On-Off Cycle probability

"The cycle may or may not restart right away." implies unknown duty cycles which would seem to make the problem unsolvable. Otherwise, are the cycle parameters the same for all processes?

OK. So we can say the cycle parameters are the same for all cases except for when they would start. In a practical sense an unlimited number of processes could not start simultaneously so at some theoretical steady state I image you could say each would restart right away but the position at any time of each case would be random. Is this then solvable?

If I think of the limiting cases: zero quantity would mean zero probability, a quantity of one would be easy - the "on" time divided by the total cycle, and an infinite number would have a probability of unity that some number of them would be "on".

Is this an accurate restatement of the problem?

There are N processes. Each is on for X counts and off for Y counts, repeating this pattern forever. Each process starts at a random time (random phase shift). At some time, after all processes have been started, we wish to know the probability that M (at least or exactly?) of the N process are currently (at this instant) on.

I think so, yes. I note "at least or exactly". How would that choice change the problem?

The probability is greater for M or more than for exactly M.

Exactly M:

C(N,M) * [X/(X+Y)]^M * [Y/(X+Y)]^(N-M)

where C(N,M) = N!/[M! * (N-M)!]

Ok, thanks, so that's pretty concise, and quite impressive. I'm curious - as a 'teach a man to fish' kind of guy - to understand the approach to come up with this? I recognize the C(N,M) as the conditional probability of N given M. I'm really interested how you think your way into it?

Actually, C(N,M) is the number of combinations of N things taken M at at time.

Perhaps this explanation using your example will help. There are C(20,5) = 15,504 ways to select exactly 5 of the 20 processes. The probability that all 5 of the selected processes are on is [X/(X+Y)]^5. The probability that the remaining 15 are off is [Y/(X+Y)]^15.

If you wanted 5 or more, you could sum the results for M equals 5 through 20, or you could subtract the sum of the results for M equals 0 through 4 from 1 since the sum of the results for M equals 0 through 20 must be 1.

What I plan to do is crunch some scenarios using this result to convince myself I understand it, and attempt to construct a probability distribution. thanks very much for your help.