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Hi Bob ~ 6 numbers chosen from 49 numbers ?

There are 13,983,816 [49! / (6! × (49 - 6)!)] ways to choose 6/49

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** IF you can select your 6 numbers for each ticket, then obviously you will select a different combination for each ticket:

You will have to buy 13,983,816 / 2 =

6,991,908 tickets to have a 50% chance of hitting the winning combination

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** IF you can't select your 6 numbers for each ticket (you buy the tickets from a randomising machine):

If you buy y random tickets the chance that you don't win is 13,983,815/13,983,816 multiplied by itself y times = p^y

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TO RESTATE YOUR QUESTION:

How often you have to multiply 0.99999992848876 by itself to reach 0.5 ?

log (0.5) / log (0.99999992848876) = 9,692,842

The answer is 9,692,843 lottery tickets

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Michael

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Hi again Bob ~ This is how I've interpreted your question:

** There are 365 days in the year

** I invite strangers randomly from the street into a party

** I write down their birthdays (the birth year doesn't matter)

QUESTION: How many people would I need at the party to have a 50% chance that ALL the dates of the year are written down ?

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My guesswork:

If 365 people were at the party it would be very unlikely that I would have 365 different birthdays

To get 365 different birthdays (the year of birth doesn't matter) I estimate off the top of my head that I would need 1,200 people to have a 50% chance that ALL 365 possible birthdays are present

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However I asked the question on a mathematical forum & I got this reply:

365 [1 + 1/2 + 1/3 + ... + 1/365] = 2,364.6 guests needed at the party

See here for the full details:

http://www.mymathforum.com/viewtopic.php?f=24&t=18519

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Michael (I don't think my guess was too bad, but thanks must go to a.s.woods at the above link)

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You can check all the probabilities on the lottery checker website. You can aslo find related info on probabilities, odds and stuff like that