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.
 A: If there are lots of entries (so the number $n$ of your entries doesn't much change the probability of a number winning) and the probability of one ticket winning is $p$ so that $np$ is small then:
If you enter all at once, the probability of winning is $np$
If you enter in $n$ separate weeks the probability of winning is $1-(1-p)^n=np-\binom n2p^2+\dots$
Since the trailing terms in the binomial expansion are small, because $p$ is small, you do better to enter all at once.
The reason for this is that each ticket has the same chance of winning, but when you split your numbers, there is a chance of winning more than once. When you enter all in the blame week (given only one prize) you can only win once.
There is a $\frac 16$ chance of throwing a $6$ with a fair cubical die. With two dice the chance of throwing at least one $6$ is $\frac {11}{36}$ rather than $\frac 13$. Where did that extra $\frac 1{36}$ go? Well the throw of double six counts one throw but two sixes. The principle is the same.
A: Let's say there are $n$ draws and you have exactly enough lids to enter one in each draw. Each draw, there are $m$ other lids entered. Then, your chances are


*

*$$P_1=\frac{n}{m+n}$$

*$$P_2=1-\left(1-\frac{1}{m+1}\right)^n=1-\left(\frac{m}{m+1}\right)^n$$


Explanation:
$_{\text{chance of winning at least once} = 1 - \text{chance of never winning}}$
$_{\text{chance of never winning} = (\text{chance of not winning the draw})^\text{number of draws}}$
$_{\text{chance of not winning a draw} = 1 - \text{chance of winning a draw}}$

Now let's plug in the numbers. With $5$ other lids every draw and $10$ of your lids (so $n=10$ and $m=5$), this becomes


*

*$$P_1=\frac{10}{5+10} = \frac{2}{3}$$

*$$P_2=1-\left(\frac{5}{5+1}\right)^{10}\approx0,84$$


So be patient and enter a single lid each draw.
Unless you can conspire with the other 5 contestants to each have a go with just their tickets for 6 draws.
A: 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. 
