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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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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UPDATE: see the comments for a link to my solution.
posted by Phil Birnbaum @ 4/29/2012 11:36:00 AM
11 comments
11 Comments:
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.
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.
Anxious to hear the non-math answer, too. I tried my best to come up with an answer without doing math, but was unable :-)
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.
I'll post my answer tonight!
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.
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...
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.
I've posted my "almost-math-free" solution here.
Fair enough. Yes, when there's zeros the probability of winning is zero.
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.
I'm not sure I buy your solution.
First, a critique of your solution:
For any number from 0-999, there is a greater than 1/1000 chance being a winner, since one or more numbers will be chosen.
Now, the fact that you picked your number reduces the chance that the fewest people picked that number, since all the other piles are unknown, and yours definitely has 1 in it, but with a sufficiently large number of players, that variable will approach zero.
I think the weakness of the thought experiment is that it is missing one possible outcome-the final purchase could push the lowest number into a tie with another number, thus giving a bunch of other people a payout. The last n can only hurt her pile, but she can help others.
But I think this is flawed reasoning anyway. My explanation is this: the whole thing is a version of the Monty Hall problem. even though each number is chosen randomly by contestants, seducing you into thinking you've got a 1/1000 chance, the prize goes to the smallest pile, which distorts the probability of your number coming up. 'you' don't have a 1 in 1000 chance of winning, because you'd only have a 1/1000 chance if each pile was the same size-you are mostly in a big pile. Since the smallest pile always wins, your winning chance is below 1/1000 in relationship to how far below average the smallest pile ends up being (since with a significantly sized pool, the chances of a tie should be close to zero, and even then would need to happen pretty frequently to overwhelm the disadvantage gained by picking the smallest pile).
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