Hi everyone, I'm quite new to PD and I've been stuck on this problem for ages. I have three random object boxes that send random numbers between 1 and 7 each time they are banged and I'm trying to connect them up in such a way that when all of the random numbers happen to be identical (i.e. equal each other) a new bang is sent out. So for example, if the random numbers of the three boxes all happened to be the 4 at the same time, a bang would be sent out. I'm trying to invoke a slot machine effect so that when the numbers equal one another another bang flashes repeatedly indicating that the player has won and matched all the numbers up. If anyone has any suggestions at all I'd really appreciate them, thanks so much!

@GDavids103 Welcome to the forum!
Try this......... slot.pd
David.

i did this one!
slot machine.pd

Thank you both so much, I've got it working now! Really appreciate the help

I'm just trying to add a pipe delay of 1 second between each number reveal, but I can't seem do be able to get it correct. Would anyone have an idea of how to do this, whilst also keeping the flashing bang when all the numbers equal eachother? Thank you!

@GDavids103 Should do that........ slot_machine.pd
Because I have set the numbers to appear from left to right the right number appears last.
But [expr] ...... like all Pd objects has its "hot" inlet at the left....... it does its action when that inlet is banged.
So I put a trigger [t b f] to make that happen when the last number appears.
If they appeared right to left that would not be necessary.

Of course your next question will be how to make them spin, with one above and below as well......
I will leave that for you to solve for now.....
David.

Ah I see, thank you very much!

@GDavids103 There was an error because of course the first number would bang through while the previous other two numbers were present......... and an error in [expr]
I have fixed it in my last post.
The [f] holds the left most value and it is only banged through when the right most inlet has received its value.
But it is much more pleasing visually to solve that like this.......

David.