This problem has three variables to consider.
tickets(T) - The number of tickets sold for the given drawing
winnings(W)- The prize availible to win
probability(P) - The probability your numbers being the winnning numbers
The P is constant for a given game, it onlt depends on how they draw the numbers. So for this game:
P = 1/(52*CHOOSE(52,5)) = 1/135155920 = 7.399x10^-9
From P and T you can calculate the average number of winners (AvgN) for a given drawing
AvgN = SUM(i=0,T, i * CHOOSE(T,i) * P^i * (1-P)^(T-i) )
So if you have 135155920 Tickets sold the average number of winners is exactly 1. This doesn't mean that if you sell 135155920 tickets that you are guaranteed a winner however.
p(0) = 36.79%
p(1) = 36.79%
p(2) = 18.39%
p(3) = 6.13%
p(4) = 1.53%
p(5) = 0.31%
p(6) = 0.05%
There is still a 36.79% chance nobody will win and a 0.00832% chance that >6 people will win.
For this game your expected return for participating (EV) is this
EV = W*P/AvgW
By this equation if W=135155920 and T=135155920 then the lottery is an even game.
Of course none of this takes volitility in to account. Since the lottery is a highly volitile game you would be required play a massive amount of games before you could expect to converge to the EV. In other words you would need to play about 1,000,000 games before you could expect to be +-0.05 from the EV. I don't know about you, but I'm sure not ready to sink that kind of money in to the lottery.