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Suppose there is a lottery where a random natural number between 1 and n = 1000 is drawn in secret. They sell two different tickets to customers:

  • A ticket with k = 10 random numbers on it (not sorted, no repetitions)
  • A ticket with an interval of k = 10 succeeding numbers starting at a random offset <= 991

A lot of tickets are sold, so there are expected to be several people with the secret random number put on their ticket. To resolve this, there are two procedures:

  • The order of the numbers on the ticket matters. Whoever has the winning number at the lowest position, wins.
  • If this did not single out a winner, the prize money is split among all those tickets with the winning number at the lowest position.

Which of the tickets should you buy? Does it matter? I figure that the chance of hitting the winning number is 1/100 in both cases. But what is the probability of taking the money home with you? Intuitively, I would say that both tickets have the same probability to win the same amount of money, i.e. hitting the winning number at a low enough position to get cashed out, but I am a layman in this field.

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You should definitely choose the tickets randomly distributed because there are way more such tickets which are different. If you take the ticket with a sequence of 10 numbers, there are only 991 different arrangements of the numbers on those so you're much more likely to be sharing your prize with a large number of people if you do win.

This is a fun question because it looks on the face of it like the problem is to choose the ticket most likely to have the winning number in the lowest position. But in reality it's about choosing the ticket least likely to share the prize once conditions one and two are met.

If all the tickets were unique (and therefore it was unlikely to share) then it wouldn't matter which you chose as both methods of choosing numbers are symmetric with respect to the ordering of the numbers, so neither is more likely to pick the winning number in its lowest position.

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