Below is the Jackpot in my home country. Based on my calculation, the probability of Jackpot 2 is higher than Jackpot 1 although Jackpot 1 has higher prize, which does not make sense to me.
Is my calculation wrong?
Updated: Based on the comment and answers, it seems that the ticket bought is important on the probability calculation.
6 numbers needed to pick per ticket and no need to indicate which is the bonus number in ticket.
Ticket bought example: 3, 9, 13, 23, 33, 43 (no need to specify bonus number)
Draw result: 3, 9, 13, 33, 40, 43 + 23 (bonus number)
Then you win jackpot 2
Jackpot 1:
Winning criteria: 6 matching number out of 50 numbers
6 matching number: 50 choose 6 = 15890700
Probability: 1 in 15,890,700
Jackpot 2:
Winning criteria: 5 matching number out of 50 numbers + 1 bonus number from the 50 numbers
5 matching number: 50 choose 5 = 2,118,760
1 bonus number: 50 - 6 = 44 numbers remaining
5 matching number + 1 bonus number = 2,118,760 x 44 = 93,225,440
Probability: 1 in 93,225,440
Your probability for Jackpot 1 is correct: $\frac{1}{50 \choose 6}=\frac{1}{15890700}$.
Your probability for Jackpot 2 is wrong.
There are ${50 \choose 7} = 99884400$ ways of choosing the six main balls and the bonus ball if you ignore their status ($7$ times more if you take account of the status). Of these, $50-6=44$ will have the six numbers on the ticket, so the probability of winning either Jackpot 1 or Jackpot 2 is $\frac{44}{{50 \choose 7}}$ which is seven times the probability of winning Jackpot 1.
By subtraction, this make the probability of winning Jackpot 2 be six times the probability of winning Jackpot 1, i.e. $\frac{6}{50 \choose 6} = \frac1{2648450}$.
Approached a different way, the probability that the first six balls drawn have five of the numbers on the ticket and that the seventh ball matches the missing number is $\frac{{6 \choose 5}{50-6 \choose 1}}{50 \choose 6} \times \frac1{50-6 \choose 1}=\frac{6}{50 \choose 6}$ again.