Any Math Geeks out there that like to mess with Dice averages?

[tarchon]

Re: Re: Re: Re: Correction of my previous result

Addition to these results:

The possibility to get a total of 108 (i.e. rolling an 18 all six times) is

0.2 * 10^(-10).

For example if your program ran with 9 digit precision, it wouldn't even have noticed this value...

For 7 x 4d6 I get 1.24942e-10 for including hopeless chars and 1.40709e-10 for dumping hopeless chars. Those are the probabilities for getting an 18 as your lowest score, BTW. Analytically, you get 1.81003E-11 = (21/1296)^6 for 6 x 4d6 and 1.24942E-10 =
(21/1296)^7+7*(21/1296)^6*(1275/1296) for 7 x 4d6 without dropping hopeless characters, so I think my results check out by that test too. I used 64-bit floats so the numbers are pretty accurate.

For means I get 12.91 and 13.11 in the 7 x 4d6 keep hopeless and drop hopeless cases.

 

[CRGreathouse]

The new calc looks better to me; my result: 73.66466458

This is an average of 12.2774441.

 

[Rayston]

Wow!!! I need to take some math courses.

Very very cool. I didnt realize this would spark quite this much reaction.

;-)

Anyways, so the average mean attribute of my system is 13.11?

I ran accross a site on google ( Yes I did try to find this answer before I asked, but nothing on google, and although I like math, I have forgotten most of it ) that had the characters rolled up and worked up to a point buy system. I believe the normal 4d6 drop lowest system came out to an average point buy of right around 31. Anyone else seen this site? as I cant find it againfor the life of me. It also had nice matrix's worked out and everything, and I believe the average character if each player were allowed to roll up 5 sets of stats, and then choose the one with the highest point buy. ( I think it increased the average point buy equivalent to 39.XX)

On a related note, since I am asking you all to do my silly math problems I dont even really "need" the answer for, any good sights or programs out there that would help in relearning this stuff?


Thanx

Rayston

BTW

I really do appreciate the help, while not necessarily....well...necessary...the stats are intersting. ;-)

 

[tarchon]

Re: Wow!!! I need to take some math courses.

Very very cool. I didnt realize this would spark quite this much reaction.

;-)

Anyways, so the average mean attribute of my system is 13.11?

Yeah, if you're rerolling helpless characters, anyway. Your method tends to increase the low stats more than it does the high stats, and the majority of characters (around 70%) won't have any negative modifers with it. The PHB method gives a mean of 12.5049, for comparison. It exceeds the 4d6-drop-lowest mean of 12.2446 because of the helpless character rule.

 

[Nathan]

Re: Re: Wow!!! I need to take some math courses.

The PHB method gives a mean of 12.5049, for comparison. It exceeds the 4d6-drop-lowest mean of 12.2446 because of the helpless character rule.

CRGreathouse, this would mean that my 75.03 ( = 6 x 12.505) is the correct answer compared with your new 73.664...

However, where is the error?

 

[Rayston]

*bump*

 

[CRGreathouse]

Re: Re: Re: Wow!!! I need to take some math courses.

CRGreathouse, this would mean that my 75.03 ( = 6 x 12.505) is the correct answer compared with your new 73.664...

However, where is the error?

The last step of my calculation was in error - that's why I blacked it. I was going to go back and edit in the correct answer, but never had the chance. (That number is the sum of the total rolls, not counting illegal rolls, divided by total rolls instead of legal rolls.)

Unfortunately, the 'correct' number I came to is different from your result: 78.17955351 (ave 13.02992559). To get my final numbers, multiply each of the below numbers by 10^14:

61- 0
62 5334.701117
63 43520.61754
64 157232.4819
65 345790.3828
66 558391.8515
67 776551.9868
68 1025721.648
69 1318248.846
70 1648242.939
71 2004928.554
72 2372824.352
73 2732662.624
74 3063135.499
75 3343882.731
76 3557523.799
77 3691083.26
78 3737104.931
79 3694392.3
80 3566228.067
81 3360458.879
82 3088876.544
83 2767368.158
84 2414401.494
85 2049261.083
86 1690261.409
87 1353166.854
88 1050026.842
89 788559.8795
90 572138.4333
91 400266.6358
92 269413.2096
93 174028.0124
94 107570.809
95 63412.62796
96 35508.57747
97 18798.61265
98 9357.019834
99 4349.533769
100 1872.401603
101 738.499892
102 263.1938685
103 83.22872712
104 22.77495403
105 5.192669136
106 0.927073447
107 0.115599009
108 0.007672637

 

[Nathan]

Re: Re: Re: Re: Wow!!! I need to take some math courses.

The last step of my calculation was in error - that's why I blacked it. I was going to go back and edit in the correct answer, but never had the chance. (That number is the sum of the total rolls, not counting illegal rolls, divided by total rolls instead of legal rolls.)

Unfortunately, the 'correct' number I came to is different from your result: 78.17955351 (ave 13.02992559). To get my final numbers, multiply each of the below numbers by 10^14:

61- 0
62 5334.701117
63 43520.61754
64 157232.4819
65 345790.3828
66 558391.8515
67 776551.9868
68 1025721.648
69 1318248.846
70 1648242.939
71 2004928.554
72 2372824.352
73 2732662.624
74 3063135.499
75 3343882.731
76 3557523.799
77 3691083.26
78 3737104.931
79 3694392.3
80 3566228.067
81 3360458.879
82 3088876.544
83 2767368.158
84 2414401.494
85 2049261.083
86 1690261.409
87 1353166.854
88 1050026.842
89 788559.8795
90 572138.4333
91 400266.6358
92 269413.2096
93 174028.0124
94 107570.809
95 63412.62796
96 35508.57747
97 18798.61265
98 9357.019834
99 4349.533769
100 1872.401603
101 738.499892
102 263.1938685
103 83.22872712
104 22.77495403
105 5.192669136
106 0.927073447
107 0.115599009
108 0.007672637

These are my results for the number of possibilities to get a specific stat total with the standard PHB method:

62, 1758002851779120
63, 13067562306614040
64, 42504488361874560
65, 83366785260726432
66, 119317385435361838
67, 146703491711454048
68, 171557948806228932
69, 195842290079752700
70, 218271126449107848
71, 237556961730109872
72, 252538917849145962
73, 262303538499701028
74, 266278157788793580
75, 264281734615212252
76, 256526875796414748
77, 243576563774681628
78, 226268210788111323
79, 205621686925367784
80, 182747514027825732
81, 158764927015493034
82, 134735665709136276
83, 111611073291869448
84, 90170401635963117
85, 70981787902311492
86, 54388079316496092
87, 40516222817346144
88, 29305985194245564
89, 20551770003405060
90, 13950482138219964
91, 9148751839925640
92, 5784144222402864
93, 3516878375780068
94, 2050638663715788
95, 1142896970826192
96, 606507608137294
97, 305063878524732
98, 144639647733744
99, 64213196762196
100, 26473282400736
101, 10029824562924
102, 3444963710553
103, 1053364723236
104, 279526350600
105, 61991152146
106, 10810031904
107, 1323248724
108, 85766121

These numbers differ quite a lot from yours... is the factor 10^14 really the correct one?

Let us look at 108 for example: There are 21 possibilities to get an 18 with 4d6, drop lowest. So there are 21^6 = 85766121 possibilities to get an 108 with the standard PHB method.

 

[Elric]

The experiment is the following:

We roll an s-sided die n times and add up the k highest results.

We ask for the average outcome, say E(n, k, s), of this experiment.

The answer is:

E(n, k, s) = the sum of

n! / (l! * (n - l - k + r)! * (k - r)!) * (t - 1)^l * (s - t)^(k - r) * (r * t + (k - r) * (s + t + 1)/2)

over t = 1,..., k; l = 0, ... n-k; r = 1... k and then divided by
s^k.'t

I'm not sure if I'm interpreting this right. How do T, L, and R work? Do you have to run every permutation of those three variables through the formula and sum all of them. Does this formula work for the more simple case (n=2, k=1)? With any set of fixed n and k values, you should be able to come up with a formula only based on s. I can't see how to work that out using the formula. Can you go into a simple example, like E(s,2,1)? Brackets would also help Thanks!

 

[CRGreathouse]

Re: Re: Re: Re: Re: Wow!!! I need to take some math courses.

These are my results for the number of possibilities to get a specific stat total with the standard PHB method:

What's the sum of your totals for each?