA)
The fourth table is with all three dice as trick dice. (output 3dA)


Wow. That was above and beyond.

Thanks you

[Larkas]

No, what you want is the unweighted average. You don't want to weight by how common each result is, because that's already taken into account by doing the average to begin with. I mean, you could introduce an additional weighting factor that happened to be equal to how common each number was, since you can after all use any weighting you like, but that wouldn't really give you anything meaningful.

I'm pretty sure I understand what it is you actually want; it's just that you (and several other posters in the thread) are using muddled terminology for what that is.

That can happen when you're not a native English-speaker trying to use English technical terms. Anyways, I'm not so sure you do understand. Let's use this example from Wikipedia (http://en.wikipedia.org/wiki/Weighted_arithmetic_mean#Basic_example) as an analogy.

It explains quite simply what a weighted average is. It teaches you how to find a "proper" average by summing up the average grade of each class multiplied by the number of students in that class, dividing by the total number of students. You could just as well find that "proper" average by taking into account each student's grade, summing all those up and dividing by the total number of students, thus ignoring from which class each student is and using an unweighted average.

Each student's grade is the "roll";
Each class' average grade is the "value";
The number of students in each class is the "weighting factor" of their respective "value".

You can find a "proper" average by multiplying each "value" by its "weighting factor", summing up all these results and dividing that by the sum of all the different "weighting factors". You can find the same "proper" average by eschewing "values" and "weighting factors" altogether and simply summing up all the "rolls", dividing that by the total number of "rolls".

The first is a weighted average. The second is not. Both, at least in this simple case we're dealing with, reach the same results. Both are equally relevant, since the end result is the same. But they are not the same thing.

I already admitted that trying to use a weighted average here wasn't exactly the easiest way to do it. It doesn't mean it can't be properly used.

EDIT: Just to make things clear: I'm not trying to confront you here, Chronos! I'm just trying to explain what I meant before! :smallredface:

[prufock]

Okay, sorry, I've been very unclear about the concepts I'm using here, and maybe ended up confusing people in the process. Let me try to explain.

"Average" usually means the arithmetic mean. You add up all the outcomes and divide the sum by the number of outcomes. You're dealing with a population, not just a sample of outcomes, so with 4d6b3, this gives you 12.24. There is no actual weighting involved.

Where I think you might be getting confused is that you're considering "the plain average" is actually the average of a sample of outcomes - the sample that includes each sum exactly once (3+4+5...+18). This is an incorrect way to calculate the population average, though. In this case, the average for 4d6b3 would be the same as 3d6 (because they have the same set of unique sums).

Hope this helps!

EDIT: I should point out why the wiki example is different, in case that is confusing you. In the wiki example, they're combining two different groups, morning and afternoon. The weighted average between the two groups by number of students is the same as the overall average ignoring groupings.

As analogy, this would be like if you were taking the average of 3d6 and 4d6b3, which would be (10.5+12.24)=11.37. Weighting it would mean you multiply 10.5 by 216 (the number of 3d6 results) and 12.24 by 1296, giving you 11.99.

[Larkas]

"Average" usually means the arithmetic mean. You add up all the outcomes and divide the sum by the number of outcomes. You're dealing with a population, not just a sample of outcomes, so with 4d6b3, this gives you 12.24. There is no actual weighting involved.

Where I think you might be getting confused is that you're considering "the plain average" is actually the average of a sample of outcomes - the sample that includes each sum exactly once (3+4+5...+18). This is an incorrect way to calculate the population average, though. In this case, the average for 4d6b3 would be the same as 3d6 (because they have the same set of unique sums).

Hope this helps!

I think it actually did! In Portuguese, average and mean are pretty much the same word, so I assumed it would be the same for English. So help me out here, how would one call these three things:

- The mean for the possible outcomes of 4d6b3?
- The mean of the samples outcomes of 4d6b3?
- The mean of the samples outcomes of 4d6b3, taking into account how many times each of these outcomes appear in the universe of possible results?

This might help dispel any further confusion regarding this theme! :smallredface:

[prufock]

I think it actually did! In Portuguese, average and mean are pretty much the same word, so I assumed it would be the same for English. So help me out here, how would one call these three things:

- The mean for the possible outcomes of 4d6b3?
- The mean of the samples outcomes of 4d6b3?
- The mean of the samples outcomes of 4d6b3, taking into account how many times each of these outcomes appear in the universe of possible results?

This might help dispel any further confusion regarding this theme! :smallredface:

English uses them mostly interchangeably too, but "average" technically can mean any central tendancy in English.

1 - Arithmetic mean, also known as just "mean" or "average."
2 - Not sure there's any specific term other than "sample mean."
3 - Also arithmetic mean - note that 1 and 3 will give the same result, though the calculation method is different. They are, however, the same thing.

[Eldan]

Thank you for explaining. I could not figure the color scheme out.

Another nice visual, if you were bored enough to do it, would be a color for each result number (all 15s are blue, etc) and comparing the same to a chart of 3d6.

Sure. Let's see...

Conditional formating makes that quite quick, really.
Tab 2. (https://docs.google.com/spreadsheet/ccc?key=0ApCuoSxQMcvddEtoTWtwNWZLZXItZWVSd0N0VFhPQ 0E&usp=sharing)

[Larkas]

English uses them mostly interchangeably too, but "average" technically can mean any central tendancy in English.

1 - Arithmetic mean, also known as just "mean" or "average."
2 - Not sure there's any specific term other than "sample mean."
3 - Also arithmetic mean - note that 1 and 3 will give the same result, though the calculation method is different. They are, however, the same thing.

Got it! And what would a weighted arithmetic mean be?

[prufock]

Got it! And what would a weighted arithmetic mean be?

If you were combining two samples or populations of different size into a single score, for example. Or if one type of score is more important than another - for example, instead of taking the average of all a student's test scores, perhaps the final exam is weighted more strongly than the others - ie, the final exam is worth 50% of your total grade, while each other exam is only worth 10%.

[Larkas]

If you were combining two samples or populations of different size into a single score, for example. Or if one type of score is more important than another - for example, instead of taking the average of all a student's test scores, perhaps the final exam is weighted more strongly than the others - ie, the final exam is worth 50% of your total grade, while each other exam is only worth 10%.

And wouldn't (3) also be this? I mean, isn't a 3 less populous than a 12 in 4d6b3, since there is only one possible result of 3, but several for 12? Or am I simply misusing "population" here?

[prufock]

And wouldn't (3) also be this? I mean, isn't a 3 less populous than a 12 in 4d6b3, since there is only one possible result of 3, but several for 12? Or am I simply misusing "population" here?

In a way, you're right, but this type of calculation is only necessary if you don't have the original data. What's happening in the wiki example we've been using is that you're take the average of two averages to get 85 (ie the average of the overall class scores), not the actual average of all test scores.

[Story]

As everyone knows, the standard way to generate ability scores in 3.X is to roll 4d6 six times, always discarding the lowest roll in a each group of die.

According to the Player's Handbook, the average PC is above average, with mean scores between 12 and 13.

That this generation method provides above average results is quite obvious - at least, above the regular arithmetic mean (which is the same as with 3d6, that is, 10,5).

I've been trying to find the "average that matters" - that is, the weighted arithmetic mean. If I can plot a graphic showing the probability of each result, all the better.

However, the way to do so is not obvious at all. I have to find a way to filter each of the 1,296 results of 4d6 (6^4), taking away the lowest die and summing up the other three for the result. And I have no idea how to do it!

Would a kind, mathematical soul please help me out?

I like to use Python for this. In fact I concidently was doing similar calculations right now. Anydice is probably faster if you don't want to do anything complicated though (like factoring in the rules for rerolling 'hopeless' arrays, which will skew things up slightly).

For example this gets you the distribution for 4d6 drop 1.


>>> import itertools as it
>>> d4s = map(sorted, it.product(range(1,7), repeat=4))
>>> d4v = [sum(x[1:]) for x in d4s]
>>> col.Counter(d4v)
Counter({13: 172, 12: 167, 14: 160, 11: 148, 15: 131, 10: 122, 16: 94, 9: 91, 8:
62, 17: 54, 7: 38, 6: 21, 18: 21, 5: 10, 4: 4, 3: 1})

[Larkas]

In a way, you're right, but this type of calculation is only necessary if you don't have the original data. What's happening in the wiki example we've been using is that you're take the average of two averages to get 85 (ie the average of the overall class scores), not the actual average of all test scores.

Got it! I was under the impression that my idea for calculating this stuff was really dumb, and it just so happens I was right! :smallbiggrin:


I like to use Python for this. In fact I concidently was doing similar calculations right now. Anydice is probably faster if you don't want to do anything complicated though (like factoring in the rules for rerolling 'hopeless' arrays, which will skew things up slightly).

For example this gets you the distribution for 4d6 drop 1.


>>> import itertools as it
>>> d4s = map(sorted, it.product(range(1,7), repeat=4))
>>> d4v = [sum(x[1:]) for x in d4s]
>>> col.Counter(d4v)
Counter({13: 172, 12: 167, 14: 160, 11: 148, 15: 131, 10: 122, 16: 94, 9: 91, 8:
62, 17: 54, 7: 38, 6: 21, 18: 21, 5: 10, 4: 4, 3: 1})

Ooooh, interesting! But how do I run this? Is there any interpreter or do I have to compile it? I've programmed very little in my life, and never using Python!

[Maginomicon]

As everyone knows, the standard way to generate ability scores in 3.X is to roll 4d6 six times, always discarding the lowest roll in a each group of die.

According to the Player's Handbook, the average PC is above average, with mean scores between 12 and 13.

That this generation method provides above average results is quite obvious - at least, above the regular arithmetic mean (which is the same as with 3d6, that is, 10,5).

I've been trying to find the "average that matters" - that is, the weighted arithmetic mean. If I can plot a graphic showing the probability of each result, all the better.

However, the way to do so is not obvious at all. I have to find a way to filter each of the 1,296 results of 4d6 (6^4), taking away the lowest die and summing up the other three for the result. And I have no idea how to do it!

Would a kind, mathematical soul please help me out?

Go here (http://anydice.com/program/13e) to see a pre-programmed example of "4d6 drop lowest". Look into the function library and documentation links on the left if you have questions.
If the link doesn't work, the following text is what I put into AnyDice.com (http://www.anydice.com/)
output [highest 3 of 4d6]

[Segev]

In statistics, in English, the actual terminology for the three main "central" values are:

(Arithmetic) Mean,
Median
Mode

The "mean" is to what most people refer when they say "average:" it is the sum of all values divided by the number of values. In our case, it would be the sum of all possible results of our die rolls divided by the number of possible unique die rolls.

The "median" is the literal "middle value." To find it, you line up every possible result from least to greatest, including all duplicates, and then you literally take the central value in that list. So if you have a set that is {1, 1, 10}, the median is "1" because it's the middle value. If you have an even number of items in the list, the median is generally taken to be the arithmetic mean of the two central ones. So {1, 2, 4, 12} would have a median that is the mean of {2,4}, or "3."

The "mode" is the entry that shows up the most times in the list. So, {1, 1, 1, 2, 3, 4, 4, 5, 6, 7} has a mode of "1," because it shows up 3 times, while the others only show up one or two times.


The word "average" in English generally means "typical" or "middle-of-the-road" result. It most commonly is used interchangeably with "mean," but in theory could refer to any of the three. "An average day" is more likely going to be the mode of your days' qualities, not the mean or median. The mode is also what people will typically picture if told "this man makes an average living." That's because we expect "average" to also mean "common."

It's actually relatively rare for us to use "median" in any context. I've heard people refer to "the median income" of an area, but it's rarely used because it can be so deceptive. It's kind-of the compromise, though, between "mean" and "mode," because it is a number that is guaranteed to actually exist in the population (or to be halfway between the two middle-point numbers, at least), and it is likely to hit, if not the mode, then at least a plurality range.

The median is never skewed by outliers, but by the same token, it can be deceptively rare:

{$10,000, $11,000, $12,000, $400,000, $1,000,000, $1,000,000, $1,000,000}

The median of this is $400,000. The mode is $1,000,000. The mean is $490,428.57.

The "average" conversationally would probably be reported as that last number, but might be reported as the median. They're "somewhat" close...except the gap between them is still greater than the values at the lower end.

"The average income" of the 7-person population in this sample, however, could be listed as any of those.

[Barstro]

Sure. Let's see...

Conditional formating makes that quite quick, really.
Tab 2. (https://docs.google.com/spreadsheet/ccc?key=0ApCuoSxQMcvddEtoTWtwNWZLZXItZWVSd0N0VFhPQ 0E&usp=sharing)

Perfect, and exactly the color scheme I would have used. (Figures there was an easy formatting formula. I didn't have time to mess with that).

[Chronos]

The median is also especially valuable for continuous distributions, since it's the only one that's guaranteed to exist. There are many distributions which don't even have a mean. And in some ways, it can be less deceptive than the mean: If, for instance, you have a bunch of pairs of pants that cost between $10 and $50, and then one pair of pants from some fancy Paris designer that's intended only to be worn on the fashion runway and costs $500,000 , then the mean of that set is going to be huge due to the outlier, but the median is still going to be somewhere in the "normal" range.

[Story]

Ooooh, interesting! But how do I run this? Is there any interpreter or do I have to compile it? I've programmed very little in my life, and never using Python!

For Windows there's a convenient installer. If you're on Linux you probably already have it installed.

After that open a command prompt (On Windows, go to start menu, type in "cmd" in the box and hit enter). Then type python, enter, and type in the program.

[Yogibear41]

How about all this awesome info previously given but on 5d6 instead? :smallsmile: lol

[faircoin]

I posted a math analysis of 4d6 best 3 before on these boards.


What you're asking, in math formulation, expressed in TeX is:

\[x=\frac{\sum_{i=1}^{n}(\textup{rand}_i[x,y])-\min(\textup{rand}_i[x,y])}{n-1}\]

http://i.imgur.com/seiKoJ8.gif

edit

Sorry, minor mistake in the image above. The leftmost x should be some other letter, doesn't matter which, it has nothing to do with the x's on the right.

We'll start with a simple, roll four times, drop lowest scenario.

Let a be the smallest roll, and b the sum of the largest rolls.

E(x) is the expected value function. We know the expected value E(a+b)=E(a)+E(b)=14.

If p_1, p_2, all the way to p_6 are the probabilities that x >= 1, 2, all the way to 6 respectively. P(x) is the probability function. Thus, P(a=1)=p_1-p_2, P(a=2)=p_2-p_3, and all the way to P(a=6)=p_6. Hence, E(a)=\sum_{i=1}^{6}(i(p_i-p_{i+1}))=\sum(p_i), where p_7 is set to be 0. p_i is (7-i)^4/(6^4). So, sum that baby up, and you should get E(a). If you know E(a), you can find E(b) very easily.

In this case it's 12.25.

edit

Crap, I got the above number wrong. This is why you don't calculate in your head. It's actually 12.25, not 12.5.

[Fax Celestis]

How about all this awesome info previously given but on 5d6 instead? :smallsmile: lol

http://img.photobucket.com/albums/v495/IronShark/prob.jpg

[Yogibear41]

That graph can't be accurate since it goes over 18 even on the 4d6 its not taking into consideration the dropped numbers.

[Fax Celestis]

That graph can't be accurate since it goes over 18 even on the 4d6 its not taking into consideration the dropped numbers.

You just said 5d6, you didn't say 5d6b3.

[Ekul]

5d6b3 here is compared to my method (no 1s), 4d6b3 and 3d6:
http://anydice.com/program/295b

(Best visualized by going to the graph tab)

Very close in expectation value to my method, but with a different distribution.