Nope. You are multiplying two statistical factors that are not independent factors. Consider this: suppose the lottery asks the entrant to pick a number from 1 to 1000. Suppose exactly one person enters, and guesses 12 as his entry. What are the odds of someone winning this lottery? One in a thousand. What are the odds of you being that winner? The same: one in a thousand. It is not P/N.
If it was possible to know, out of all the people who entered the lottery, how many different possible winning numbers were chosen by at least one person and that number was n, and the total possible winning combinations was N, then the odds of *someone* winning would be n/N. The odds that you would be one of them, if you put in only a single ticket, would be 1/n. The actual odds of you being the winner would then be (1/n) * (n/N). Notice that ends up being 1/N, which is also the odds of winning the lottery computed directly: you put in one entry, and the odds of that one entry matching the winning combination is 1/N.
You math is still correctly calculated but misapplied. Somehow, you are invoking the set n, the total nunber of *different* combinations entered out of all the people who enter P. That quantity has nothing to do with the discussion at hand, but its an improperly defined value in your calculation assumptions.
Arcana, I already spotted my own error. Of course I multiplied the factors. The percent chance OF a winning draw is applied to the chance OF a given person being the one to draw it. The word "of" indicates multiplication.
There is a number games played, G.
In each game the players act to create a pool of possible winning numbers of size V each game. For the sake of simplification I will treat this as a constant and it will be the same value each game.
In each pool there will be a certain percentage of duplicates, D. Also being treated as a constant for the same reasons as above.
The pool of unique numbers, U is equal to V*(1-D).
p is the percent chance of a specific number sequence being generated in the game drawing.
The chance of a player having picked a specific entry in the pool is 1/V for each game.
The chance of one of the numbers in the pool being drawn is 1-(1-p)^(U*G)
I'm not going to compute the odds of one of the duplicate values being drawn. This is complicated enough as is.
Let's recap. The chance of a given pool producing a wining entry is random with a binomial distribution, meaning the chances of a winning entry being produced increases with each game.
The chance of a given player having that winning entry is random with a flat distribution.
Both random selections are independent of each other.
The odds of two random selections occurring together is the product of their individual probabilities. R=(1-(1-p)^(U*G))/V
The odds of a random individual winning the lottery is 1-((1-(1-p)^(U*G))/V)^G if the same people play each game. The odds of an individual winning changes from game to game within these constraints.