# What's Wrong With This Lottery Strategy?

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.

It’s a long time since I did probability, but here goes:

First since you effectively buy 8 tickets, you have 8 times more chances of winning. (Don’t get excited, it’s still a remote chance!).

Now the first number that comes up is bound to be on one of your tickets. But the next 5 numbers are not a 12.5% shot - because there are many different ways they can NOT be on the same ticket, but only one way for the jackpot.

Try this. The first number is on ticket no. 1. The second and third numbers could be on tickets 2+3; 2+4; 2+5; 2+6…5+6; 5+7; etc. etc. - or they could both be on ticket 1. No way is that a 12.5% chance! (and you still have 3 more numbers to go).

The ways to do well on a lottery are:

run it yourself

put your entry money in a savings account

He’s not saying the next five numbers in toto are a 12.5% chance; he’s saying that EACH of the next five numbers is a 12.5% chance.

This is wrong, but not for the reason given above.

The first number drawn is on the first ticket (if it’s not, we switch so that it is).

The chances of the next number being on that ticket are not 1 in 8, though… they’re 5 in 47, which is significantly less (why? Because there are only five numbers left on ticket one, and 47 balls left in the lotto ball machine). The math error comes in assuming that you always have six good (i.e. unused) numbers on the ticket, but actually the number goes down each time there’s a hit (and if there’s not a hit, it’s not a winner anyway).

Assume you’re lucky (it’s over a 10% chance). The odds of the next number being on the same ticket are 4 in 46.

Following this all the way out assuming you’re lucky each time:

1 in 1 (you can always switch the first ticket)
5 in 47
4 in 46
3 in 45
2 in 44
1 in 43

Muliplying that out I get a grand total of 1 in 1,533,939, which is basically what you’d expect for buying eight tickets.

Yes, I expressed myself badly. I was trying to explain why although you have every winning number, there’s far more ways for them to be scattered around the wrong tickets, than all gathered together on one ticket.

Ah, statistics!

Actually, that should be “Ah, probability.” But it’s Friday; I’ll cut you some slack.

My statistics suck too, but the prob is pretty much the same. The number of combinations is 48x47x46…I forgot how high it went, some thing like several Billions.

You got 8 tickets, so 8/gazillion chances

I think this kind of series is called “factorial”, comes up very often in probability, and using 48 as the example is written (48!). My question: is there any way of calculating the value without actually multiplying out all the numbers?

Yeah, you don’t have to multiply all the numbers out:

(484746454443)/(654321)

is the number of lottery tickets in this case.

Thanks folks, I feel much better now.

“A lottery is a tax on stupidity.”

Lumpy:

No. There are approximations involving exponential terms, but there is no exact calculation other than successive multiplication.

Wasn’t it Cecil who said that the best way to make money on the Lottery is to write one of those stupid computer programs that are supposed to help you pick your numbers, and then selling it for \$39.99 a whack.

Enright3

calottery.com states the odds of getting 6 numbers is 1 in 18M. Biggest prize is about \$38M. Does that mean if you bought every ticket you would win \$20M?

Yes it would, assuming you were the only winner. If someone else also had the winning numbers you would split the winnings with them (and lose 1 million).

Also, the 38 million is the amount you get if you take payments over a period of time (usually 20-25 years). So you would get 1.9 million a year for 20 years. You spent the 20 million up front. If you have 20 million to throw around there are investments with much better returns.

“You can’t run away forever; but there’s nothing wrong with getting a good head start.” — Jim Steinman

Dennis Matheson — Dennis@mountaindiver.com
Hike, Dive, Ski, Climb — www.mountaindiver.com

AuraSeer,
What are the payoffs for getting less than 6 numbers right? If two tickets are worth anything, there’s at least some justification.

I did an iterative simulation (my probability math skills are stuffed up with a cold) for 300,000 lottory drawings with 8 tickets picked somewhat like you described. (Basically, I made sure that all 48 numbers were used across the 8 tickets.)

Here’s the results I got:
<table border=“2”><tr><th>Tickets:matches</th><th>Percentage</th></tr>
<tr><td>1:6</td><td>0</td></tr>
<tr><td>1:5 & 1:1</td><td>0.02</td></tr>
<tr><td>1:4 & 1:2</td><td>0.11</td></tr>
<tr><td>1:4 & 2:1</td><td>0.72</td></tr>
<tr><td>2:3</td><td>0.09</td></tr>
<tr><td>1:3 1:2 1:1</td><td>4.99</td></tr>
<tr><td>1:3 3:1</td><td>9.94</td></tr>
<tr><td>3:2</td><td>1.53</td></tr>
<tr><td>2:2 2:1</td><td>27.72</td></tr>
<tr><td>1:2 4:1</td><td>44.24</td></tr>
<tr><td>6:1</td><td>10.64</td></tr>
</table>

The x:y figures stand for x tickets with y winning numbers.

Some years ago in Virginia, some people actually tried the strategy of buying EVERY ticket in the lottery. A large group of people in Australia got together and tried it–they kind of screwed up and only covered about 3/4 of the tickets. They got lucky, though, and did manage to win without splitting it with anyone. I think now there’s a law in Virginia against buying tickets in bulk.