MegaMillions Overlay?

[BigFishSmallCardRoom]

I was wondering if someone could do the calculations and tell me when is it +EV to play the MegaMillions lottery? What does the jackpot have to be? What adjustments need to be made to factor in the chance that you will have to split the jackpot? What about the payouts for hitting some of the numbers but not all of them?

Thanks,

Biggs

P.S. I'm not being lazy, I'm just not sure how to do the calculations.

[Delphin]

If you post all the numbers, I'll do the calculations. How much is a ticket. How many numbers do you pick and from how many choices? How much is the payout for getting 3,4,5 numbers correct?

[JinX11]

I believe it's something like this:

You pick 6 numbers, the first 5 numbers from a pool of 1-56 (no duplicates), the last number (the "Mega" ball) from a pool of 1-46.

$1 per ticket. Payout schedule:

5 + Mega Ball = Jackpot (currently $225 mill.)
5 (no Mega Ball) = $250,000
4 + Mega Ball = $10,000
4 (no Mega Ball) = $150
3 + Mega Ball = $150
2 + Mega Ball = $10
3 (no Mega Ball) = $7
1 + Mega Ball = $3
0 + Mega Ball = $2

[JinX11]

If I managed to do the math correctly, the jackpot value I get for 0 EV is a little less than $144 million.

EDIT to add: I have neglected the possibility of splitting the jackpot with one or more other people who chose the same correct numbers as you. If you were to split the jackpot with one other person, the jackpot would need to be twice as large so that the effect of splitting it with another would be nil (i.e., about $288 mil). If split between three people, it'd need to be three times as large (i.e., about $432 mil).

[BigFishSmallCardRoom]

How do you get these numbers?

And for the possibility of duplicate numbers, i figure you only have to multiply the chance of hitting the jackpot times itself then multiply the amount of the jackpot to get your -EV for that event. Should only discount it by not that much.

Bigs

[BigFishSmallCardRoom]

Straight from the MegaMillions website:

http://www.megamillions.com/images/odds_062205.gif

[BigFishSmallCardRoom]

I don't understand how they get the chances they post. Why is the chance to hit the Mega number 1 in 75 when there are only 46 Mega numbers?

[AndysDaddy]

[ QUOTE ]
I don't understand how they get the chances they post. Why is the chance to hit the Mega number 1 in 75 when there are only 46 Mega numbers?

[/ QUOTE ]

Becuase to win, you must not only pick the mega number correctly (1 in 46) but also have none of your five normal numbers hit (about 3 chances in 5).
--
Scott

[JinX11]

[ QUOTE ]
How do you get these numbers?

[/ QUOTE ]

I went to the megamillions (http://www.megamillions.com/) website and found the chances listed of hitting each of the prize levels (this is what another poster has placed in this thread) - I could have tried to calculate this, but I was lazy; since they are published, I figured they are accurate.

Then, for each prize level, I multiplied the chance of winning the prize times the amount of payout (e.g., at the easiest prize level, $2, this would be $2 * (1/75) = 0.02666666....). Finally, I summed up the output of these prize-level multiplications.

At $144 million, the resulting sum is $1.00141, which is slightly more than the $1.00 ticket price. With the current estimated payout for Friday night's drawing at $262 million, this value becomes about $1.6730.

[ QUOTE ]

And for the possibility of duplicate numbers, i figure you only have to multiply the chance of hitting the jackpot times itself then multiply the amount of the jackpot to get your -EV for that event. Should only discount it by not that much.


[/ QUOTE ]

Hmmm...not sure if I agree here - I do agree that you need to perform some multiplication on your chance of winning the jackpot, but I do NOT think you should multiply it times itself. It seems that you should multiply it times the chance that two people choose the same 6 numbers, the probability of which is NOT the same as the chance of you getting the correct 6 numbers and winning the jackpot. Obviously, this chance goes up as more people play and it goes up pretty fast, actually (google: "Birthday Problem" for a similar situation - the chance of an event occurring goes up fairly dramatically the larger your sample) - I need some time to calculate the chance that two people choose the same set of numbers.

[JinX11]

Ok, I gaffed on my last post - this situation isn't really comparable to the "Birthday Problem", which seeks to determine out of a certain population of people, what is the chance that any two members share the same birthday. This scenario is different - we don't want to know what is the chance that two people share the same numbers - we want to know what is the chance that one other person shares our (winning) number. It's still a factor of the population size, but unlike the "Birthday Problem", it doesn't scale as rapidly as the population size grows.

I have no idea how many entries are sold in the megamillions, so ultimately, I have no idea what the chance is someone will share your set of numbers. But, since there are P = 175711536 combinations (according to the odds posted), if the number of entries sold is given by n, I believe the probability formula would be given by:

1 - ((P-1)/P)^n.

So, at 100,000 entries, there is less than a 0.06% chance of someone sharing your winning numbers. At 1,000,000 entries, 0.57%; at 10,000,000 entries, it goes to 5.53%. I have no idea how many entries are typically sold in each drawing.

[DoomSlice]

I was under the impression that lotteries were designed so that there was always a negative EV.

[chunk]

this problem has been mentioned several times.

common concerns that are often neglected are the distribution of payments and taxes. Typically the actual breakeven EV number needs to be calculated as 25-35% of what the advertised jackpot amount is.

[JinX11]

[ QUOTE ]
this problem has been mentioned several times.

[/ QUOTE ]

Yes; and there are a number of sites out there on how to analyze the lottery, this being one of them. (http://www.stat.umn.edu/~charlie/lottery/)

[ QUOTE ]
common concerns that are often neglected are the distribution of payments and taxes. Typically the actual breakeven EV number needs to be calculated as 25-35% of what the advertised jackpot amount is.

[/ QUOTE ]

Good point. I missed these, as well.

[pzhon]

[ QUOTE ]

Yes; and there are a number of sites out there on how to analyze the lottery, this being one of them. (http://www.stat.umn.edu/~charlie/lottery/)


[/ QUOTE ]
That site makes the derivation far too complicated. You don't need to use the Poisson distribution and then sum an infinite series as that site does. You can calculate your expected share of the jackpot by computing the expected jackpot money paid out divided by the number of tickets sold.

The expected total jackpot money paid out is the jackpot times the probability at least one person wins. This is easy to compute if you assume the tickets are independent.

If you have one of N tickets which win independently with probability p, your expected share of the jackpot J is J*(1-(1-p)^N)/N.

[MikeSmith]

I hope you guys are buying lotto tickets for better reasons than the EV, thats all im gonna say.

[BigFishSmallCardRoom]

Thanks for all the input and calculations. I was the winner of the $313 million jackpot a few weeks ago.