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The birthday problem and the lottery
02-14-2011, 09:28 PM (This post was last modified: 02-14-2011 09:51 PM by Frank.)
Post: #2
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.

[Image: ncjzoo.gif]


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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RE: The birthday problem and the lottery - Frank - 02-14-2011 09:28 PM

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