Good Math, Bad Math

A simple, silly, but entertaining example of mathematical illiteracy by way of the Associated Press:

OMAHA, Neb. (AP) — The odds are against something this odd. But a Nebraska Lottery official says there was no mistake: The same three numbers in Nebraska’s Pick 3 lottery were drawn two nights in a row this week.

Lottery spokesman Brian Rockey said one of two lottery computers that randomly generate numbers produced the numbers 1, 9 and 6 — in that order — for Monday night’s Pick 3 drawing. Rockey says the next night, the lottery’s other computer produced the same three numbers in the same sequence.

The odds of such an occurrence? One in a million.

Close… Only off by three orders of magnitude…

Assuming a fair system, and assuming that the drawing system
can produce the same number in multiple positions, the odds of drawing the numbers “1,9,6” twice in a row are, indeed, one in one million. But the odds of drawing the same number two nights in a row are just one in 1/1000.

An easy way to think of it: the first days draw doesn’t matter. Whatever number it produces is unimportant – any value for the first draw results is fine. The question is, what are the odds of the second draw producing the same result as the first?

Three digits means 1000 possibilities – so 1 in 1000.

Of course, it looks like they got even more than that wrong. From what I can find in a quick search, the Nebraska lottery draw produces three
different digits – once a number has been drawn, it can’t be drawn again. So you’ll never get a result like “998”, because you can only draw one 9. So in fact, the total number of possible draws isn’t 1000. It’s 720. So the odds of a duplicate draw on a given pair of nights is just 1 in 720.

And, of course, if you really wanted to figure out how likely it was for this to happen, you wouldn’t choose one particular pair of nights. The real question of probability here isn’t “How likely is it that tonight and tomorrow, a three-draw lottery will produce the same draw results?” The real question is, “If I operate a three-draw lottery for a year, what are the odds that I’ll draw the same result two nights in a row?”. And the answer to that is: not terribly unlikely.

It’s like the old birthday game: in a class of 30 people, what are the odds that two of them will have the same birthday? According to the reasoning
of the AP, it would be something like 1 in 133,000. In reality, it’s around 2 in 3.

Comments

  1. #1 Uncle Al
    January 23, 2009

    The second best lotto strategy is to use the previous day’s winning number, for you know it can arise if the second occurance is not excluded. The best strategy is not to play, since the numerical expectation of winning is about the same either way, but the cost is not.

  2. #2 Jackal
    January 23, 2009

    Uncle Al, the preveous day’s winning number is no more likely to occure than, in NE’s case, any three digit number with no repeated digits.

    With only 720 possibilities, we’d expect to see a two-day run like this about once every two years, assuming the lotto runs every day. (I did not bother to check.)

  3. #3 arbeck
    January 23, 2009

    Jackal,

    It is true that the previous day’s winning number is no more likely to occur than any other. However, in most of these lotteries, you split the winnings with any other people who picked the same numbers as you. Therefore, if you always pick yesterdays numbers; you are just as likely to win, but less likely to split any winnings because most people would never pick the same numbers.

  4. #4 SteveM
    January 23, 2009

    However, in most of these lotteries, you split the winnings with any other people who picked the same numbers as you.

    This is the thing that really compicates calculating expected value of a lottery. Even when those megajackpots get to be larger than the inverse probability of winning (e.g. $5.1million * 1/5million – $1ticket > $0) but when it gets that high, so many people play that the odds of having to split the winnings go up, so reducing the EV once more to less than $0.

  5. #5 Jackal
    January 23, 2009

    Dang, SteveM, I never thought of that. Now I’m glad I don’t bother trying to catch the lotto when the EV is suposedly greater than $0.

  6. #6 Ian Preston
    January 23, 2009

    @Mark: From what I can find in a quick search, the Nebraska lottery draw produces three different digits – once a number has been drawn, it can’t be drawn again.

    According to the lottery site:

    Players will pick three numbers, each from a separate set of ten digits (0 through 9). The numbers can be unique such as 1, 2, 3 or identical such as 1, 1, 1.

    They also give the probability of matching all three in order as 1 in 1000 so what you say seems like it can’t be right.

    @Jackal: However, in most of these lotteries, you split the winnings with any other people who picked the same numbers as you.

    That doesn’t seem to be true here either. Even if it were true then your argument for choosing yesterday’s numbers would only be convincing if the number of people having the same thought as you were small relative to the number put off choosing these numbers by the gambler’s fallacy. If the numbers of winners were reported somewhere then Tuesday’s draw would offer an ideal opportunity to actually test the extent to which an unusually high or low number of players do choose the previous day’s numbers.

  7. #7 REBoho
    January 23, 2009

    A strange game. The only winning move is not to play. How about a nice game of chess?
    Joshua

  8. #8 Jackal
    January 23, 2009

    @Ian Preston: @Jackal: However, in most…

    Those were arbeck’s words, not mine.

  9. #9 Ian Preston
    January 24, 2009

    @Jackal: Those were arbeck’s words, not mine.

    Sorry, you’re right.

  10. #10 EJ
    January 24, 2009

    The AP is standing by their innumeracy in this case! This according to Derek Donovan, the reader representative at the Kansas City Star:

    Now, I spoke to the Associated Press in Omaha, and they stand by the figure, because they interpret it according to the latter explanation.
    http://adastrum.kansascity.com/?q=node/512

    Donovan’s column explains the APs mistake and gives 1/1000 as a better answer, but he’s too charitable.

  11. #11 Jon Hendry
    February 2, 2009

    “but when it gets that high, so many people play that the odds of having to split the winnings go up, so reducing the EV once more to less than $0.”

    Apparently people are becoming a bit jaded, so the mass participation doesn’t happen until the prize gets really huge, like over $200 million.

    Aside from that, it seems as though when multiple people share a prize, it’s most often a group of coworkers who pooled their money to buy a lot of tickets, rather than multiple unrelated players who happened to pick the same number.

  12. #12 Daithi
    February 3, 2009

    I remember having a discussion about whether or not to spend more than $1 for a lottery ticket.

    His position was that you should buy more than $1 because if you spent $2 you would be twice as likely to win.

    My position was that if my chance of winning was 1 in 80 million then my chance of losing was more than 99.999% and if I bought $1 in lottery tickets or $100 in lottery tickets my chance of losing was still more than 99.999%.

    His counter position was that I should just shut up.

  13. #13 Mike Lotto
    March 16, 2010

    Hi, love your blog, I’ve discovered it last week and read it these past few days, but I have a problem with your RSS feed. It throws out an error when I click on it. Maybe other people have the same problem, so can you tell us if anything special needs to be done or just clicking on the RSS feed buttin should do the job?

  14. #14 James Sweet
    March 16, 2010

    “If I operate a three-draw lottery for a year, what are the odds that I’ll draw the same result two nights in a row?”. And the answer to that is: not terribly unlikely.

    It would be 1-((719/720)^(n-1)), where n is the number of nights you operate the lottery, right? So 5 days a week for 52 weeks, that would be n=260, or about 30%.

    Oh whoops, I just saw that the digits can repeat, so it’s (1-((999/1000)^(n-1)), so for n=260, the odds are about 23%.

    Either way, yeah, that’s not terrible unlikely at all.

  15. #15 James Sweet
    March 16, 2010

    The article linked to by EJ makes me want to weep.

The site is currently under maintenance and will be back shortly. New comments have been disabled during this time, please check back soon.