To maximise my chance of winning one prize should I put all my entries in a single draw? Every week there's a prize draw. It's free to enter using the code from a soup tin lid. You can enter as many times as you like during the week until Monday's draw and then it starts all over.
The prizes are experiences and the value to me is not especially relevant. Winning more than once might be worth more in value terms, but I only want to win once.
So my question is; in order to maximise my chance of winning once should I;


*

*batch up my lids and enter as many of them as I can in one go, probably during the last week of the prize draws?

*enter lids as I go along.


It feels like (1) to me, but the maths to explain it is beyond me.
To be clear, I'm not actually interested in trying to gain a winning advantage. I realise that in reality the odds are small and about even either way, but as a maths puzzle I'm interested.
Say for instance that I have 10 lids. Lets assume that the number of other entries are an even number (say 5 each week), and lets say there are 10 weeks and one prize available per week.
Do I have more chance of winning a prize by entering one lid over 10 weeks, or 10 lids in any other week.

Let's assume that you eat $k$ soups per week and want to maximise your chance of winning at least once during next $n$ weeks. The question is: Is it better to (1) enter the game each week with one lid or (2) collect lids and enter only once? In order to estimate the probability in each case we have to make some additional (reasonable) assumptions: (a) total number of lids in the game is (approximately) the same each week and equalls $m$, (b) $m$ is much bigger than $n$, (c) every week only one lid wins. Then the probabilities of winning are: 
(1) $P_1=1-(1-\frac{1}{m})^n$
(2) $P_2=\frac{n}{m}$
Because $m$ is much larger than $n$ the probabilities are almost the same. 
