01-22-2021, 05:29 PM
Your P(0x) were what I computed initially. The complicating factor is that the last pattern wins if multiple rows have their FX1 evaluated.
So for pattern 01 to be jumped to, we don't only have to get lucky on our 25% chance on the first row, but also we have to avoid any of the other rows getting lucky with theirs. So I still believe the probabilities in my "EDIT" above, P(01) = 0.25 * 0.75 * 0.75 * 0.75 = 10.5%
In any case, as I said initially, the goal is to add a bit of structured randomness to the snare pattern, so it doesn't matter exactly what the chances are as long as the net result sounds good
So for pattern 01 to be jumped to, we don't only have to get lucky on our 25% chance on the first row, but also we have to avoid any of the other rows getting lucky with theirs. So I still believe the probabilities in my "EDIT" above, P(01) = 0.25 * 0.75 * 0.75 * 0.75 = 10.5%
In any case, as I said initially, the goal is to add a bit of structured randomness to the snare pattern, so it doesn't matter exactly what the chances are as long as the net result sounds good