I'm not afraid of math so please explain :)
Generally speaking the free ticket counts as ticket price * RTP for the EV. Unless there are weird rules limiting you in some way.
OK if the return is 50% and a ticket cost $1 then the EV for the free ticket would be 0.50....wouldn't that inherently make the return of a ticket greater than 50%? I'm kinda looking for a mathematical explanation
The EV of that result is equal to the EV of the ticket. So you would take the EV of the other prizes, add them up, subtract the ticket price, call that x, and then you get x' = x + px' => x' = x/(1-p)
"subtract the ticket price" I think you mean do this at the very end? Otherwise that comes out to -0.558 which makes no sense if the total other prize EV is -0.50. So if someone wants to check my math for NY Take 5 it would be total 55.8% rtp or -0.442 EV.
When I gave you my instructions, I was assuming the game had fixed payouts, as well. What I probably should have said was "work out the expected return (not the EV) as if 'free tickets' were a loss, then multiply by 1/(1-p)."
So in the case of the Take 5, that would indeed be 50% * 1.116 = 55.8%.