Yes. The house advantage in most lotto-based lottery games is close to 50%. So, a hypothetical $2 game with a $1,000,000 jackpot, with no smaller prizes, would need to have a probability of winning of 0.5*(2/1000000) = 1 in one million to retain a 50% house edge.
Here is my strategy to turn $2 into $1,000,000 with better odds than that.
First bet $2 on any number in double-zero roulette. You can find $2 minimums at some Vegas casinos, like the El Cortez and South Point. If you win, you'll be up to $72.
Next, parlay your $72 onto another single-number bet. If you win, you'll be at $2,592.
Next, carry that $2,592 into one of the high-end casinos on the Strip, like the Wynn, Venetian, or Bellagio. Bet your $2,592 from roulette on the Banker bet in baccarat. Do this a total of nine times, letting everything ride each time. After your ninth win, you'll be up to $1,056,687. Your ninth bet would be for $541,891, which I'm sure any of these casinos would take if they saw you win it in front of their noses.
The probability of winning a single-number bet in double-zero roulette is 1/38. The probability of winning the Banker bet in baccarat is 50.6825%, not counting ties. So, the probability of two roulette wins and nine Banker wins is (1/38)^2 × 0.506825^9 = 1 in 654,404. Those are better odds than the 1 in a million you would get with a lottery and you'll have a little more than a million dollars too.