July 19th, 2014 at 11:17:36 AM
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I've noticed that in some lottery games, instead of the bottom prize being ~cost of a ticket, they give you a free ticket for the next draw. How does this factor into the total EV? I wonder because, if every ticket has equal EV, then a free ticket would actually increase the EV, and so on. You have chance of winning on the free ticket (even another free ticket on a free ticket), so is it treated separately? Example, NY Take 5 game. I also curious how it affects rolling jackpot situations. I'm just trying to figure out how you would calculate the EV on these types of bets.

I'm not afraid of math so please explain :)

Thanks

I'm not afraid of math so please explain :)

Thanks

July 19th, 2014 at 12:08:21 PM
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Generally speaking the free ticket counts as ticket price * RTP for the EV. Unless there are weird rules limiting you in some way.

July 19th, 2014 at 12:24:33 PM
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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)

The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.

July 19th, 2014 at 12:26:48 PM
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Quote:NeutrinoGenerally 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

July 19th, 2014 at 12:40:36 PM
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Quote:24BingoThe 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.

July 19th, 2014 at 3:49:23 PM
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You're right, just add up the prizes, don't subtract the ticket price yet, because that math takes into account the return only. (Or add the ticket price back in.)

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%.

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%.

The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.