## Sunday, April 29, 2012

### Puzzle

(Non-sports post.)

Here's a puzzle that occurred to me a few days ago.  I don't know what to do with it, so I might as well post it here.

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In a certain country, the "daily numbers" lottery works like this: you buy a \$1 ticket with your choice of 3-digit number.  The winning 3-digit number is announced.  If you match exactly, you win \$1000.

However, the draw isn't random.  Instead, the winning number is always the *least popular* number chosen by ticket-buyers -- that is, the number that is on the fewest tickets.

There's one catch: if there is a "tie" for least popular number, every number in the tie counts as a winning number.  For instance, if there are 1,359 tickets with number "000", 1,359 tickets with number "844", and every other number is on more than 1,359 tickets, that's a two-way tie for least popular, so there are two winning numbers.  Holders of either number win the full \$1000.

That means, in theory, that *every* ticket could win.  For instance, if all 1,000 numbers are bought exactly 3,453 times, it's a 1000-way tie, and every ticket buyer wins \$1000.  (Presumably, the lottery authority has a large cash reserve to cover this possibility.)

Assume that every ticket buyer chooses his or her number randomly.  Is the chance of winning less than 1 in 1000, more than 1 in 1000, or exactly 1 in 1000?

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I think there's an answer that requires almost no math, just logic and a very basic common-sense understanding of how probability works.  (There might be other answers, and it could be that I missed an answer so obvious that the question isn't very interesting.)

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At Sunday, April 29, 2012 4:21:00 PM,  j holz said...

this does seem fairly easy, but i think it could be used to illustrate a point or two in a statistics class. you might also want to specify that a large number of tickets are purchased, to cover trivial cases where some number combinations have zero tickets bought.

At Monday, April 30, 2012 10:23:00 AM,  Mike said...

Depends on how many tickets are sold. If it's less than 1000, then of course your chance of winning is > 1/1000. Even if it's 2000 people buying tickets, I'd say it's > 1/1000.

At large numbers of ticket purchases though (like in your example of a few thousand of some numbers being bought), I ran a quick experiment in Excel and it looks like less than 1/1000.

What would also be interesting is if there was some way to figure out where that breakeven point is.

At Monday, April 30, 2012 10:25:00 AM,  Mike said...

Anxious to hear the non-math answer, too. I tried my best to come up with an answer without doing math, but was unable :-)

At Monday, April 30, 2012 10:28:00 AM,  Phil Birnbaum said...

I should have mentioned that under the terms of the puzzle, if there's a number where NO tickets are bought, nobody wins (because that's the least popular number).

But, if you don't like that, then I can stipulate (as j holz suggested) that there are so many tickets sold that the probability of a number having no tickets is pretty much zero.

At Monday, April 30, 2012 11:11:00 AM,  Jim A said...

Reminds me of a similar thought experiment I read about several years ago where the winning number is defined as the one that is closest to 0.67 times the average of all the numbers chosen.

At Monday, April 30, 2012 2:45:00 PM,  David said...

Definitely less than 1/1000 (provided no zeros).

Though I think as N gets large, it should approach 1/1000. It would probably take some math to prove that part though...

At Monday, April 30, 2012 3:48:00 PM,  Phil Birnbaum said...

David,

With zeroes, the chance is also less than 1/1000, no? It's zero.

I think you're right about the asymptote, and that you'd need math for it.

At Monday, April 30, 2012 5:53:00 PM,  Phil Birnbaum said...

I've posted my "almost-math-free" solution here.

At Monday, May 07, 2012 11:44:00 AM,  David said...

Fair enough. Yes, when there's zeros the probability of winning is zero.

At Wednesday, May 09, 2012 5:35:00 AM,  reinis said...

I thing posted math-free solution might be flawed, this statement for instance:
"If it's a single number, the Nth ticket wins if and only if it's that exact number. The chance of that is 1/1000."

Yes, but this Nth ticket can also cause effect, that there is a tie at that moment and chance of winning is greater than 1/1000 other tickets.

When N is <= 1000, obviously chance is greater than 1/1000, intuition tells me that chance of winning drops below 1/1000 at some particual value of N and continues to drop to 0. But I cannot prove that.

At Sunday, May 13, 2012 12:39:00 AM,  StLhawk said...