Good evening, how do I make a random number with percentage? I want to draw a number between 1 and 5, but it has to have a 50% chance of getting number 2, how do I do that?

RandomInFloatRange(1,5) thats a lot of changes

I know how to do a random draw, just open a random(5)+1 Okay, there will be a number between 1 and 5, but what about the percentage%? is there a way?

Simple rules if 1=0% then 5=100% and 2.5=50% so do a randomInFloatRange(1,5) and if the result is nearly 2.4,2,5 or 2,6 you can consider that as 50% then you just show a message with a condition

if result >2.3

if result <2.7

Set text messsage you got 50%

This might work.

Pick a number between 1 and 9. If it’s greater than 5 then make it 2. You have the normal chance of 1-5 plus 4 additional chances in 6,7,8,9.

Variable R=randomInRange(1,9)

R >5 | variable r=2

That would give you a 5/9 which is more like 55%. A true 50% would be a little more complicated. Is that close enough?

The way I’ve done this:

```
First event:
Your conditions here >>> Set a variable to the value of RandomInRange(1,100)
Separate event (or subevent):
Value of the variable is > 50 >>> set a result variable as 5
Then say you want 20% for 4, 10% for 3, 15% for 2, 5% for 1.
You'd have another separate event (or subevent)
Value of the variable is <= 50 (no actions)
(Subevent 1) Value of variable is >30 >>> Set a result variable as 4
(Subevent 2) Value of variable is >20 >>> Set a result variable as 3
Value of variable is <=30
(Subevent 3) Value of variable is > 5 >>> Set result variable for 2
Value of variable is <= 20
(Subevent 4) Value of variable <= 5 >>> Set result variable for 1
```

You’re now working with a normal percentage (1-100), and have weighted the results.

Thanks, problem solved