I’ve heard with a system that seems to increase one’s probability of winning the lottery. I’m quite sure that there are problems with the reasoning, but I can’t put my finger on why; maybe one of the probability specialists can tell me what’s wrong.

(For the sake of illustration I’ll talk about New York State Lotto, which involves 6 numbers drawn from a pool of 48.)

To try this system, get yourself eight jars, and a set of 48 numbered balls like the ones they use in the drawings. Mix the balls up and divide them evenly among the jars, so that each jar contains six randomly-chosen balls.

Pull the numbers out of first jar, and buy a Lotto ticket with those numbers. Then buy another ticket based on the numbers from the second jar, and the third, and so on until you’ve emptied all the jars. You will of course end up with 8 tickets, which collectively use each possible number exactly once.

Now it comes time for the weekly drawing. The first number pops up, and you look at your tickets. You are certain to have picked this number exactly once. It has a 12.5% chance of being on ticket 1, a 12.5% chance of being on ticket 2, and so on down the line.

The second ball pops up, and you have picked it exactly once. Again, for each ticket there is a 12.5% chance that the number is contained on that ticket.

So, the chance that all six balls are listed on the same ticket is: (12.5% ^ 6), which is about 3.81E-6, or a one-in-262,000 shot.

By contrast, if you pick your numbers at random, your chance of winning on any given ticket is the reciprocal of the number of combinations of numbers; if my figures are right, you have a 8.15E-8 or one-in-12,271,000 chance per ticket. Multiplying this by eight tickets gives your total odds as one-in-1,534,000, far worse than those given by the above system.

Where is the error in my math?

Laugh hard; it’s a long way to the bank.