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Some of you guys are over thinking this.
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You may be right.
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Since the numbers are listed from low to high, just choose one of the numbers (say the 3rd lowest) and let odd/even determine who wins.
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You seem to be under thinking this. You didn't bother to check whether your solution was valid.
This same technique would apply in variants with no solution, when the number of possible tickets is odd, such as if the numbers were 1-47 instead of 1-49. That doesn't mean it is wrong, but it should make you worry, particularly if you mean to correct my over thinking.
On 6990896 tickets (49.9928%), the third number is even.
On 6992920 tickets (50.0072%), the third number is odd.
Mathematica code:
<ul type="square"> Sum[Binomial[i, 2]If[EvenQ[i + 1], 1, x]Binomial[49 - i - 1, 3], {i, 2, 45}]
6990896 + 6992920 x[/list]