You'll be surprised. The correct probability of winning at least one ticket is around $0.2242$.
Assuming exactly one prize is given, your answer of $\frac{1}{160}$ is the probability of winning is correct. That is, you go home empty-handed with probability $\frac{159}{160}$. However, $40$ tickets are chosen for prizes, not just one. So even if you miss out on a prize the first time, you could still end up with the second winning ticket; or the third; or the $40^{th}$. What we need to calculate is the chance of winning at least one of those tickets.
For the moment, assume that the prizes are drawn with replacement. Then in order for you to not get a prize, you need to miss the first time, and the second time, and the third time, and so on, until the $40^{th}$ time. Under our assumption that these are drawn with replacement, all these $40$ events are independent. Therefore, the probability that you miss out on a prize is simply the probability that miss out in any given trial, raised to the power of $40$; i.e.,
$$
\left(\frac{159}{160} \right)^{40} \approx 0.7782.
$$
Hence, the chance that you win a prize is $1 - 0.7782 \approx 0.2218$.
When the prizes are drawn without replacement. Now we are going to compute the exact answer without any assumptions. There are $1600$ tickets, out of which you bought the first ten (say). The judges pick $40$ winners out of the $1600$ tickets; this can be done in $\binom{1600}{40}$ ways. [See binomial coefficients in Wikipedia.] Of these, you will not win a prize if those $40$ tickets are drawn from the $1590$ tickets that you did not buy. That is, there are $\binom{1590}{40}$ possible outcomes in which you will go home empty-handed. That is, you go home empty-handed with probability
$$
\frac{\binom{1590}{40}}{\binom{1600}{40}}.
$$
Therefore, you will win a prize with the complementary probability
$$
1 - \frac{\binom{1590}{40}}{\binom{1600}{40}} = \frac{1420730930795547} {6335978517846620} \approx 0.2242.
$$
As you can see, that the approximate answer is quite close to the exact one.
[I did these calculations in Wolfram Alpha.]
No - there is no difference in the probability of a particular ticket winning Prize 1.
In scenario A the probability is obviously $\dfrac{1}{100}$.
In scenario B it is $\dfrac{99}{100}\cdot\dfrac{98}{99}\cdot\dfrac{97}{98}\cdot\dfrac{96}{97}\cdot\dfrac{95}{96}\cdot\dfrac{94}{95}\cdot\dfrac{93}{94}\cdot\dfrac{92}{93}\cdot\dfrac{91}{92}\cdot\dfrac{1}{91}=\dfrac{1}{100}$
From personal experience Scenario A allows the top prize winner to choose their prize from the set of ten. It also allows people with several tickets have who won earlier prizes to waive further prizes if they have already won one, leading to a redraw. But it reduces the tension in the draw compared with Scenario B.
Best Answer
There are $\binom31\binom{97}3$ ways Jason could have exactly $1$ winning and $3$ non-winning tickets.
Against this, Jason could have got any $4$ tickets in $\binom{100}{4}$ combos.
Thus $Pr = \dfrac{\binom31\binom{97}{3}}{\binom{100}4}$