A friend of mine just alerted me to a story on NPR describing a prize on offer from Warren Buffett and Quicken Loans. The prize is a billion dollars (1B USD) for correctly predicting all 63 games in the men’s Division I college basketball tournament this March. The facebook page announcing the contest puts the odds at 1:9,223,372,036,854,775,808, which they note “may vary depending upon the knowledge and skill of entrant”.

Being curious, I thought I’d see what the assumptions were that went into that number. It would make sense to start with the assumption that you don’t know a lick about college basketball and you just guess using a coin flip for every match-up. In this scenario you’re pretty bad, but you are no worse than random. If we take this assumption, we can calculate the odds as 1/(0.5)^63. To get precision down to a whole integer I pulled out trusty bc for the heavy lifting:

$ echo "scale=50; 1/(0.5^63)" | bc 9223372036854775808.000000

Well, that was easy. So if you were to just guess randomly, your odds of winning the big prize would be those published on the contest page. We can easily calculate the expected value of entering the contest as P(win)*prize, or 9,223,372,036ths of a dollar (that’s 9 nano dollars, if you’re paying attention). You’ve literally already spent that (and then some) in opportunity cost sunk into the time you are spending thinking about this contest and reading this post (but read on, ’cause it’s fun!).

But of course, you’re cleverer than that. You know everything about college basketball – or more likely if you are reading this blog – you have a kickass predictive model that is going to up your game and get your hands into the pocket of the Oracle from Omaha.

What level of predictiveness would you need to make this bet worth while? Let’s have a look at the expected value as a function of our individual game probability of being correct.

And if you think that you’re *really* good, we can look at the 0.75 to 0.85 range:

So it’s starting to look enticing, you might even be willing to take off work for a while if you thought you could get your model up to a consistent 85% correct game predictions, giving you an expected return of ~$35,000. A recent paper found that even after observing the first 40 scoring events, the outcome of NBA games is only predictable at 80%. In order to be eligible to win, you’ve obviously got to submit your picks *before* the playoff games begin, but even at this herculean level of accuracy, the expected value of an entry in the contest plummets down to $785.

Those are the odds for an individual entrant, but what are the chances that Buffet and co will have to pay out? That, of course, depends on the number of entrants. Lets assume that the skill of all entrants is the same, though they all have unique models which make different predictions. In this case we can get the probability of at *least one of them* hitting it big. It will be the complement of *no one* winning. We already know the odds for a single entrant with a given level of accuracy, so we can just take the probability that each one doesn’t win, then take 1 minus that value.

Just as we saw that the expected value is very sensitive to the predictive accuracy of the participant, so too is the probability that the prize will be awarded at all. If 1 million super talented sporting sages with 80% game-level accuracy enter the contest, there will only be a slightly greater than 50% chance of anyone actually winning. If we substitute in a more reasonable (but let’s face it, still wildly high) figure for participants’ accuracy of 70%, the chance becomes only 1 in 5739 (0.017%) that the top prize will even be awarded even with a 1 million strong entrant pool.

tl;dr You’re not going to win, but you’re still going to play.

If you want to reproduce the numbers and plots in this post, check out this gist.

It seems they’ve set a ceiling of 10 million participants in the pool but have curiously left themselves the right to increase that number, “Sponsor reserves the right to expand the entry pool for the Grand Prize to a larger number ofregistrants”. Either way, I’d say it implies they are expecting much lower than 70% mean accuracy.

My analysis actually made a sneaky assumption when estimating the probability that the grand prize will be awarded. It actually assumes independence of outcomes (ie every submission is predicting the outcomes in their own unique tournaments – or alternate realities, if you prefer.) As the predictions draw on information (p(win game) > 0.5), the non-random entries will tend to be similar since they are all predicting the same 63 games. This will effectively reduce the probability that anyone will win since the space of possible outcomes is not being sampled with independence. The organizers could look at the submissions so far and evaluate how much of the outcome space (ie the 2^63 possible brackets) is represented. Depending on how large it is compared to the independence model, they could safely (from a probabilistic perspective) expand the entry pool.

The PV of the prize isn’t $1BB.

Of course, an entrant has a much higher probability of picking a 1-seed vs 16-seed round game than, say, an 8 vs 7 seed, or a later round game. Using the average accuracy of picks yields a very biased result. Using probabilities based on individual games (for example, a model based on the relative seeding of each team) suggests a much lower probability of of winning than using an average accuracy. Note that, for example (.7 * .7) < .9 * .5).

Absolutely! 0.7 * 0.7 is actually greater than 0.9 * 0.5, but I’m sure that was a typo. I was going to go into the implications of heterogeneity in prediction probabilities but tried to keep it simple. You’re absolutely right though.

So for extra credit🙂, Assume it takes 30 seconds and $1 to lay down a Powerball bet. For simplicity, assume the Powerball jackpot remains constant at $100million. Generate curves showing at what level of predictive accuracy you’re better off spending X hours filling in the hoops bracket. Remember to charge yourself an hourly rate (and thus 1/120 an hour’s pay while buying the Powerball ticket).