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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?

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Almost surely not. They are not in the business of losing money. –  Alex Becker Mar 30 '12 at 17:58
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@AlexBecker They wouldn't lose money overall... the jackpots accumulate from previous weeks. –  user23784 Mar 30 '12 at 17:59
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True. I suppose it would hurt their reputation to change their formula when the jackpot got large. +1 for applying math to making money. –  Alex Becker Mar 30 '12 at 18:04
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If you factor in the discount for taking the lump sum up front (or the discount of inflation eroding the value of the payout if you take it in installments) as well as the amount taken out for taxes, there's a good chance it won't be +EV. One strategy that people have employed in the past is picking number combinations that most others would not (e.g. numbers > 31 b/c they aren't birthdays) in the hope of increasing the likelihood that they are the sole winner if they do win, thereby increasing their EV. But I suspect nowadays that most tickets purchased have the numbers randomly generated. –  Michael Joyce Mar 30 '12 at 19:09
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5 Answers

up vote 19 down vote accepted

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+.

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Great analysis. Of course, as I think I commented elsewhere, I would expect faster than linear growth to abate at some point (with 600mm+ ticket sales that point is obviously not this week!). Certainly super linear growth cannot be sustained indefinitely. Obviously we have no data upon which to base any analysis though... –  user23784 Mar 31 '12 at 1:51
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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.

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The big problem of course is the huge number of tickets sold when the lottery is this big, and gets tremendous free publicity. When my woman went in to buy some tickets earlier this evening (she didn't win by the way), there were seven other people there buying. This was the first time she had bought Mega Millions lottery tickets, and all but one of the other people there were also buying Mega Millions lottery tickets for the first time. –  Mark Adler Mar 31 '12 at 5:18
    
I'm neither an accountant nor a lawyer, but I believe you should be able to deduct your losses up to your winnings for federal tax purposes. For normal people this is a negligible difference, but Money Bags will have about 175 million losing tickets. This shouldn't make it into a winning strategy, but it should have a significant impact on the expected ROI. –  Michael McGowan Mar 31 '12 at 17:18
    
Right you are @michael. That is a big effect. I have updated my answer. –  Mark Adler Mar 31 '12 at 21:39
    
I have corrected the results for the fact that you cannot deduct gambling losses against other income -- only against gambling income. –  Mark Adler Apr 1 '12 at 16:26
    
Beautiful answer. –  Carl Mummert Apr 14 '12 at 11:56
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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.

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Great, thanks! Interesting how high the probability of a large number of winners is. –  user23784 Mar 30 '12 at 20:59
    
@rar Right...higher than most would probably guess (way higher than I would have guessed). You can use that distribution to approximate the expected jackpot payout by doing a weighted sum. It might take some extra effort though to avoid overflowing your calculating device though. –  Michael McGowan Mar 30 '12 at 21:04
    
You can safely ignore any amount of winners more than 1 std. deviation away from the mean: the contribution to your EV is so small, it won't change the dollar value. –  user641 Mar 30 '12 at 21:24
    
My brother is buying tickets while I check MSE from the car. If we win, I'll buy you dinner! –  The Chaz 2.0 Mar 31 '12 at 0:47
    
@MichaelMcGowan Thanks for this answer. I incorporated this info into the attempt at a more general formula for when there is an $EV > 0$ in the answer I've posted. –  user23784 Mar 31 '12 at 1:49
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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.

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$640mm is the "advertised" amount. The actual cash amount is around 70% of that. –  user23784 Mar 30 '12 at 18:10
    
@rar: If there is no winner this week, the carryover will be large. We could have positive EV for small enough numbers of tickets sold. If the carryover is 448M, you need less than 2*448M tickets sold to have positive EV. I presume the difference between \$640M and \$448M is the difference between immediate cash and the payout over 20 or so years. –  Ross Millikan Mar 30 '12 at 18:12
    
Yes, if it is not positive EV now then this drawing will not be since each additional ticket sold lowers EV. However, if there is no winner for this drawing, the jackpot could be positive EV next week depending on the number of tickets sold. So is there some point at which it becomes positive EV? I wouln'd think that ticket sales will climb so far from these levels as to offset the increasing carried over jackpot. –  user23784 Mar 30 '12 at 18:14
    
Right. I suppose this would rely on a better understanding of the "jackpot size" to "ticket sales" function. –  user23784 Mar 30 '12 at 18:16
    
@rar: Following your friend, the value of a ticket is carryover/tix sold + 0.50, so as long as the carryover per ticket is greater than 0.50 you are ahead. –  Ross Millikan Mar 30 '12 at 18:28
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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."

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Although I believe your conclusion is correct, you simply assert it at the beginning without proof. Your "probably split" argument is vital to your conclusion, but you omit any calculations to back it up (but like I said, I think it's right). –  Michael McGowan Mar 30 '12 at 20:37
    
I do not think that is the only way to see positive EV. If fewer tickets were sold for this drawing it would have been, so maybe at some slightly smaller jackpot there would be a "sweet spot" in which it turned positive. Additionally, the hypothesis that ticket sales will continue to climb faster than linearly with increasing jackpot size is unlikely to be true. It seems more likely that they level off at some point (since I doubt people will go from buying 10 tickets each to 20). –  user23784 Mar 30 '12 at 20:47
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