Re: Formula for nobody winning is ((N-1)/N)^T where N is ...Prev TopicNext Topic

• Gamma Ray

  

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  April 10, 2002
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  Apr 19, 2002, 1:08 pm

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  Reply to Joe Schmoe's 4/17/2002 8:02:05 PM ET message
  
  Your formula is creative, but I don't think it is correct!
  
  Assume there is a lottery with 1/10 odd of win (9/10 Odd of lose), and suppose 10 different tickets sold (all combinations). With your formula, chance of nobody win
  
  = ((N - 1)/N)^T
  = ((10 -1)/10^10
  = 0.347
  
  In simple logic, the winning ticket is sold. It is because all possible patterns sold. Therefore, the formula is false!
  
  Please participate and solve this math problem!
  
  
  Original Message:
  
  A few messages ago I was wondering how to calculate the odds that nobody would win. Looks like it's ((N - 1)/N)^T where N is the total pool of numbers and T is the number of tickets sold. Let's apply this to the drawing just held: 2,744,447 won $1 with the odds of 1 in 62. So there were approximately 170,155,714 tickets sold. (76,275,359/76,275,360)^170,155,714 gives 10.75% chance that nobody wins.
• CASH Only

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  June 5, 2002
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  Apr 19, 2002, 3:16 pm

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  and/or quick-picks are available.
• Guest

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  May 31, 2000
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  Apr 19, 2002, 4:16 pm

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  Ok. Let's assume that 10 tickets are sold and the game is that you have to pick a number between 1 and 10 so that the odds for each ticket are 1 in 10. You're assuming that, say, somebody picks the number 5, then none of the other players can then pick that number. When you go in to buy lottery tickets, the computer doesn't spit back the ticket because that combination of numbers has already been selected.