Is Mega Millions Positive Expected Value? Given the rapid rise of the Mega Millions jackpot in the US (now advertised at \$640 million and equivalent to a "cash" prize of about \$448 million), I was wondering if there was ever a point at which the lottery became positive expected value (EV), and, if so, what is that point or range?
Also, a friend and I came up with two different ways of looking at the problem, and I'm curious if they are both valid.
First, it is simple to calculate the expected value of the "fixed" prizes. The first five numbers are selected from a pool of 56, the final "mega" ball from a pool of 46. (Let us ignore taxes in all of our calculations... one can adjust later for one's own tax rate which will vary by state). The expected value of all these fixed prizes is \$0.183.
So, then you are paying \$0.817 for the jackpot prize. My plan was then to calculate the expected number of winners of the jackpot (multiple winners split the prize) to get an expected jackpot amount and multiply by the probability of selecting the winning numbers (given by $\binom{56}{5} * 46 = 1 \text{ in } 175,711,536$). The number of tickets sold can be easily estimated since \$0.32 of each ticket is added to the prize, so: 
(Current Cash Jackpot - Previous Cash Jackpot) / 0.32 = Tickets Sold
$(448 - 252) / 0.32 = 612.5$ million tickets sold (!!).
(The cash prizes are lower than the advertised jackpot. Currently, they are about 70% of the advertised jackpot.) Obviously, one expects multiple winners, but I can't figure out how to get a precise estimate, and various web sources seem to be getting different numbers.
Alternative methodology: My friend's methodology, which is far simpler, is to say 50% of this drawing's sales will be paid out in prizes (\$0.18 to fixed prizes and \$0.32 to the jackpot). Add to that the carried over jackpot amount (\$250 million cash prize from the unwon previous jackpot) that will also be paid out. So, your expected value is $\$250$ million / 612.5 million tickets sold = \$0.40 from the previous drawing + \$0.50 from this drawing = \$0.90 total expected value for each \$1 ticket purchased (before taxes). Is this a valid approach or is it missing something? It's far simpler than anything I found while searching the web for this.
Added: After considering the answer below, this is why I don't think my friend's methodology can be correct: it neglects the probability that no one will win. For instance, if a $1$ ticket was sold, the expected value of that ticket would not be $250 million + 0.50 since one has to consider the probability of the jackpot not being paid out at all. So, additional question: what is this probability and how do we find it? (Obviously it is quite small when $612.5$ million tickets are sold and the odds of each one winning is $1:175.7$ million.) Would this allow us to salvage this methodology?
So, is there a point that the lottery will be positive EV? And, what is the EV this week, and the methodology for calculating it?
 A: As you say, the chance of a winning ticket is about 1 in 176M.  The expected number of  winners is therefore 612.5/176 or about 3.5.  The expected prize for a winner is not simply \$448M/3.5 as the average of inverses is not the inverse of the average.  You quote \$640M as the jackpot early, but then seem to use \$448M for your calculation.
Your friend's approach is fine, too.  I haven't checked whether your numbers are consistent.  The EV only goes down as more tickets are sold, as the \$0.40 per ticket goes down, so this time it will never be above \$1.
Added:  the value of the carryover $C$ is $\frac C{176M}$ for the first ticket bought as the chance of winning.  In general, it is $\frac {C\cdot P_{win}}{n}$ where $P_{win}$ is the probability of somebody winning =$1-(\frac {176M-1}{176M})^n$ and $n$ is the number of new tickets bought.  For $n$ large, $P_{win} \approx 1$ and we get the earlier formula.
A: Regarding the last part of your question about the probability of nobody winning, I think we can approach it in the following way.  Assume for the sake of argument that all tickets are randomly generated and independent of one another.  A given ticket will lose the jackpot with probability $p = \frac{175711535}{175711536}$.  Since we said the tickets were independent, the probability of two tickets losing $p^2$, and in general, the probability of $n$ tickets all losing is $p^n$.  With 612.5 million tickets sold, this equates to roughly a 3% chance that nobody wins $(p^{612500000})$.
Now it is not strictly the case that the tickets in play are random and independent from one another.  I suspect that could lead to more duplicate tickets (birthdays, etc.), and consequently a slightly higher chance of nobody winning.  I don't have a good framework for estimating that, but I doubt it affects results that much.
Edit: As a related aside, these ticket assumptions allow us to use the binomial distribution to find the probability of exactly $n$ winners for our choice of $n$.  Here is a random sampling of that distribution for 50 drawings:
$$7, 3, 3, 4, 0, 4, 1, 3, 5, 1, 7, 4, 3, 5, 5, 6, 5, 6, 3, 5, 2, 2, 2, 1, 5, 3, 5, 2, 3, 6, 9, 1, 4, 1, 4, 3, 4, 2, 2, 5, 3, 3, 4, 0, 2, 5, 6, 5, 1, 3$$
As you can see, there's a decent chance of 4, 5, 6, or more jackpot winners.  
A: I did a fairly extensive analysis of this question last year.  The short answer is that by modeling the relationship of past jackpots to ticket sales we find that ticket sales grow super-linearly with jackpot size.  Eventually, the positive expectation of a larger jackpot is outweighed by the negative expectation of ties.  For MegaMillions, this happens before a ticket ever becomes EV+.
A: An interesting thought experiment is whether it would be a good investment for a rich person to buy every possible number for \$175,711,536.  This person is then guaranteed to win!  Then you consider the resulting size of the pot (now a bit larger), the probability of splitting it with other winners, and the fact that you get to deduct the \$175.7M spent from your winnings before taxes.  (Thanks to Michael McGowan for pointing out that last one.)
The current pot is \$640M, with a \$462M cash payout.  The previous pot was \$252M cash payout, so using \$0.32 into the cash pot per ticket, we have 656,250,000 tickets sold.  I, the rich person (who has enough servants already employed that I can send them all out to buy these tickets at no additional labor cost) will add about \$56M to the pot.  So the cash pot is now \$518M.
If I am the only winner, then I net (\$518M + \$32M (my approximate winnings from smaller prizes)) * 0.65 (federal taxes) + 0.35 * \$176M (I get to deduct what I paid for the tickets) = \$419M.  I live in California (of course), so I pay no state taxes on lottery winnings.  I get a 138% return on my investment!  Pretty good.  Even if I did have to pay all those servants overtime for three days.
If I split the grand prize with just one other winner, I net \$250M.  A 42% return on my investment.  Still good.  With two other winners, I net $194M for about a 10% gain.
If I have to split it with three other winners, then I lose.  Now I pay no taxes, but I do not get to deduct my gambling losses against my other income.  I net \$161M, about an 8% loss on my investment.  If I split it with four other winners, I net \$135M, a 23% loss.  Ouch.
So how many will win?  Given the 656,250,000 other tickets sold, the expected number of other winners (assuming a random distribution of choices, so I'm ignoring the picking-birthdays-in-your-numbers problem) is 3.735.  Hmm.  This might not turn out well for Mr. Money Bags.  Using Poisson, $p(n)={\mu^n e^{-\mu}\over n!}$, where $\mu$ is the expected number (3.735) and $n$ is the number of other winners, there is only a 2.4% chance of me being the only winner, a 9% chance of one other winner, a 17% chance of two, a 21% chance of three, and then it starts going down with a 19% chance of four, 14% for five, 9% for six, and so on.
Summing over those, my expected return after taxes is \$159M.  Close.  But about a 10% loss on average.
Oh well.  Time to call those servants back and have them make me a sandwich instead.
Update for October 23, 2018 Mega Millions jackpot:
Same calculation.
The game has gotten harder to win, where there are now 302,575,350 possible numbers, and each ticket costs \$2. So now it would cost $605,150,700 to assure a winning ticket. Also the maximum federal tax rate has gone up to 39.6%.
The current pot (as of Saturday morning -- it will probably go up more) has a cash value of \$904,900,000. The previous cash pot was \$565,600,000. So about 530 million more tickets have been or are expected to be purchased, using my previous assumption of 32% of the cost of a ticket going into the cash pot. Then the mean number of winning tickets, besides the one assured for Mr. Money Bags, is about 1.752. Not too bad actually.
Summing over the possible numbers of winners, I get a net win of $\approx$\$60M! So if you can afford to buy all of the tickets, and can figure out how to do that in next three days, go for it! Though that win is only a 10% return on investment, so you could very well do better in the stock market. Also that win is a slim margin, and is dependent on the details in the calculation, which would need to be more carefully checked. Small changes in the assumptions can make the return negative.
Keep in mind that if you're not buying all of the possible tickets, this is not an indication that the expected value of one ticket is more than \$2. Buying all of the possible ticket values is $e\over e-1$ times as efficient as buying 302,575,350 random tickets, where you would have many duplicated tickets, and would have less than a 2 in 3 chance of winning.
A: The only way to see positive expected value is for no one to win tonight and for there to be some kind of major attention-grabbing popular culture or other kind of phenomenon which causes people to uncharacteristically forget that the lottery exists so that ticket sales are unusually low at the same time that a jackpot is high.
Short of this, ticket sales will tend to grow dramatically as the jackpot grows causing the expected number of winners splitting the jackpot to substantially diminish the expected size of the payout given a correctly picked entry.
Listening to people's rationalizations around this is dizzying: A: "Look, the jackpot is 640M and the odds are only 1:175M that sounds like +EV!"
B: "The present value of the jackpot is more like 448M..."A: "That still sounds like +EV"
B: "And you have to pay significant taxes..."A: "Still +EV"
B: "And lots of people are playing so even if you do win you will probably split with a number of others."A: "Yeah, but in that case I will win at least 100M which is more than I can imagine."
B: "Which is the same as any other ordinary Mega Millions week; massively -EV."
