Inspired by Vaibhav Ojha

Since there is only one winner, If you buy two tickets it is guaranteed that at least one of the tickets is a loser. Thus stating that one of your tickets is a loser is a tautology, and hence provides no "useful" information regarding what the "other" ticket might be. Thus the desired probability is just the probability that both tickets are in fact losers, which is the product of the probabilities of choosing one loser followed by another, (without replacement), i.e.,

$\dfrac{99}{100} * \dfrac{98}{99} = \dfrac{98}{100} = \boxed{\dfrac{49}{50}}.$

We can do this using laplace law:

Favorable cases: $\binom{99}{2}$
Posible cases: $\binom{100}{2}$

So, our probability is:

$\frac{\binom{99}{2}}{\binom{100}{2}}=\frac{49}{50}$

First step: $(\frac{1}{100} W$ , $\frac{99}{100} L)$ of total $\frac{1}{1}$ alone;

Second step: $(\frac{0}{99} W$ , $\frac {99}{99} L)$ + $(\frac{1}{99} W$ , $\frac{98}{99} L)$ of total $\frac{2}{2}$ alone.

Hence $\frac{99}{100} L \times \frac{98}{99} L = \frac{98}{100} L² = \frac{49}{50} L²$

Verified that $\frac{100 - N}{100}$ shall be the chance of having all losers, particularly when N = 99 and N = 100, checked that all branches add up to 1, we can be confident that the answer is $\frac{49}{50}$ particularly when all other options given are not likely to be correct.

Answer: $\boxed{\frac{49}{50}}$