Need my math gurus: 10d6 probability work

[Stalker0]

I'm doing some modeling for a debate I'm having with some other people, so I'm gaining as many facts to back my argument up.

What I need (if it exists) is a formula that can tell me the probability for a certain specific amount of damage to occur. If a roll 10d6, what's the probabily of doing 31 damage for example?

Then I could also use a similar formula, except this is the probability of that number or above. So the probability of doing 31 damage or higher.

 

[Kurotowa]

I can't tell you the exact complex formula. But the simple formula is that the average result of a d6 is 3.5, so the average damage of 10d6 is 35. Dice pools have a bell curve result, so there's a high probability of getting 35 damage and an increasingly lower probability as you go higher or lower.

 

[Land Outcast]

Well...

(1/6^10)*24= Probability of getting four 6's, five 1's, and a 2...

Hum... 'tis somehow complex... sry, don't have the knowledge to provide a formula capable of taking into account all the different number combinations which will give 31...

Basically the probability of getting a result of "31" is equal to (1/6^10)*24*X; where "X" is the number of number combinations (of ten numbers, all of them being Natural numbers greater or equal to 1 and smaller or equal to 6) which add up to 31.

 

[Altamont Ravenard]

Argh I have a bachelor's degree in that (which I don't use and I have pretty much forgotten) but the basic way to do it is to count the number of possible combinations of dice that amount to 31 and divide by the total number of possible combinations (in case of 10d6, it's 6^10, which is 60,466,176.

For your second formula, you'd need to add, for example, the probabilities of getting 31, 32, 33, 34, 35 and 36, calculating each indenpendently before.

Now, a simpler, less accurate way to do it would be to simulate, say, 1,000 or 10,000 rolls in a excel sheet and estimate the probability from there.

I'll whip something up and attach it.

AR

EDIT: What's in the file:
Prob Estimator sheet: Enter a number of dice to roll (1-20), a maximum for each die (1-100) and a modifier to each die (1-20). Below appears the range of sums possible, and the calculated occurances of each sums. For example, if you wish to roll 10d6+0, you enter 10; 6 and 0, and below will appear a table with the possible sums of 10 to 60, with the number of occurances found in the "Rolls and Sums" sheet, the % of occurances for each sum, and the % of occurances of each sum high or equal to the sum in the corresponding line.
Rolls & Sums sheet: 1000 random rolls of the die of your choice. I went with 1000 because at 10000, the sheet kind of bogged down. At 1000, it's still pretty fast.

Note: If you want to print the table out, make sure to set the print area accordingly, since there are 2000 lines in that sheet (most of them empty) and you don't want to be printing out a great number of blank pages.

Note 2: I didn't calculate the real probability, but a number of occurances in 1000 random rolls. The results will vary from computation to computation even if the variables stay the same. It should nevertheless give you a good idea of the probability (within a few %'s, I'd say).

Hope someone finds this useful!

AR

 

[ThoughtfulOwl]

Here. The columns from left to right are the results, their probability as number of combinations and their probability on a 0 to 1 scale (# of combinations divided by total combinations).

To get the probability of a range of results, add together their probabilities.

---------------------------
 v |         prob          
---------------------------
10 |        1 | 1.65382e-08
11 |       10 | 1.65382e-07
12 |       55 | 9.09599e-07
13 |      220 | 3.63840e-06
14 |      715 | 1.18248e-05
15 |     2002 | 3.31094e-05
16 |     4995 | 8.26082e-05
17 |    11340 | 0.000187543
18 |    23760 | 0.000392947
19 |    46420 | 0.000767702
20 |    85228 | 0.00140952
21 |   147940 | 0.00244666
22 |   243925 | 0.00403407
23 |   383470 | 0.00634189
24 |   576565 | 0.00953533
25 |   831204 | 0.0137466
26 |  1151370 | 0.0190416
27 |  1535040 | 0.0253868
28 |  1972630 | 0.0326237
29 |  2446300 | 0.0404573
30 |  2930455 | 0.0484644
31 |  3393610 | 0.0561241
32 |  3801535 | 0.0628704
33 |  4121260 | 0.0681581
34 |  4325310 | 0.0715327
35 |  4395456 | 0.0726928
36 |  4325310 | 0.0715327
37 |  4121260 | 0.0681581
38 |  3801535 | 0.0628704
39 |  3393610 | 0.0561241
40 |  2930455 | 0.0484644
41 |  2446300 | 0.0404573
42 |  1972630 | 0.0326237
43 |  1535040 | 0.0253868
44 |  1151370 | 0.0190416
45 |   831204 | 0.0137466
46 |   576565 | 0.00953533
47 |   383470 | 0.00634189
48 |   243925 | 0.00403407
49 |   147940 | 0.00244666
50 |    85228 | 0.00140952
51 |    46420 | 0.000767702
52 |    23760 | 0.000392947
53 |    11340 | 0.000187543
54 |     4995 | 8.26082e-05
55 |     2002 | 3.31094e-05
56 |      715 | 1.18248e-05
57 |      220 | 3.63840e-06
58 |       55 | 9.09599e-07
59 |       10 | 1.65382e-07
60 |        1 | 1.65382e-08
---------------------------
tot. 60466176
---------------------------

 

[Pinotage]

There is a formula that can do that. You should be able to track it down with google. I once used it to extensively create probability rules for Mass Combat, i.e. to determine that if 100 creatures were in a group, what is the probability for a given number of hits. I then contrsucted a table on which you rolled d100, which would then give a number of hits based on the probability for each number of hits. In any event, while I could dig out the formula, here's an online calculator that's already done everything for you:

http://www.anwu.org/games/dice_calc.html

Pinotage

 

[Rothe]

I'm doing some modeling for a debate I'm having with some other people, so I'm gaining as many facts to back my argument up.

What I need (if it exists) is a formula that can tell me the probability for a certain specific amount of damage to occur. If a roll 10d6, what's the probabily of doing 31 damage for example?

Then I could also use a similar formula, except this is the probability of that number or above. So the probability of doing 31 damage or higher.

I can tell you the exact complex formula It was also floating around here before the crash. However if you have some programing acumen, the quickest way to do it is probably simulating to just 10,000 rolls. Since I have no prgramming skills anymore, and still like algebra, here is the precise formula or generating function for the distribution.

For n dice expand out the expression (aX +aX^2 + aX^3 + aX^4 + aX^5 +aX^6)^n
where for a fair d6, a=1/6. Unlike what you might be used to, the X is a dummy variable, you don't need to solve for it. The number that you want will be the coeficient in front of a specific X. If you are looking for the probability of rolling 10 on 10 d6 look at the coeficient in front of the X^10. If looking for the probability of rolling 31, it is the coeficient in front of the X^31 term. As you can see, once you get to high n the algebra becomes fairly tedious unless you can program it into Mathematica.

One thing I like about generating functions is you can use them to get the probability distributions for mixed dice, unfair dice, and even mixed unfair dice.

 

[nogray]

For those interested in the results and without access to Mathematica or similar programs, this link should be helpful. For the exact problem from Stalker0 and solution given by Rothe, enter (x/6+(x^2)/6+(x^3)/6+(x^4)/6+(x^5)/6+(x^6)/6)^10. You will, upon clicking expand, see exact fraction values for each probability. If you instead enter (x+x^2+x^3+x^4+x^5+x^6)^10, you will get coefficients that mimic the second column in ThoughtfulOwl's table. Each number could then be divided by 6^10 to get a probability.

Hope that helps some.