Odds of winning lottery with multiple unique tickets If I buy one lottery ticket with odds of $14M$ to one if I buy another ticket with different numbers does this slash the odds to $7M$ to one? if so it follows that if I double my tickets again to $4$ Tickets it will half again to $3.5M$ to one? 
So $8$ tickets will be $1.75M$ to one. 
$16$ Tickets $\rightarrow.875M$ to one
$32 \rightarrow .4475$M to one 
$64 \rightarrow 218750$ to one 
$128 \rightarrow 109375$ to one
$256 \rightarrow 54587$ to one 
Something seems wrong here, i am not looking for a exact answer just ballpark working out.
 A: 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! ;)
A: Nearly but not quite.  Start with a simpler case e.g. two dice and you are trying to get at least one 6.  What are the odds?  A common naive answer is 2/6.  However, this is obviously wrong, consider 3, 4, 5, 6, and 7 dice.  With 6 dice, getting at least one 6 is likely but certain.  With 7 dice, it is not more than certain.  
The correct answer can be seen by listing all 36 ways that two dice could fall.  You will see that only 11 out of 36 contain a 6 so the odds are only 11/36.  The overcounting is due to one of the 36 cases being double 6.  
Back to the lottery, this approach is not so practical but fortunately there is an alternative.  It is easier to calculate the chances that you don't win with $n$ tickets.  
Don't win with 1 ticket: $\frac{13999999}{14000000}$.
Don't win with 2 tickets: $(\frac{13999999}{14000000})^2$.
Don't win with 3 tickets: $(\frac{13999999}{14000000})^3$.
Don't win with n tickets: $(\frac{13999999}{14000000})^n$.
Subtract from 1 to get the odds of winning.  
With a very low probability such as $\frac{1}{14000000}$, the first few will be very close to the naive answers.  As $n$ gets bigger, the difference will become more obvious.  
Thanks to comments by Losvos and TMM: note that I have accidentally assumed that each ticket is an independent random choice as would be typical with dice.  If you are careful not to repeat your selections then the naive logic will be correct.  Buy all possible 14000000 tickets and you are sure to win.  However, it is very likely that you will win less than the cost of the tickets.  Even if the jackpot is higher, due to rollovers, be aware that someone else might win and you will share the jackpot.
