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Picking Lottery Numbers

Games and gaming, Math and Science, Outdoors

Bleah, enough strike crap… Time for something nerdier…

I was having a discussion on a science-related message board recently and the discussion came up regarding the selection of lottery numbers and how some people really don’t understand what they’re doing when it comes to buying a lottery ticket.

In particular, this person said that he often picks 1, 2, 3, 4, 5, 6, 7 as his lottery numbers, and that he gets chided for it by other people who see his choice because, in their words “that’s not random” or “that will never come up.” In fact, 1, 2, 3, 4, 5, 6, 7 is as likely to come up as any other combination of seven numbers, and here’s the proof.

Since the discussion was around a 7-number choice, I’ll use Canada’s “Super 7” lottery. In this lottery, 7 numbered balls are drawn at random from a field of 47. An eighth ball is drawn as a bonus number but for these calculations that really doesn’t matter – all we care about is the jackpot 🙂 Players choose 7 numbers and pay their $2. For every play, you also get two sets of “quick-pick” numbers generated randomly by the computer.

How many possible choices of 7 numbers are there? There is a simple way to calculate this.

If order matters (i.e if the choice of 1, 2, 3 has a different meaning than 2, 3, 1) it is called a permutation and the formula is n! / (n-r)! where n is the total size of the field, and r is the number of terms being chosen. In the little example of a field of 1, 2, 3 where we’re choosing all three, this works out to 3! / (3-3)!. This evaluates to 6. If you want to know about the mathematical operation of factorial (the ! sign), click here. You can easily write out the possible combinations in this case:

1 2 3, 1 3 2, 2 1 3, 2 3 1, 3 1 2, 3 2 1

Notice that each combination is equally possible: 1 2 3 occurs only once, just the same as 3 2 1 and the more “random-looking” 2 3 1 and 1 3 2.

For lottery numbers, however, order doesn’t matter. Picking 3, 2, 1 is the same as picking 2, 1, 3 if we’re talking about balls drawn from a machine. To calculate the number of possibilities, you have to take into account the number of ways to arrange what you’ve picked. The example above happens to show how to calculate the number of ways of arranging a field of objects: n! This is so because the number of permutations when you’re selecting all the objects always results in a term of 0! (which equals 1) on the divisor. Thus if order doesn’t matter in the group you’re choosing, you have to divide out the number of ways you can arrange the group you’re choosing. The new formula is called combination and is n! / ((n-r)! * r!) where n is the size of your field, and r is the number you’re choosing from it. This formula is, quite literally, the number of permutations divided by the number of ways of arranging what you’ve chosen.

So how many possible number choices are there for the Super 7? You choose 7 numbers from a field of 47, and order doesn’t matter:

47! / ((47-7)! * 7!)

Now, 47! is an impossibly huge number, approximately 259000000000000000000000000000000000000000000000000000000000 (259 octodecillion – kudos if you can use that in a conversation), or roughly 50 times the total number of atoms in the sun. This is one of the places where it pays to have paid attention in math class and learned all those boring, long-hand expansions. It turns out that if you expand that formula, it distills down to this:

(41*42*43*44*45*46*47) / (1*2*3*4*5*6*7)

This simplifies further to:

3*11*23*41*43*47

The total is 62,891,499. That’s all there is. There are just under 63 million ways to choose 7 numbers from a field of 47… and it doesn’t matter the order, nor what the specific numbers are. At no point in the calculation did we calculate “except sequential picks” or anything like that because the order doesn’t matter. Nor does the formula require input like “doesn’t start at 1”, or “skips prime numbers” because all the picks have equal value – it’s a pick, nothing more. There’s no magic here.

Now back to the lottery. With Super 7 you pick a set of numbers and you get 2 more sets. So a $2 ticket has three sets of numbers. The odds of winning the jackpot (based on the Lottery Corp. selecting 7 numbers at random from the 49) are, therefore 3 in 62,891,499 or 1 in 20,963,833 exactly. Now you know how they get the numbers on the Super 7 page.

The important point here is that the calculations don’t care what order the balls are in. Order is typically something we perceive. The human brain is an incredible pattern matching engine, and it looks for order whenever and wherever it can. As a result, we see “1, 2, 3, 4, 5, 6, 7” and think “that’s not random.” And it may not be – if we were looking at the sequence of whole numbers, it certainly wouldn’t be a random pattern… but consider this:

…63566581170270089376123456776393345484018493336…

In that jumble of numbers is an excerpt from the decimal places of pi starting at 9,470,325 places after the decimal point, and there, staring you in the face is 1234567. Arguably, any small slice of the decimal places of pi is a random number but there’s a group of seven sequential digits. It’s sequential, and it’s just as likely as any other. About 200,000 positions before that nicely ordered bunch you can find 8675309 if you’re so inclined. Jenny should be proud to be immortalized in one of the fundamental constants of the universe.

I hope I have established by now that 1, 2, 3, 4, 5, 6, 7 is just as good a set of lottery numbers as any other set you can think of. From the lottery’s point of view, there is nothing special or “more random” about some combination of your children’s birthdays, the day you lost your virginity, the number of cigarettes you smoked each day last week or anything else.

As final proof, I offer this… If you had chosen 1, 2, 3, 4, 5, 6, 7 since the beginning of the Super 7 Lotto Max (which replaced Super 7) lottery, what would you have won:

1234567

In going through the number checker, I noticed that if you picked the seemingly-not-random-but-really-as-random-as-anything-else set of 2, 4, 6, 8, 10, 12, 14 as your numbers for the 2 Feb 2007 Super 7 draw, you’d have matched 5/7 and got a decent prize.

Still not convinced? Prove it to yourself at the OLG Number Checker.

But I’ve played every week since the beginning… my numbers are due!

No, they aren’t. The numbers are random. They aren’t “due”. The little numbered balls don’t gather before the draw and decide “Geez, #11 hasn’t been out for a while, better let him go tonight.” and they certainly don’t come up with “numbers 1, 2, 3, 4, 5, 6, 7 you can’t go out together because that’s not random.” Previous draws do not affect future outcome.

If you’d played a ticket each week since 1 Jan 1994 until 31 Dec 2009, your odds of having won a jackpot in all that time are approximately 0.000013 (about 1 in 75000, similar to the odds of you scoring a date with a supermodel, although if do win the jackpot those latter odds probably go up a whole lot), and your odds of winning the jackpot on the 16 Jan 2009 are still 1 in 20,963,833. These odds are worse than the odds of being struck by lightning in the same time period (1 in 44000).

Enjoy! and game responsibly.

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