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 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.
