For every N occurrences of E, E is = V. So V has a probability of 1/N of happening.
What does that mean??
I think you are misunderstanding the word "event". Probably what you are thinking is that you do something (this might be called an
experiment, rather than an event), and 1/N of the time the outcome of the experiment is an
event V, so that P(V) = 1/N.
Let 'N^m' represent multiples of N.
An exponent represents multiplying N by itself, which is not the same as a multiple of N. I'm not at all sure what you mean here.
It seems to me that the odds of V happening at exactly the ratio of 1/N increase in proportion to 'm'; it seems that there is an underlying probability that determines how likely it is that V occurs exactly according to 1/N, and that this underlying probability is proportional to 'm'.
You probably mean "probability", not "odds", which is a different way to measure likelihood of an event.
And I doubt that you are really talking about "V happening at exactly the ratio of 1/N"; that would mean that out of mN trials, V happens exactly m times, which would be rather unlikely. Probability deals with approximation; it says that
on the average, V will happen
about 1/N of the time, not exactly that often.
To illustrate. Say the probability of encountering the Pokemon Swinub in the wild is 1/10; if I were to encounter 100 different Pokemon it seems to me that the odds that 10 of them would be Swinub are higher than those odds of one of them out of only 10 encounters being Swinub.
Here's how probability really works: If we suppose that 1/10 of all Pokemons in the world are Swinub, and they are randomly distributed, then out of 100 Pokemon you find, you would expect that
about 10 will be Swinub. But we can use the binomial distribution to calculate the probability that
exactly 10 of 100, or exactly 1 of 10 will be Swinub:
P(exactly 1 Swinub out of 10) = C(10,1) (0.1)^1 (0.9)^9 = 0.3874
P(exactly 10 Swinub out of 100) = C(100,10) (0.1)^10 (0.9)^90 = 0.1319
So in fact, it is considerably less likely to get
exactly 10 out of 100 than to get
exactly 1 out of 10. And that makes good sense, because there are many more possible numbers to get in the former case!
In any case, what I have demonstrated is that whatever effect you are looking for, it will not be some extra something underlying probability, but something
on top of the basic probabilities (in this case, the binomial distribution)! There is no mystical force, just an application of the laws of probability to a particular question.
But I imagine what I've done here is not what you have in mind, and you will have to restate your idea using proper terminology to make it clear.
