Le Phoenix - where you can SCORE points off each other...
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- Senior Member
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I think it's 23.
logic being that the probability of something happening + probability of it not happening always equals 1. so work out the likelyhood of people not sharing the same birthday and go from there until your answer dips below 0.5. god bless the calculator!!
365/365 * 364/365 * 363/365 * 362/365 * 361/365 etc
logic being that the probability of something happening + probability of it not happening always equals 1. so work out the likelyhood of people not sharing the same birthday and go from there until your answer dips below 0.5. god bless the calculator!!
365/365 * 364/365 * 363/365 * 362/365 * 361/365 etc
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- Senior Member
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- Joined: Wed Jul 20, 2005 3:33 pm
YES! Well done thehut! Even if you'd got it wrong, I'd have given you something for showing your working!
Indeed, all you need is 23 random people in a room and you can bet your mathematically-challenged friend that 2 people share a birthday and you're more likely than not to win. Most people wouldn't guess that the number of people is so low.
Anyway, take the plate, thehut!
Indeed, all you need is 23 random people in a room and you can bet your mathematically-challenged friend that 2 people share a birthday and you're more likely than not to win. Most people wouldn't guess that the number of people is so low.
Anyway, take the plate, thehut!
Okay, let me try and understand this here...
You have one person, who's birthday is on a specific date. If there is another person in the room with them, then these is a 1/365 chance that they have the same birthday as the first person. Or 1/366, I'm not sure how the whole "leap years don't matter" thing works.
If there's two other people, there's a 2 / 365 chance, and so on until you get to better than 50%, which is 183/365.
Bah, I'm stuck. 10.
Edit: That'll teach me to take so long typing!
You have one person, who's birthday is on a specific date. If there is another person in the room with them, then these is a 1/365 chance that they have the same birthday as the first person. Or 1/366, I'm not sure how the whole "leap years don't matter" thing works.
If there's two other people, there's a 2 / 365 chance, and so on until you get to better than 50%, which is 183/365.
Bah, I'm stuck. 10.
Edit: That'll teach me to take so long typing!
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- Senior Member
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- Joined: Wed Jul 20, 2005 3:33 pm
If you want more of an explanation, check this out
http://en.wikipedia.org/wiki/Birthday_problem
I got to the 3rd paragraph before I gave up trying to understand it!
http://en.wikipedia.org/wiki/Birthday_problem
I got to the 3rd paragraph before I gave up trying to understand it!
based on the principle of people in the room NOT sharing the same birthday:
you are person number 1, your birthday can be any one of 365 days, so the first prob is 365/365 which =1.
the second person can have their birthday on 364 out of 365 days to NOT have the same birthday as you so the sum is now 365/365*364/365.
a third person can have their birthday on 363 out of 365 so as NOT to have the same birthday as anyone else in the room. now (365/365)*(364/365)*(363/365). keep working on these lines until your answer is less than 0.5 and conversely, the chance of people sharing the same birthday is greater than 0.5. then just add up how many 'sums' you had to do.
you are person number 1, your birthday can be any one of 365 days, so the first prob is 365/365 which =1.
the second person can have their birthday on 364 out of 365 days to NOT have the same birthday as you so the sum is now 365/365*364/365.
a third person can have their birthday on 363 out of 365 so as NOT to have the same birthday as anyone else in the room. now (365/365)*(364/365)*(363/365). keep working on these lines until your answer is less than 0.5 and conversely, the chance of people sharing the same birthday is greater than 0.5. then just add up how many 'sums' you had to do.
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- Senior Member
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- Joined: Wed Jul 20, 2005 3:33 pm
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- Senior Member
- Posts: 4166
- Joined: Wed Jul 20, 2005 3:33 pm