Is there a formula to determine the cumulative distribution associated with multiple different kinds of events?

Question

Similar to this question, but with more than one different sided die.

Is there a way to calculate probability distributions associated with dice, where the probability distributions for each die are not all the same?

Example: Using dice, find the probability of rolling 1 once in a set when rolling 2d4 and 1d6.

As I understand it, the long way is to calculate each event separately and then add up the events where a 1 is rolled once:

 #  result #of 1s

1: |{1,1,1}   3  
2: |{1,1,2}   2  
3: |{1,1,3}   2  
4: |{1,1,4}   2  
5: |{1,1,5}   2  
6: |{1,1,6}   2  
7: |{1,2,1}   2  
8: |{1,2,2}   1  
9: |{1,2,3}   1  
10:|{1,2,4}   1  
11:|{1,2,5}   1  
12:|{1,2,6}   1  
13:|{1,3,1}   2  
14:|{1,3,2}   1  
15:|{1,3,3}   1  
16:|{1,3,4}   1  
17:|{1,3,5}   1  
18:|{1,3,6}   1  
19:|{1,4,1}   2  
20:|{1,4,2}   1  
21:|{1,4,3}   1  
22:|{1,4,4}   1  
23:|{1,4,5}   1  
24:|{1,4,6}   1  
25:|{2,1,1}   2  
26:|{2,1,2}   1  
27:|{2,1,3}   1  
28:|{2,1,4}   1  
29:|{2,1,5}   1  
30:|{2,1,6}   1  
31:|{2,2,1}   1  
32:|{2,2,2}   0  
33:|{2,2,3}   0  
34:|{2,2,4}   0  
35:|{2,2,5}   0  
36:|{2,2,6}   0  
37:|{2,3,1}   1  
38:|{2,3,2}   0  
39:|{2,3,3}   0  
40:|{2,3,4}   0  
41:|{2,3,5}   0  
42:|{2,3,6}   0  
43:|{2,4,1}   1  
44:|{2,4,2}   0  
45:|{2,4,3}   0  
46:|{2,4,4}   0  
47:|{2,4,5}   0  
48:|{2,4,6}   0  
49:|{3,1,1}   2  
50:|{3,1,2}   1  
51:|{3,1,3}   1  
52:|{3,1,4}   1  
53:|{3,1,5}   1  
54:|{3,1,6}   1  
55:|{3,2,1}   1  
56:|{3,2,2}   0  
57:|{3,2,3}   0  
58:|{3,2,4}   0  
59:|{3,2,5}   0  
60:|{3,2,6}   0  
61:|{3,3,1}   1  
62:|{3,3,2}   0  
63:|{3,3,3}   0  
64:|{3,3,4}   0  
65:|{3,3,5}   0  
66:|{3,3,6}   0  
67:|{3,4,1}   1  
68:|{3,4,2}   0  
69:|{3,4,3}   0  
70:|{3,4,4}   0  
71:|{3,4,5}   0  
72:|{3,4,6}   0  
73:|{4,1,1}   2  
74:|{4,1,2}   1  
75:|{4,1,3}   1  
76:|{4,1,4}   1  
77:|{4,1,5}   1  
78:|{4,1,6}   1  
79:|{4,2,1}   1  
80:|{4,2,2}   0  
81:|{4,2,3}   0  
82:|{4,2,4}   0  
83:|{4,2,5}   0  
84:|{4,2,6}   0  
85:|{4,3,1}   1  
86:|{4,3,2}   0  
87:|{4,3,3}   0  
88:|{4,3,4}   0  
89:|{4,3,5}   0  
90:|{4,3,6}   0  
91:|{4,4,1}   1  
92:|{4,4,2}   0  
93:|{4,4,3}   0  
94:|{4,4,4}   0  
95:|{4,4,5}   0  
96:|{4,4,6}   0  

Which totals 39 out of 96 possibilities that a 1 will be rolled only once, or 40.625%

But I know there has to be formula that does this because if I can tediously count it out manually (as above), then there must be some solid math function behind it.

Through the past few days of research, I've learned that this can also be done using the cumulative distribution function. However, so far as I understand it, the cdf requires you to have the same probability for all three rolls - so it would work if I did one formula for 2d4 and then one for 1d6 but not cumulatively.

Delving deeper, it seems that the Poisson distribution may provide the probability distributions that I want, but honestly, my capacity for learning this from a formula on wikipedia is rapidly approaching zero.

If this is the correct formula (or if the answer lies with a different one), could you provide a proof using my example from above? Similarly, could you provide an example where there are several different probabilities (ex: 1d2, 2d4, 1d6, 3d10)?

Answer 1

To answer your question, there is sometimes a formula you can use to work out probabilities, but not always. If the thing you're trying to find the probability for is quite complicated, and the total number of outcomes is relatively small, you are probably better off counting. Tediously counting doesn't necessarily mean that there must be some elegant mathematical alternative!

In your case, there is a formula for the probability you derived by couting. We need to consider that there are three different ways of getting exactly one 1, i.e. one of the three dice has to be 1. So the probability of getting exactly one 1 is given by

(1d4 = 1 AND 1d4 ≠ 1 AND 1d6 ≠ 1) OR
(1d4 ≠ 1 AND 1d4 = 1 AND 1d6 ≠ 1) OR
(1d4 ≠ 1 AND 1d4 ≠ 1 AND 1d6 = 1)

The dice are independent, so we multiply probabilities for the ANDs, and the events in brackets are mutually exclusive (i.e. they can't happen at the same time), so we can add these probabilities together. This gives

> (1/4 * 3/4 * 5/6) + (3/4 * 1/4 * 5/6) + (3/4*3/4*1/6)
0.40625

confirming your result.

As you've been learning about probabilities, it might help to comment on some of the things you wrote.

First of all, in your example, the cumulative distribution function F(x) for rolling 1s is the probability of getting less than or equal to x 1s. So F(0) is the probability that there are no 1s, F(1) is the probability that there are no 1s or one 1. F(2) is the probability that there are no 1s or one 1 or two 1s. And F(3) is the probability that there are at most three 1s. Of course, as there are three dice, you can only have at most 3 ones, and so F(3) = 1.

For example, you can work out F(0) as follows: no 1 on d4 AND no 1 on d4 AND no 1 on d6. Because these dice are independent you can multiply probabilities to get:

> .75*.75*5/6
0.46875

If you want to get F(1), you can add together F(0), 0.40625, and the probability of getting exactly one 1, 0.46875, to get F(1) = 0.875. The probability of getting exactly two 1s can be calculated by adding together the three ways you can get exactly one die that is not 1, and from this you can get F(2).

Second, you mention the Poisson distribution. This is definitely worth considering when you are counting things, because the Poisson distribution is suitable for whole numbers, unlike other distributions which describe quanities like heights or speeds which vary continuously and don't have to be whole numbers. However the Poisson distribution is used when there isn't any limit on the count you can get, even if the chance of getting a high count might be very low, and the chance of getting zero votes might be high. So you might use the Poisson distribution for the number of votes on a stackoverflow answer, because the answer could get any number of votes, at least in theory!