The expected value of a Mega Millions ticket

As I type this post, the Mega Millions jackpot projects to be $500,000,000 (annuity value – after-tax cash value is more like $220,000,000). A single ticket costs $1. Is it worth it?

By “worth it,” I mean that the expected value of a ticket is greater than zero. The expected value of a random variable \(W\) which represents the event of winning the Mega Millions lottery with the ticket is equal to, roughly, (Probability of winning)*(Value of winning) + (Probability of losing)*(value of losing). In shorter math notation, this looks like:

\(E[W]=P[W]*V+(1-P[W])*C\)

Here, \(E[W]\) is the expected value of \(W\), \(P[W]\) is the probability of \(W\), and therefore \(1-P[W]\) is the probability of NOT \(W\). Also, \(V\) is the value of winning – in this case, $500,000,000 or $220,000,000, and \(C\) is the value of losing: -$1.

What is this probability? The players are asked to pick 5 balls from a first set of 56, then 1 ball from a second set of 46, the first set and second sets being independent. The number of possibilities is \(46 \binom {56}{5} = 175,711,536\). The probability \(P[W]\) is then the reciprocal of this number.

Although this probability is indeed small, it would seem that a sufficiently large jackpot would make buying a ticket worth it. And, indeed, even with the after-tax cash value, the expected value is greater than zero:

\(E[W]=\displaystyle \frac{220,000,000}{175,711,536} – \displaystyle \frac{175,711,535}{175,711,536} \approx 0.25\).

This could be taken as an encouragement to purchase Mega Millions tickets; after all, you can expect about $0.25 of value for each ticket you buy. What a bargain! With a return on investment like that, why don’t investment houses just put their client’s money into Mega Millions tickets?

The reason is that return is wrong and based on a false assumption: no matter how many tickets got the correct number, the payout is the same. In fact, the payout is split evenly among the winners. This will of course lower the expected value of a ticket. But by how much?

Consider a population of \(N\) tickets, where yours is among the \(k\) winners. (Mazeltov.) That means there are \(k-1\) winners besides yourself among the rest of the remaining \(N-1\) tickets. What is the probability that there are \(k-1\) winners among \(N-1\) tickets?

We consider each event of a ticket being a winning ticket being independent from any other ticket being a winning or losing ticket. (This is clearly the case.) The probability of a particular set of \(k-1\) winners among the \(N-1\) tickets is then \(p^{k-1} (1-p)^{(N-1)-(k-1)}\). That is, when there are \(k-1\) winners, there are also \((N-1) – (k-1) = N – k\) losers.

That said, there are also many different ways to arrange the other \(k-1\) winners among the remaining \(N-1\) tickets. To be precise, there are \(\binom {N-1}{k-1}\) different arrangements of winners and losers, on top of you as a winner. (Not literally!) I imagine that my friends literate in statistics recognize the random variable \(W\) as having a binomial distribution.

Note that, when there are \(k\) winners, the jackpot is split evenly among the winners; that is, each winner gets \(V/k\), and not \(V\). The expected value of a ticket is then a sum over all possible values of \(k\):

\(E[W]= \displaystyle V \sum_{k=1}^{N} \frac{1}{k} \binom{N-1}{k-1} p^{k} (1-p)^{N-k} – C (1-p)\),

where \(p = P[W]\) as defined above. Note that there is an extra factor of \(p\) coming from your own ticket.

The above sum is very difficult to evaluate numerically for \(N \approx 100,000,000\), and approximations to normal or Poisson distributions do not apply. However, we can observe that we are comparing the value of the above sum to \(p\). We can see that the value of the sum is less than \(p\) because of the fact, well-known from binomial distributions, that

\(\displaystyle \sum_{k=1}^{N} \binom{N-1}{k-1} p^{k} (1-p)^{N-k} = p\).

Therefore,

\(\displaystyle \sum_{k=1}^{N} \frac{1}{k} \binom{N-1}{k-1} p^{k} (1-p)^{N-k} < p[/latex], because the factor [latex,size="-2"]\frac{1}{k} < 1[/latex]. In fact, a rough calculation (it turns out you can neglect anything beyond the 6th term) gives a value of about 0.374 for the above sum. Again, compare this to 1 using the crude (and incorrect) estimate. So, in multiplying the first term in the first expected value equation by about 0.374, we get an expected value of about -$0.53. That is, rather than gaining 25 cents, you should expect to lose about 47 cents for every dollar you spend on a Mega Millions ticket. Then again, we're all kind of suckers for that galactically small chance we could win, and I don't blame any of you for throwing away a few pennies chasing the dream. Update (3/31/12): I need to correct an assertion I made, and some numbers. The conclusion stands - in fact, in light of what has happened over the past 24 hours, the conclusion is even more stark. First of all, I posted some incorrect numbers that I have since corrected. For the above values of the number of tickets in circulation and probability of winning: [latex]\displaystyle \sum_{k=1}^{N} \frac{1}{k} \binom{N-1}{k-1} p^{k-1} (1-p)^{N-k} \approx 0.763[/latex]. This is the reduction factor on the jackpot, not 0.374 as I published before. The expected value of a ticket is then about -$0.05. Again, still a loser. That said, it turns out that 100,000,000 tickets was a gross underestimate; rather, 1,500,000,000 tickets were sold! For this number, the reduction factor is about 0.117, which gives an expected value of a ticket as -$0.85, an even worse value than my incorrect previous numbers show. Second, although I was right in asserting that the multiple winners are not governed by a Poisson distribution (a great explanation is here), I was incorrect in ignoring the Law of Rare Events, which states that, under a certain limiting behavior, the binomial sum approaches the Poisson sum that results from assuming a Poisson distribution. Further, the limiting behavior need not be rigorously enforced: a large enough sample and a small enough probability does the trick.

The mathematical statement of the Law of Rare Events in this context is

[latex]\displaystyle \sum_{k=1}^{N} \frac{1}{k} \binom{N-1}{k-1} p^{k-1} (1-p)^{N-k} \approx \sum_{k=1}^{\infty} \frac{(N p)^{k-1}}{k!} e^{-N p} = \frac{1 – e^{-N p}}{N p}\).

It turns out that this is a very good approximation out to many decimal places. So, a very simple formula for the expected value of a Mega Millions ticket is

\(\displaystyle E[W] = V \frac{1 – e^{-N p}}{N} – C (1-p)\).

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Ron Gordon

Math nerd in his early 40's who seems to have an opinion about everything and an inability to keep it to himself.