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 JackpotPrevious 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?

Answer

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+.

Attribution
Source : Link , Question Author : Community , Answer Author : Jeremy Elson

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