So I am making a simple Dice that has 6 sides, but I want to modify the chances of those 6 sides.

Now my dice can have N sides, it grows, so you start with 6 sided dice and you may get up to 10 sided dice. The chances for a specific side to come up on a roll depends on its value. Chances should decrease depending on the value on a side so if side value is 1 its chance is higher than the side numbered 6 whose chance would be much lower.

Example (6 Sided):

```Side   :   Chance
1     :   35  %
2     :   25  %
3     :   20  %
4     :   11  %
5     :   6.5 %
6     :   2.5 %
```

So as sides increase the chances should decrease never going over 100.

I tried making formula depend on the side and divide the current chance by number of sides but did not work.

Edit:

Side 6 should have 6 times less probability than side 1 and 5 times less probability than side 2 and 4 times less probability than side 3 etc... My example does not match this because I could not come up with numbers so they would add up to 100 and qualify the conditions. Mar 15 33 views

## 1 answer to this question.

If I understand you correctly, you're looking for the following equation:

The overall "weight" of a dice with N sides is (N/2)*(n+1).

1 The total "weight" for 6 sides is (6/2)*(6+1) = 3*7 = 21.

The calculation is then straightforward.

```1 -> 6 / 21 = 0.28571428571
2 -> 5 / 21 = 0.23809523809
3 -> 4 / 21 = 0.19047619047
4 -> 3 / 21 = 0.14285714285
5 -> 2 / 21 = 0.09523809523
6 -> 1 / 21 = 0.04761904761```

6/21 is obviously 6 times the size of 1/21, therefore that part is correct. Finally, here's the bottom line:

```0.28571428571         6/21
+ 0.23809523809      +5/21
+ 0.19047619047      +4/21
+ 0.14285714285      +3/21
+ 0.09523809523      +2/21
+ 0.04761904761      +1/21

--------------- -----
0.99999999996        21/21```

In any case, the left side is close to 100 percent. Being what it is, rounding is inevitable. The right side demonstrates that this is a rounding issue rather than an error.

*This equation (and its variation (N/2)*(N-1)) are quite useful. It's a shortened version of 1+2+3+4+5+6... answered Mar 17 by
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