Answer depends on who's asking the question. If it's a good 'math' teacher the answer chance at least one ticket wins, 1 - (1 - q)^2, which fellow algebra fans can simplify to compute in their head by recalling (a - b)^2 = a^ - 2ab + b^2 here, with a=1 and b=q=-10^-6, (1 - q)^2 expands to 1 - 2*10^-6 + 10^-12. Subtracting that from the initial 1 leaves Bob's chances of walking at 2*10^-6 - 10^-12, which converts to decimal percentage as 0.0001999999%. Note that is LESS than double the chance of one ticket winning because purely mathematically winning both tickets is redundant towards getting your legs out of hock. Math is FUN and thinking the above through was easier than tapping it into my iPad!
However a great teacher would point out that in the real world government taxes are withheld from lottery payouts. Winning just one ticket will leave Bob's legs short by that amount. So he'll need to win both tickets to cover the loan shark, a 10^-12 = 0.0000000001% chance. And even that chance depends on not being in a delayed pay jurisdiction (loan sharks are too smart to accept Illinois IOUs) or if someone like Bernie wins and decides to withhold more than 50% gambling taxes so even a double win wouldn't leave you with enough!
