Ok let's say I have 2 switches that randomize. That means I have 4 combinations. If I only use 2 switches for event A and 1 switch for event B theoretically it would mean event A would have 66% chance to fire and B have a 33% to fire, as long as don't use the fourth combination right?
am confoozed...
so if you have no trigger for the fourth combination...
wait
the chances are:
1=25%
2=25%
3=25%
4=25%
they all have the equal percentage of firing
but am confoozed so this may be wrong...
i have no clue what you are talking about
EDIT!!!
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ok so:
1=A
2=A+B
3=B
4=[none]
lets say you onlyhave triggers for if A is set w/o b, a is set with b, and b is set w/o a.
then, when none of them are set, you have no trigger. still, it has an equal chance of happening. am I correct?
This is done, well known among people who know binary, and explained. We have no need for this topic. Ig uess you wou;dn't know since this took place a while ago so I don't blame you.
I'm not saying this to say it's a new thing. I'm asking if i have 4 combinations and remove 1 will each choice represent 33%?
Oh then yes if you have so that it picks the 4th choice it re-picks then yes it a 33% chance.
Great that's all I needed to know...
Time to close the thread.