# How to calculate dice roll probability with more dices? [duplicate]

I'm not math graduate, so be patient.

For two dices it is easy, because of small number of possibilities, also it can be easy for three, but how is it works for ten (some formula)?

How to calculate probability:

A) Rolling three dices and sum of them have to be greater than 8

B) How to do it with more, let's say 10 dices, where sum have to be greater than 27

Can you explain it and you can also send some useful resources.

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• Check this. You can approach the probabilities using a normal distribution too, when more dice you throw then the approximation is better. – Masacroso Jan 8 '17 at 12:04

Here's an approach that I haven't found before, but you can build up from it quite easily without remembering formulae. We start from the distribution of a single die. It's easy, and we cross it with itself.

1 | 1  1  1  1  1  1
2 | 1  1  1  1  1  1
3 | 1  1  1  1  1  1
4 | 1  1  1  1  1  1
5 | 1  1  1  1  1  1
6 | 1  1  1  1  1  1
—————————————————
1  2  3  4  5  6


Then we count the diagonals, which are $12$ in total. This gives the distribution for 2 dice:

2  | 1
3  | 2
4  | 3
5  | 4
6  | 5
7  | 6
8  | 5
9  | 4
10 | 3
11 | 2
12 | 1


For three dice we cross the count for $1$ die and $2$ dice. For each entry we multiply the number of combinations that are possible for $x$ and $y$.

  x
1 | 1  2  3  4  5  6  5  4   3   2   1
2 | 1  2  3  4  5  6  5  4   3   2   1
3 | 1  2  3  4  5  6  5  4   3   2   1
4 | 1  2  3  4  5  6  5  4   3   2   1
5 | 1  2  3  4  5  6  5  4   3   2   1
6 | 1  2  3  4  5  6  5  4   3   2   1
——————————————————————————————————— y
2  3  4  5  6  7  8  9  10  11  12


When we add columns we get the counts for $3$ dice, again, by summing over diagonals. That is a total of $16$ diagonals. The count looks like:

3  | 1
4  | 3
5  | 6
6  | 10
7  | 15
8  | 21
9  | 25
10 | 27
11 | 27
12 | 25
13 | 21
14 | 15
15 | 10
16 | 6
17 | 3
18 | 1


This is evidently not the most efficient way to do it, but you don't really need to remember anything as long as you can compute the tables. Want you can optimize is the choice of which tables to calculate e.g. for $13$ dice, calculate the tables for $(1,2,4,6,12,13)$ in that order. Again, this isn't the surgical solution, but it's memory-proof.

Earlier, I showed how to make a computer spreadsheet to calculate the probabilities of the sums of multiple dice using Microsoft Excel, OpenOffice Calc, or similar software.

The spreadsheet shown in that answer is for $20$-sided dice rather than six-sided dice, but it can easily be adapted. (It should take sums of ranges of six cells instead of twenty, and divide by $6$ rather than multiplying by $0.05$.)

The spreadsheet for six-sided dice would look like this:

\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline & A & B & C & D & E & F & G & H & I \\ \hline 1 & & & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 2 & \\ \vdots & \\ 7 & \\ \hline 8 & 0 & 1 & \\ \hline 9 & \text{=A8+1} & & \text{SUM(B3:B8)/6}& \\ \hline 10 & \\ \vdots & \\ \hline \end{array}

Leave rows $2$ through $21$ and all the other cells of row $22$ blank. (Alternatively, you can put zeros in any of those cells. But do not put anything else there.)

Copy cell C9 to cells D9, E9, and so forth in the same row, until you have as many copies of the formula (includeing C9) as the number of dice you want to roll. Then copy row 9 to rows 10, 11, 12, and so forth, filling as many rows as you need until all the sums you want to calculate have shown up in column A.

The numbers in the block of cells from cell C9 to the right and below will be a table of probabilities of rolling exactly the sum in column A using the number of dice in row 1. To answer questions such as "probability of rolling a sum greater than $27$" you can use a spare cell of the spreadsheet (below the table, or to the right) to add up the probabilities in a range of the corresponding column. For example, rolling greater than $27$ on $10$ dice is =SUM(L36:L68). You can also get this result by taking =1-SUM(L18:L35) ($1$ minus the probabilities of the sums from $10$ to $27$ inclusive).