# Lotto question: 2 weeks with same numbers or 1 week with 2 different numbers

We are playing $90$ number lotto, where they draw $5$ number out of $90$. Is it better to play two different combination for one week or play same combination for $2$ weeks to win the lottery?

My idea is that: In the first case, since the optimal event is only the $5$ winning number, It doesnt matter, if you have another lotto ticket with different numbers, they cant be the lucky numbers or if they do, then the first ticket was useless. On the other hand, in the second case you can win the lottery in both weeks. Any suggestion/help is appreciated.

The odds of winning when you pick two different tickets in the same draw, the probability that you win is:

$$\frac{2}{\binom{90}{5}}$$

If you buy one ticket in each of two events (whether those tickets are the same or different doesn't matter) the probability that at least one of them wins is:

$$\frac{2}{\binom{90}{5}}-\frac{1}{\binom{90}{5}^2}$$

So you are very slightly less likely to win if you buy one ticket in each of two draws.

It turns out the expected amount that you win, however, is exactly the same. That's because you can win more, potentially, by buying tickets in two draws. Specifically, you can win in both draws.

If the expected winnings of a single draw is $W$, then with two tickets in one draw, your expected winnings are:

$$\frac{2}{\binom{90}{5}}\cdot W$$

On the other hand, with one ticket from each of two draws, you expect to win both of them with probability $\frac{1}{\binom{90}{5}^2}$. So the expected value is:

$$\left(\frac{2}{\binom{90}{5}}-\frac{2}{\binom{90}{5}^2}\right)\cdot W + \frac{1}{\binom{90}5^2}\cdot2W = \frac{2}{\binom{90}5}\cdot W$$

Consider the extreme case that you buy $\binom{90}5$ different tickets in one draw, or you buy on ticket in each of $\binom{90}5$ draws. In the first case, you definitely win in that one draw. In the second case, you can actually lose all the draws. On the other hand, again the expected winnings in both cases are the same.