MegaMillions or PowerBall

So, here in Virginia we're now going to have the option to play MegaMillions OR PowerBall. I already only play MegaMillions when the payout is over $165 Million (which is somewhere close to the positive expectation point).

In general: Which is better, MegaMillions or PowerBall. What's the actual positive expectation point for these games? Looking at payouts for them both, how do I decide which one to play?

Lets start with the JP odds: Mega-Money is about 175 million to 1, and PowerBall is 195 million to 1. As a rule of thumb, i figure that 2/3 of the Customer contribution goes towards the JP, and 1/3 goes to the minor prizes. The House will take 50%... so that means 50 cents is the Customer contribution. Roughly 33 cents on the dollar paid goes to the JP.

As another rule of thumb, a 20-year annuity will pay double its cash value upon completion of payout. Therefore, the annuity value of the JP represents 2/3 of the dollar paid. All of this boils down to a simple rule: when the JP is 3/4 of the odds, throw a dollar at it, (50% of the odds have been sold) if you dare. $130 Million for Mega-Money, and $145 million for PowerBall. If equal or greater than the odds try two tickets. You're on your own if any JP reaches $300 million.

The problem with this way to play is that for any one draw, the aamount of tickets sold for that drawing is less than the odds (else the JP might increase by $100 million per draw!). That alone accounts for why the JP rolls over: insufficient sales per draw. Combined, there might be enough tickets sold that its probable to win the JP. If 40 million tickets are sold for one draw, the odds are still about 1 in 5 to win the JP, thats why the roll-over.

Edited to clarify points made.

Here is a return table for PowerBall assuming a $20 million Jackpot. The jackpot would have to be $162 million for a return greater than 1. At $62 million or less, the return of the PowerPlay option is greater than a ticket without it. These numbers don't match wiz of odds lottery because PowerBall is now a 5/59 1/39 game, and Powerplay is always 5x on matching all white balls, but not the powerball.

white	red	combinations	probability	pays	return	power play return
5	1	1	0.000000005121664	20000000	0.10243327478555	0
5	0	38	0.000000194623222	200000	0.038924644418508	0.15569857767403
4	1	270	0.00000138284921	10000	0.013828492096049	0.034571230240122
4	0	10260	0.000052548269965	100	0.005254826996499	0.013137067491246
3	1	14310	0.000073291008109	100	0.007329100810906	0.018322752027265
3	0	543780	0.002785058308144	7	0.01949540815701	0.048738520392524
2	1	248040	0.00127037747389	7	0.008892642317232	0.022231605793081
2	0	9425520	0.048274344007833	0	0	0
1	1	1581255	0.008098656396051	4	0.032394625584204	0.08098656396051
1	0	60087690	0.30774894304994	0	0	0
0	1	3162510	0.016197312792102	3	0.048591938376306	0.12147984594076
0	0	120175380	0.61549788609987	0	0	0
		195249054	1		0.27714495354226	0.49516616351954

Here is a return table for MegaMillions assuming a $12 million Jackpot. The jackpot would have to be $144 million for a return greater than 1. At $46 million or less, the return of the Megaplier option is greater than a ticket without it.

white	gold	combinations	probability	pays	return	megaplier return
5	1	1	0.000000005691146	12000000	0.068293751640757	0
5	0	45	0.000000256101569	250000	0.06402539216321	0.1585390663089
4	1	255	0.000001451242222	10000	0.014512422223661	0.035935521696684
4	0	11,475	0.000065305900006	150	0.009795885000971	0.024256477145262
3	1	12,750	0.000072562111118	150	0.010884316667746	0.026951641272513
3	0	573,750	0.003265295000324	7	0.022857065002266	0.056598446672278
2	1	208,250	0.001185181148266	10	0.011851811482656	0.029347342718959
2	0	9,371,250	0.053333151671954	0	0	0
1	1	1,249,500	0.007111086889594	3	0.021333260668782	0.052825216894126
1	0	56,227,500	0.31999891003172	0	0	0
0	1	2,349,060	0.013368843352436	2	0.026737686704873	0.066207605173971
0	0	105,707,700	0.60159795085964	0	0	0
		175,711,536	1		0.25029159155492	0.45066131788269

Thanks, Miplet! This is just the information I was looking for!

Quote: miplet

The jackpot would have to be $162 million for a return greater than 1.

Does this calculation take into account that the jackpot might be split between multiple players with winning tickets? Considering that the number of winning tickets is unpredictable, how can a mathematician determine when a lottery becomes positive expectation?

Quote: reno

Considering that the number of winning tickets is unpredictable, how can a mathematician determine when a lottery becomes positive expectation?

Here's how I do it:

Let P be the probability of winning the jackpot (= 1 / the number of possible ticket combinations),
Q = 1 - P,
N = the number of tickets,
and P(K) the probability that there are exactly K winners
P(0) = Q^N
P(1) = N x P x Q^N-1
P(2) = N(N-1)/2 x P² Q^N-2
...
P(K) = (N)C(K) x P^K * Q^N-K
where (N)C(K) is the number of combinations of N items taken K at a time

The expected number of jackpot winners = the sum of:
0 x Q^N
1 x N x P x Q^N-1
2 x N (N-1) / 2 x P² * Q^N-2
3 x N (N-1) (N-2) / 6 x P³ * Q^N-3
4 x N (N-1) (N-2) (N-3) / 24 x P⁴ * Q^N-4
...
N x P^N

This can be rewritten as NP x the sum of:
Q^(N-1)
(N-1) x P x Q^(N-1)-1
(N-1) (N-2) / 2 x P² x Q^(N-1)-2
(N-1) (N-2) (N-3) / 6 x P³ x Q^(N-1)-3
...
P^(N-1)

By the binomial theorem, this equals NP x (P + Q)^N-1
Since P + Q = 1, the expected number of jackpot winners = NP.
(At one time, I think I calculated that the "most likely" number of winners was NP rounded down, except that if NP was an integer, then NP and NP + 1 were equally likely.)

The EV of each share of the jackpot = the jackpot amount / the expected the number of winners
= the jackpot amount / (the number of tickets x the probability of winning)
= the jackpot amount / (the number of tickets / the number of combinations)
= the jackpot amount x the number of combinations / the number of tickets.

Note that the "expected value" of a jackpot share can be - and, except when the jackpot is rather large, usually is - greater than the value of the jackpot itself. This means that the expected number of winners is "between 0 and 1."

Also note that the jackpot amount = the previous draw's jackpot amount + (some amount per ticket) x the number of tickets.
This makes the EV = # of combinations x (previous jackpot / # of tickets for the current game + amount of each ticket that goes into the jackpot).

The problem is, there's no way of knowing this without knowing in advance how many tickets will be sold.

Don't forget that since this thread was started, the cost of a PowerBall ticket has gone from $1 to $2. And, as mentioned in other threads on the topic, you may want to include the effect of taxes and/or the fact that the jackpot is paid as an annuity in your EV calculations.