Sunday, April 8, 2012

Expected Winnings From Mega Millions Lottery

I suppose I'm late to the game on the recent Mega Millions jackpot hype.  Still I'm intrigued by the possibility that there could be a net positive value of a lottery ticket.  The Mega Millions jackpot is different from most lotteries in that it rolls over if it's not won.  Because of this, there can be occasional net positive values, while the game remains profitable for the state.  In other words, it could be possible for the winnings to be so high that it makes it logical to play.

Expected value is a way of measuring how favorable the odds of a gamble are.  If there is a simple raffle with a $100 prize and$1 tickets we can then look at the expected value of each ticket based on the odds.  If the odds are 1 in 100 (this would mean 100 tickets are sold and one winner is picked), then the expected value of a single ticket would be $100 \cdot \frac{1}{100} = 1$.  This means that if you bought many tickets over a very long period of time you could expect to gain $1 back for each ticket you bought. Since the tickets cost$1 themselves this means you would expect to break even in the long run.

We can look to roulette for another example.  Roulette is simply a wheel with number 1 through 36 on it, which are colored red and black.  The player can bet on a single number or on evens, odds, reds, or blacks (as well as many other more complex bets).  If the player bets on even he has a 50% chance and will gain his bet amount on a win.  This would lead to a perfectly fair game (with no house edge).  However, in real roulette there is the addition of a green 0 (and in the US also a green 00).  The 0 is not odd, even, red, or black, and thus all players lose on 0.  This is the source of the house edge.  The payout is the same, but the odds go down from 1/2 to 16/37.  This slight change makes the expected value $2 \cdot \frac{18}{37} = 0.97$  Since each bet 'costs' $1, this means you lose about 3 cents each time you bet. It is worth pointing out here that the player can bet on a single number, which has much lower odds, but also a much higher payout. With the house edge, the expected value over time is the same for both single number bets and the odd/even bets. Now, back to the Mega Millions. This lottery uses 5 balls drawn from 56, and one extra ball drawn from a separate 46. In both cases order doesn't matter and each ball can only be drawn once. This is enough info to calculate the probabilities, but it's easier just to get them (and the winnings for each) from their site. A simple attempt at determining net value would be to take the probability (1 in 175,711,536) times the jackpot ($640,000,000) and arrive at $3.64. Since tickets are$1, this would imply an expected value of $2.64. However there are several caveats. Annuity vs Cash First off, the$640 million figure is the total value of the annuity.  An annuity is just a reoccurring payment for some number of years.  If you want all your winnings up front, you will get 'only' $462 million. The annuity amount is always higher, and thus, always the advertised amount. There doesn't seem to be any concrete way of calculating the cash payout amount from the announced annuity amount, but a rough estimate is 70%. Taxes There are federal taxes on lottery winnings, and most states also tax it. This site lists state tax rates. It ranges from 5-10%. Note it also lists federal taxes at 25%. This is the mandatory withholding, ie, how much will be subtracted from the jackpot before you get it. However, getting$400 million will push you up into the highest income bracket which is currently taxed at 35%.  I suppose you could do some sort of accounting tricks to get the rate down, but I suspect the IRS will be paying close attention to any multimillion dollar lottery winners.

A combined 7% state tax, and 35% federal tax will mean a 42% tax rate.  This is still $268 million, which gives an expected value of$1.52.  Still more than the ticket price.

Other Prizes
Before we move on to the biggest factor I should mention the effect of the other jackpots.  Missing just one number reduces the prize to at most a paltry $250,000 ($145,000 after taxes).  Since these prizes don't change, their total expected value is easy to calculate.  LibreOffice makes short work of the calculations, giving a total value of $0.15. In other words, for every dollar you spend on tickets you can expect to win about 15 cents back (excluding the jackpot). In practice this means the cost of a ticket can be viewed as$0.85.  The jackpot will have to have an expected value of more than this for ticket to have an overall positive net value.

Multiple Winners
If multiple people all win the jackpot it is split amongst them.  This is a significant factor that negates the previous calculations, which all assumed that if you won, you would win the full jackpot (the previous value of $0.15 for non jackpot prizes is still valid as they aren't split). How likely is multiple people winning? Well that depends on the number of tickets sold. Luckily I found this site, which lists jackpots and total tickets sold. Armed with that data I set out to figure out the expected winnings with the chance of multiple winners. This involved the Poisson Distribution. While I knew of the poisson distribution, I didn't actually know how to use it. Despite the Wikipedia page's attempts to make it look terrifying, some other sites made it clearer. $$p(r,n) = \frac{r^n}{(n!) (e^r)}$$ p is the probability of exactly n occurring given a general rate of r. In other words, if we take the odds of winning and multiply them by the number of tickets, we get the expected number of winners: $$5.69 \times 10^{-9} \cdot 651,915,940 = 3.71$$ Given the odds, and 650 million tickets sold, we would expect 3.7 winners on average. This is our value for r. The value for n is how many winners we want to know the odds for. In the last jackpot the odds of various numbers of winners were:  0 1 2 3 4 5 6 7 8 9 10 2.45% 9.08% 16.84% 20.83% 19.32% 14.34% 8.87% 4.70% 2.18% 0.90% 0.33% There was only a 2.5% chance that there wouldn't be a winner. In order to find the total expected value we just have to multiply those odds by the jackpot amount. There is a problem though. If we want to know the total expected value of one ticket we need to add up all the expected payouts of the various numbers of possible winners. The probability of multiple winners typically drops off pretty fast, but with this last jackpot and its 650 million tickets it remained significant for much longer. In addition, I was hesitant to discount the total value of all those many tiny probabilities combined as insignificant (we are already talking about 1 in 175,711,536 odds of winning here). This meant I had to find the sum of the series. The expected value (Vn) of any one number of winners (n), and a given jackpot (J) is: $$V_n = \frac{J}{n} \cdot p$$ To find the total we need to sum this: $$V = \sum_{n=1}^{} \frac{J}{n} \cdot p$$ Now we need to replace p with the above formula: $$V = \sum_{n=1}^{} \frac{J}{n} \cdot \frac{r^n}{(n!) (e^r)}$$ Note that for any one jackpot both J and r are constant. They can thus come out of the sum: $$V = \frac{J}{e^r}\sum_{n=1}^{} \frac{r^n}{n(n!)}$$ The factorial in the denominator tells us that the series will very quick approach 0 as n climbs. Factorials grow much faster than exponents (try it). This lends credence to the idea that the values of the higher number of winners will be insignificant. Since the most recent jackpot had the record for most tickets (and thus most significant odds of large numbers of winners), I calculated its value of $\frac{J}{e^r} = 15,663,284.42$, and then headed over to Wolfram Alpha to do the summation. http://www.wolframalpha.com/input/?i=sum+from+1+to+10+of+3.7101487748^n%2F%28%28n%29%28n!%29%29+*+%2415%2C663%2C284.42 This tells us that the expected total value of all winnings with 1 to 10 winners is$222.1 million.  Increasing to 1 to 25 gives $222.2 million. Increasing to 1 to 10000 still gives$222.2 million.  Due to the factorial, the expected value of any value above 100 is so absurdly small that even if it didn't decrease any further it would still be insignificant.

This meant I could simply calculate the values for the first 25 number of winners and sum that instead of trying to figure out how to do a series in LibreOffice.

Is it worth it?
I'm sure some of you have already noted that $222 million in expected winnings is greater than (the inverse of) the 1 in 175,711,536 odds of winning. However, we haven't taken taxes, or lump sum payments into account. Taking 70% in lump sum, and then 58% post taxes gives$90.5 million.  This multiplied by your odds gives an expected value of $0.52 (plus the$0.15 from other prizes).  This falls short of the $1 price, and thus gives an overall value of -$0.33 per ticket.  It isn't worth it.

This result is highly dependent on how many people are playing.  It's possible that when there are less people playing, even though the jackpot is less, that your expected value could be higher.  I took all the jackpots from 2010-present and looked at the expected values.  The recent jackpot had the highest ever expected value.  In other words, it has never been profitable to play the Mega Millions (at least since 2010).

Looking at the ratio of Jackpot / tickets, it generally hovers around $3. There are two times it tapers off towards$1.  First is when the jackpot resets to $12 million. There are likely a core group that buys tickets every week no matter what the jackpot. Because of them, your expected value is worst when the jackpot is smallest. Second, is when the jackpot soars to its very highest point. The last three possible jackpots where$640m, $363m, and$290m, with ratios of $0.98,$1.90, and $2.61 respectively. Note that somewhat surprisingly, despite the relatively low ratio the recent jackpot was the best expected value ever. Looking at the history, I'd say the ratio is at$3 when the jackpot is from $70m -$250m.

Final thoughts
One caveat to all this though.  The odds of multiple winners isn't strictly what I calculated here.  The reason is that people can choose their numbers.  This results in higher probability of multiple winners for sets of numbers that fit patterns (birthdays, etc).  Allowing the machine to pick your numbers for you would then slightly decrease the odds of multiple winners if you do win (as long as not everyone does that).  I'm not going to bother estimating this effect, as I don't think there will be good data about what sets of numbers are most likely to be picked.  Either way, I doubt it would change the odds enough to make it profitable.

I should also point out here that expected value only really applies if you can play a game enough times to expect to win multiple times.  Seeing as you could buy 100 tickets for every drawing for 80 years and still not even have a 1% chance of winning (and have spent \$832,000) this isn't the case with the Mega Millions lottery.

My spreadsheet with all this data is here:
http://daleswanson.org/blog/lottery.ods

I did another post about attempting to predict Mega Millions sales here:
http://daleswanson.blogspot.com/2012/07/attempting-to-predict-mega-millions.html