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TheDeath
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posted November 09, 2008 07:05 PM |
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First find out the total number of "combinations" with everyone being put on a different day:
x = C(365,23) = 365! / ((365-23)! * 23!) (let's keep it like that  )
Then you have the "combinations" when 2 people are born the same day, so we do:
y = C(365,22) = 365! / ((365-22)! * 22!) (take 22 at a time, because one of them is the same as the other one)
Further, we can have combinations when 3 people are born the same day, and so forth. Notice that all the values except 'x' are actually "a won bet" because AT LEAST 2 people are born the same day, so the denominator only has an extra +x at the end compared to the nominator.
To find this out, we can use a program like Pari to make all calculations 100% correct and write a script for it to perform it, so I ended up with 6.68% (approx.), poor chance. So I would bet against it, since I have a 93.32% chance of winning that bet
I'll get to a "traditional" simplification (no Pari) method, but if you have any tips on that I'll like to know cause I'm not very good at simplifying factorials 
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Ecoris
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posted November 09, 2008 07:23 PM |
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Edited by Ecoris at 19:27, 09 Nov 2008.
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Since I know the answer I'll not post it. Nor will I post any hints. I'll just say that TheDeath is way off.
Edit: There's a very subtle "hint" in the post in which dimis posed the problem.
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TheDeath
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posted November 09, 2008 07:45 PM |
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Damn. I knew the chances were too small and the calculation too complicated so there must have been some sort of "hint" or "trick" or something
(here's my PARI script in case anyone's interested)
Quote:
comb(x,y) = x!/((x-y)!*y!);
x = comb(365,23);
y = 0; for(i = 1, 22, y=y+comb(365,i));
y/(x+y) + 0.
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friendofgunnar
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posted November 09, 2008 07:50 PM |
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Here's my answer:
You start with the first person, his birthday is on an arbitrary date, so we'll make it January 1. The second person has a 364 in 365 chance of not having the same birthday. The third person has a 363 in 365 chance of not having the same birthday as the two people that came before.
The probability that nobody will have the same birthday as the person that came before is (364/365) * (363/365) * etc....
The probability that somebody will have the same birthday is 1 - the probability that nobody will have the same birthday as the person that came before.
1 - ( (364! / 342! ) / ( 365^22) )
I had to use a spreadsheet since I found calculators couldn't handle this problem
answer
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TheDeath
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posted November 09, 2008 07:53 PM |
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Quote:
1 - ( (364! / 342! ) / ( 365^22) )
I put that in Pari and it gives 0.5072972343239854072254172283  (that is 50%)???
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JollyJoker
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posted November 09, 2008 07:54 PM |
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Indeed, the Death is way off.
It's again quite counter-intuitive. I know the solution as well, and 23 persons is just the "critical" number.
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Asheera
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posted November 09, 2008 07:54 PM |
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Quote:
I had to use a spreadsheet since I found calculators couldn't handle this problem
With Windows Calculator you can do it and it gives: 0.50729723432... etc
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TheDeath
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posted November 09, 2008 07:56 PM |
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So let me get this straight. It is not a simple thing like "at least 2 people having the same 'day' out of 365"?
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Mamgaeater
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posted November 09, 2008 08:43 PM |
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i found this on the net...
can someone explain before my brain hemorrhages
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JollyJoker
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posted November 09, 2008 09:04 PM |
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Quote:
So let me get this straight. It is not a simple thing like "at least 2 people having the same 'day' out of 365"?
It's simply this way: if you take 23 people, you have a slightly better than 50% chance that 2 people out of those 23 will be born on the same day of the year.
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friendofgunnar
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posted November 09, 2008 09:10 PM |
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Quote:
i found this on the net...
can someone explain before my brain hemorrhages
mamga, move your eyes closer to the picture of the triangles.
closer...
closer...
now turn your head to the side so that you can look at the triangles from the left side, near the bottom left point of each triangle. (it helps if you close one eye)
examine the two triangles carefully, especially the hypoteneuse.
you will find your answer 
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Mamgaeater
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posted November 09, 2008 09:33 PM |
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one hypotenuse appears to bend inward and the other bends outward?
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Ecoris
Promising
Supreme Hero
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posted November 09, 2008 09:53 PM |
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Easy one: Find 100 consecutive numbers containing no primes.
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TheDeath
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posted November 09, 2008 10:22 PM |
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Quote:
Quote:
So let me get this straight. It is not a simple thing like "at least 2 people having the same 'day' out of 365"?
It's simply this way: if you take 23 people, you have a slightly better than 50% chance that 2 people out of those 23 will be born on the same day of the year.
Personally, I don't get it
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Asheera
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posted November 09, 2008 10:42 PM |
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What don't you get?
If you have 23 people, the chance to have two of them born on the same day of the year is a little more than 50%.
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Lexxan
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posted November 09, 2008 11:11 PM |
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lol I DO know someone who was born on the same Day...
but I know more people than 23 lmao.
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Coincidence? I think not!!!!
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Binabik
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posted November 09, 2008 11:23 PM |
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I will bet against two people in the list having the same birthday. But before you accuse me of cheating.....
My bet *IS* math based. All math starts with a given. And well, maybe the givens I start with are not quite the same as the givens others start with.
Even without seeing the solutions already provided, I can easily guess that the "subtle hint" mentioned by Ecoris was when Dimis said "@Ecoris: Thanks for participating. This gives me the opportunity for the following problem:". In other words, the addition of Ecoris to the list was JUST enough to make the odds favorable that at least two people would have the same birthday.
But that's assuming the birthdays are randomly picked. But for me at least, they aren't 100% random. I've seen the list of birthdays for many HC members. Although I remember almost no details of that list, I know a few things.
1) There is overlap between that list and the people participating in this thread. (I am on both lists, maybe others)
2) "At a glance", that birthday list had less than expected duplicate birthdays given the number of people on the list.
3) If that set had below expected duplicates, can you expect the intersection of the two sets also to have less than expected duplicates? Hmmmm, interesting question there. Intuitively I would say yes, but I might be wrong.
4) I know that I was one of the "pairs" of duplicate birthdays on the other list. And the person with the same birthday as me hasn't participated in this thread. Therefore the intersection of the two sets eliminates some people from this thread's list.
With the assumption that the odds of RANDOM birthdays BARELY puts the odds in favor of at least 1 duplicate birthday in this thread, then the above should change the odds just enough to favor NOT having duplicates.
Therefore I bet against it. The wager is a bright shiny Helm of Heavenly Enlightenment. Any takers?
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TheDeath
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posted November 09, 2008 11:44 PM |
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Quote:
What don't you get?
If you have 23 people, the chance to have two of them born on the same day of the year is a little more than 50%.
But I don't get the actual computations that lead there
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Binabik
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posted November 10, 2008 01:00 AM |
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Quote:
The second person has a 364 in 365 chance of not having the same birthday. The third person has a 363 in 365 chance of not having the same birthday as the two people that came before
OK, the second person has 364/365 chance of not having the same birthday as the person before.
But for the third person to have a 363/365 chance as stated, that would be assuming that the first two do NOT share the same birthday, right?
I'm not sure how to fix this. Would it be (363+1/365)/365? Or maybe 363/365 * (1+1/365)? My guess is the former.
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dimis
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posted November 10, 2008 03:30 AM |
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Edited by dimis at 11:54, 10 Nov 2008.
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FriendOfGunnar came up with very good justification. Very nice!
And a hint: Every occurrence of the symbol '-' in his post should be read as 'minus'. Does it make more sense this way on a second read?
We have two open problems. One by mamga in which FriendOfGunnar gave a hint. And one by Ecoris that nobody has touched so far.
Oh, please, can the participants send me an email/im with their birthday (not year)? Let's see which side of the coin we are facing ... Of course if you mind telling your birthday, just send me a "negative" email/im, so that I don't wait forever for your email/im.
Thanks in advance
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This thread is pages long: 
