Better odds with $9 bucks on lotto? I think I have a juicy one! My state's lotto game is 2 dollars, you select 6 numbers between 1 and 40, (odds to get all 6 are 1 in 3,838,380). However, if you pay an extra 1 dollar (so 3 in total) you get a second "Bonus" chance to win, meaning it's still your original 6 numbers that you selected, but they do a second "Bonus" drawing. So to clarify, it's 3 bucks, one set of 6 numbers you've selected, but two drawings on the same day/time.
Now, Let's say you have 9 big ones in cash and you want to put it all on a single game night.
So if your goal is to get a match of all 6, do you have better odds buying 3 "Bonus" tickets at 3 bucks each or is it better to buy four standard tickets at 2 dollars each.
The real question: are the odds better if you have 3 sets of 6 numbers and 2 drawings, or 4 sets of numbers against a single drawing??
Thanks in advance!!!
Sincerely, Hopeful Financial Planner
 A: The events are so unlikely that they are almost independent. Therefore a good approximation is to sum the probabilities. Since $2 \cdot 3 > 4$ it is $3/2$ times better to spend 9 than 8 and it would require 12 dollars to match the probability obtained with the bonus.

 The probability of any given number is
 $p = \binom{6}{40}$
 The probability that a second number wins depends on if there is any overlap. No overlap maximizes the probability so let's assume that.
 $$ P(\texttt{win_2}) = P(\texttt{win_2}\mid \texttt{win_1}^*)P(\texttt{win_1}^*) + 0 = p\cdot(1-p)$$
 which generalizes to $p(j) = p\cdot(1-p)^j = p + \mathcal{O}(p^2)$.
 I.e.
 $$ p(x) = P(\texttt{win}) = P(\cup_j^x \texttt{win_j}) = x p +\mathcal{O}(xp^2)$$
 The probability of winning in any of y draws is $1-(1-p(x))^y$ and the probability of winning in two with 3 tries in each is:
 $$1 - (1-p(3))^2 = 1 - (1 - 2\cdot p(3) + \mathcal{O}(p^2)) =2\cdot 3 \cdot p +\mathcal{O}(p^2) > 4\cdot p+\mathcal{O}(p^2) = p(4).$$
 So, the probability is $\frac{3}{2}$ times larger of winning and you would need to spend 12 dollars to get a similar probability.

