Here is the scenario: There is a lottery running for $n$ terms, which means that it is repeated. In each term, there are a total of $T$ tickets and one prize. You currently own $t$ tickets and your dilemma is to either use all of your tickets in one go or spread them for $n$ terms. The probability of winning at least one prize (naturally, only one in the first scenario) when you group or evenly spread your tickets can be calculated respectfully:
$P_1=\frac{t}{T}$ and $P_2=1-(\frac{T-\frac{t}{n}}{T})^n$
Let's say that $T=100$, $t=12$ and $n=2$:
$P_1=\frac{12}{100}=0.12$ and $P_2=1-(\frac{100-\frac{12}{2}}{100})^2=0.1164$, hence $P_1>P_2$.
Even if I try to spread the tickets unevenly, like 11 tickets in one term and 1 in the other, the relation is the same:
$P_1=0.12$ and $P_2=1-(\frac{100-11}{100})⋅(\frac{100-1}{100})=0.1189$, still $P_1>P_2$.
The mathematical model where the tickets are distributed unevenly becomes:
$P_2=1-\prod_{i=1}^n\frac{T-t_i}{T}$ where $t_i$ is the number of tickets spent in each term.
I tried plotting the mathematical models on Desmos and playing around with different combinations of variables, but it always seemed that using all tickets together always, even if by a minuscule margin, gives better chances of winning anything at all than spreading them in every case.
Will this always be the case; should we always use all of our tickets in one go? How can it be mathematically proved then? I think that the number of prizes shouldn't change the outcome, should it?
Thank you for reading!
Suppose $n=2$. Using $x$ tickets in the first lottery and $t-x$ in the second yields a probability of winning $1-(1-\tfrac{x}{T})(1-\tfrac{t-x}{T})$. The maximum in the range $0\leq x \leq t$ is at the bounds, so it is better to attend in only one lottery, either the first or the second.
This is true in general for $n>2$ (by induction). The idea is that you can always repeat the previous argument for two of the lotteries (say, the 4th and the 10th) keeping the number of tickets used in the others fixed, and deduce that the optimal strategy would be to use all the tickets dedicated to the 4th and the 10th lottery in only one of them.