Pattern Analysis of the PowerBall Lottery

Since you’ve only got data for about 20 drawings in there, your actual error bars are pretty bloody huge — way bigger than your formulae are computing for them.

Also, I think there’s a theoretical problem with using a fixed 90% confidence interval the way you appear to be using it.

There’s a 10% chance (using a 90% confidence interval) that a particular measurement likes outside the interval entirely by chance — which, of course, means that if you make many measurements, you expect that about 10% of them will lie outside the confidence interval, even if the effect being studied is entirely nonexistent. But, in your case, you’re presumably making measurements of the relative frequency of occurrence for each possible triple of numbers. That’s over 32,000 separate measurements.

Currently, the measurements are of very low precision — each drawing generates only 20 triples, so you’ve only got about 400 events. The actual expected value for the number of occurrences of each triple is around 0.012 — but events have to be counted with whole numbers, that alone forces the margin of error to be big.

Since you’ve got 20 drawings, and the one triple you’re currently showing says it’s occurred 10% of the time, that means it’s occurred twice. This is pretty close to being the birthday problem played on a really long calendar. A back-of-the-envelope calculation puts your odds at something a bit over 90% in favor of having at least one triples appearing twice in the dataset you have.

The “hit” you have isn’t at all as unlikely as “14.0712 standard deviations away from the mean” — it’s a near certainty.

I’m still trying to figure out how to actually compute the probability distributions related to some of the analyses done on the site, but while I’m at it, I worked out one relevant formula I thought I’d share.

It’s not especially hard to answer this question: if I played the game on every drawing, and the drawings were completely random (that is, each ball has an equal chance of being drawn on each drawing independent of everything else), on the average, how much would I make (or lose)?

For those who don’t remember the details from their probability class, let’s work out a similar problem for a simpler game. Let’s say you paid a dollar to play the game, in which you picked just one number. When they held the drawing, they drew just one number from the hopper, and if you matched it, they paid you $50.

Each number has an equal chance of being drawn, with 59 balls in the hopper, so you’ve got a 1/59 chance of winning. If you played the game a million times, you’d win about 1/59th of them, gaining $50 each time. So, afterward, you’d have paid $1,000,000 to buy tickets, and received (1,000,000 / 59) * $50, or $847,457.62. In the net, you lost $152,542.37, or about $0.15 per game. Another way of saying it is that a ticket in this game is worth about $0.85 cents, even though it costs a dollar to buy.

The payoff and probabilities in the Powerball game are more complicated. See the Wikipedia page for the details. The biggest complexity comes from the Jackpot prize. The size of the Jackpot depends on how many tickets were sold, and if nobody wins the Jackpot in a given drawing, the money for it gets rolled over into the next drawing. If more than one ticket meets the Jackpot conditions, the prize is split equally among them.

All of that means we can’t work out an exact amount for the expected payoff. But let’s just use “J” to represent the amount you actually get (including rollovers and accounting for multi-way splits) for hitting the Jackpot, and work out the formula anyway.

For a ticket that doesn’t have the “PowerPlay” option, the average ticket value is (40,243,435 / 292,837,581 + J / 4,685,977,296). That first fraction works out to a bit under fourteen cents, and the denominator in the second fraction is pretty huge. The largest payout ever was a $365 million Jackpot, so the second fraction contributes only a little less than eight cents to the ticket value, even under the best circumstances.

The picture’s a little more rosy if you choose the “PowerPlay” option — the ticket now costs $2, instead of $1, but they pick a random multiplier, from two to five, that’s applied to all the non-Jackpot prizes. If we include that in the mix, the first term in the exact ticket value becomes 40,243,435 / 83,678,166 (3.5 times bigger), or a smidgen over $0.48, with the Jackpot accounting for the same amount as before. On average, you lose fifty-two cents per game.

What’s interesting to consider, then, is just how much the kind of data in the “Pattern Analysis” site might influence the payoff. Suppose we could use the data to always pick one number correctly, but it gave us no edge on predicting the other four, or the PowerBall. How would that change the expected payoff? How about if we could always get two balls right?

Both of these questions are fairly straightforward to answer, and I’ll give the actual number tomorrow or the next day — it’s late enough right now that I’m bound to screw it up. Clearly, if we could always get at least three right, we’d be well in the black, gaining at least $6 per play, or averaging $22.50 per play by taking the PowerPlay.

Meanwhile, though, realize that being able to always predict even one of the five balls, much less two, would be an enormous amount of information to extract out of the data. In reality, knowing what we do about how the machines that draw the balls are constructed, and the physical processes involved, it seems unlikely that even the most detailed analysis, using years of data, could give us more than a tiny fraction of the amount represented by being able to always pick one ball right, so working out how much knowledge of one ball improves the payoff would give us an upper bound (though one with a whole lot of clearance) on how much we could ever get out of this kind of analysis.