On the assumption that there are lots of tickets sold (so your buying one more does not change the value of the existing tickets), each ticket has some expected value $v$. This includes all the prizes. If you buy another ticket, you get $v$. If you buy a multiplier, you get $v(\frac 12 \cdot 1 + \frac 3{10}\cdot 2 + \frac 1{10}\cdot 3 + \frac 1{10}\cdot 4)=\frac {18}{10}v$ added to your value. Your loss on buying another ticket is $C_1-v.$ Your loss on buying a multiplier is $C_2-\frac {18}{10}v.$ Compare these and you will know which way you are less badly off.
The odds are in this case given by
$$\text{number of tickets you have}:{\text{number of tickets you don't have}}.$$
The important thing is that the denominator is not $\text{total number of tickets}$. So if you have $2^n$ tickets and the total number of tickets is $T,$ the odds of you winning will be
$$2^n:T-2^n \iff 1:\frac{T}{2^n}-1,$$
which is different to your calculations. Let's take a look:
$n = 1$ gives odds of $ 1:\frac{T}{2}-1$
$n =2$ gives odds of $1:\frac{T}{4}-1,$
and so on, which grows much slower than your solution.
Edit: Here you can see a graph of your odds of winning as a function of the amount of $n$, the number of times you double your number of tickets. As you can see, it does actually grow exponentially, but it is off to an extremely slow start. At around $n=24$, you have bought all the tickets, so here the odds are $1:0,$ which gives the singularity shown.
So the bottom line is: Don't play the lottery! ;)
Best Answer
Yes, you do double your chance of winning. There are $\binom{49}{6}$ different possible draws. The probability of winning with one ticket is the chance that the 6 drawn balls are one specific combination, i.e. $\frac{1}{\binom{49}{6}}$. With two tickets, there are two winning combinations, so the probability is $\frac{2}{\binom{49}{6}}$
Basically, the reason it is doubled is because winning one ticket and winning the other ticket are mutually exclusive, i.e. you can't win both tickets (note the problem statement says "different tickets"). If there was a chance of winning both, the chance would be less than doubled.