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The birthday problem and the lottery
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02-14-2011, 09:28 PM
(This post was last modified: 02-14-2011 09:51 PM by Frank.)
Post: #2
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RE: The birthday problem and the lottery
Its a pity theres no spoiler facility in this forum, I would have used it here.
You might be tempted to think linerarly and think that about half of 365 people , say 160 need to be in the room. You'd be wong. The answer is 23 ! How can so few people when comparing birthday dates have a 50% chance of 2 people having the same birthday ? Doesn't seem possible does it ? It all comes down to comparisons, and how many comparisons are being made. Lest consider the first six people to enter the room. person A compares birthday with persons B,C,D, E and F thats 5 comparisons. person B compares birthday with persons C,D , E and F thats 4 comparisons. person C compares birthday with persons D , E and F thats 3 comparisons. person D compares birthday with persons E and F thats 2 comparisons. person E compares birthday with person F thats 1 comparison. So thats a total of 5+4+3+2+1 =15 comparisons of birthdays for 6 in the room. I think you can see a pattern forming here.... So when we get to 20 people in the room, you can work out that there are (add 1 to 19) =190 comparisons. Already thats more than half the 365 possible birthday dates. In fact a quick approximation to the answer can be obtained by taking the square root of the possibilities (365) and adding 17% to that result to give 22. A good explanation is here:- http://betterexplained.com/articles/unde...y-paradox/ For a more complicated but exact method of arriving at the answer of 23 by plotting on a graph you can read it here..http://mathforum.org/dr.math/faq/faq.birthdayprob.html The good news is that you can use a spreadsheet to do the calculations. Use each row of the sheet to work out the probability of no 'n' people sharing the same birthday. Have a different value of n on each row. This means that the probability of n people sharing the same birthday is 1- that value. ![]() You can read off when the probability of NOT sharing a birthday is 49.27% which means that the probability of at least 2 people sharing a birthday is 100-49.27%=50.73%. Okay, I'll leave you with a thought. How many lotto results need to be out such that the probability of a result repeating is 50% ? |
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The birthday problem and the lottery - Frank - 02-12-2011, 05:02 AM
RE: The birthday problem and the lottery - Frank - 02-14-2011 09:28 PM
RE: The birthday problem and the lottery - Frank - 02-16-2011, 11:26 PM
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