Le Phoenix - where you can SCORE points off each other...

The famous fruit-chat quiz!
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Weyland
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Post by Weyland »

Hmmm. Well, presumably 366 people would ensure you won, but I suspect that's not the point! ;)

Let's get it out of the way then: 366 / 2 == 183?
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Istenem
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Post by Istenem »

assuming the random people weren't chosen from the twins convention and that the year of birth is unimportant; based wholly on guesswork i'll say 92
nobody ever wins on those things.
Cardinal Sin
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Post by Cardinal Sin »

Weyland ... completely wrong! If there were 183 people in the room, the chances of me winning the bet would be something like 99.9%!

UP ... not quite as wrong but still pretty wrong! You were right that the year is unimportant though!
Weyland
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Post by Weyland »

Gah, I hate statistics. Okay then... 2
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Istenem
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Post by Istenem »

i'm probably barking in the wrong forest but according to my beermat arithmetic it should be 46. anywhere near?
nobody ever wins on those things.
Cardinal Sin
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Post by Cardinal Sin »

Weyland! Too low! (although I suspect you knew that).

UP! Still too high!
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Istenem
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Post by Istenem »

(i think) i understand the principle of this but my brain just strikes when i try to think about maths. back to guessing: 32?
nobody ever wins on those things.
thehut
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Post by thehut »

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
Cardinal Sin
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Post by Cardinal Sin »

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!
Weyland
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Post by Weyland »

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! ;)
Cardinal Sin
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Post by Cardinal Sin »

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!
thehut
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Post by thehut »

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.
Cardinal Sin
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Post by Cardinal Sin »

I understand it the way you do it. It's when they start adding all the squiggly symbols and talk about fish (poisson) that I get a bit confused!
thehut
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Post by thehut »

An easy one as i'm off home in a mo... :)

Which Shakepere play features the line:
"All the world's a stage, And all the men and women merely players"
Cardinal Sin
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Post by Cardinal Sin »

Interesting!

I'm hopelessly addicted to Civilization IV, and whenever you research the drama technology, Leonard Nimoy butts in and makes that very quote.

Not that that helps me in the slightest. I'm going to guess The Taming of the Shrew.
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