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| Posted by Princess_Neopia | 12/15/02 |
| You must have worked REALLY, REALLY hard. Your excellent at this. |
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| Posted by od-1 | 05/02/03 |
| OH MY! |
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| Posted by beanie89 | 08/30/03 |
| uh....right. |
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| Posted by Varthen | 07/02/04 |
| This teaser should be called...
HEADACHE IN A BOX!!!!!!!!!!!!!!!!!! |
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| Posted by trickster2005 | 09/10/05 |
| r u a genius? :o |
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| Posted by lessthanjake789 | 01/14/06 |
| i remember hearing about the birthday phenomenon but didnt quite understand it (i was sleeping in math class at the time) but it was quite interesting and i think this is a "practical" application of the numbers. nicely done and quite interesting. |
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| Posted by glenn222 | 04/13/06 |
| That was my next guess... |
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| Posted by NomadShadow | 04/25/06 |
| I really loved it, I got the answer using difference equations.
Anyway, what if someone had a twin brother! wouldn't be better to stand second in line with the brother standing first;) |
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| Posted by Unstumpable | 06/20/06 |
| wow, that would be annoying to type all of that, but the probibility works out all right, good teaser |
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| Posted by Dedrik | 08/09/06 |
| :D Good problem |
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| Posted by foraneagle2 | 10/06/06 |
| Classic probability question. Another way of saying it: With as few as 20 random people in a room, the chance of two having the same birthday > 50%. |
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| Posted by ztodd | 10/21/06 |
| That's not saying quite the same thing - that's just saying that the probability that either you or someone ahead of you will win is > 50%, but it doesn't necessarily mean you have the best chance. |
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| Posted by pearl5608 | 10/25/06 |
| Great teaser. i'm in AP Statistics in my high school, its my favorite class, this was an interesting problem |
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| Posted by roaddevil | 02/08/07 |
| What if there is only two people at the movies? You can't figure probability with no numbers to start with! |
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| Posted by chickwithbrains | 09/13/07 |
| uhhh.... what what what what??? Holy cow, I definitely didn't get this one!! :o :0 !!!!! |
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| Posted by ZJT_05 | 02/26/08 |
| One Word - WOW. |
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| Posted by Anushka | 04/11/08 |
| ver clever. and by the way...u must have a lot off free time on ur hands :wink: |
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| Posted by javaguru | 12/08/08 |
| Great problem!
I'm surprised given how long this problem has been posted that nobody bothered to give the probability of winning the free ticket, which is approximately 3.0628%. 19th place in line is almost as good at 3.0608%. This is a pain to calculate recursively by hand but very easy using a spreadsheet.
At 20 people the probability that two people have the same birthday is approximately 41.1% and the cumulative probability that someone in line before you wins the ticket is approximately 36.67%.
Also, the point at which the probability of two people having the same birthday exceeds 50% is with 23 people, not 20. |
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| Posted by javaguru | 05/14/09 |
| Janeys-
The answer given for the teaser is correct. You need to take into consideration the CUMULATIVE probability that someone in front of you wins the ticket before you get to the front.
The fact that at 23 people there is a better than 50% chance that two have the same birthday isn't really relevant to the problem. I just included that in my previous comment as a point of interest. |
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| Posted by rlc327 | 07/17/10 |
| i would think the last person in line at any time where position P < or = 20. |
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| Posted by normadize | 03/29/12 |
| I know this is old but it took me only a few minutes to solve and my solution is quite different, so I thought I might as well post it. In my opinion the answer given is convoluted.
Let p(k) be the probability that the first duplicate bday is at person k+1 in the line, i.e. the first k people have all different bdays and the bday of person k+1 matches one of the previous k bdays. No need for any conditional probabilities (why the Bayes equation in the answer?) since all events are independent and p(k) can be expressed directly:
p(k) = 365/365 * 364/365 * ... * (365 - k + 1)/365 * k/365
(easily verifiable thoruhg simulation for those who doubt it)
It says that the 1st person can have any bday (365/365), the 2nd person has a different bday than the 1st (364/365), and so on, and person k+1 has one of the first k bdays (k/365).
We just need to maximize p(k) as a function of k, that's it. It is easy to show it has only one maxima. The current form is ugly but we can do a trick, that is to find an equivalent, continuous function. We can write p(k) as:
p(k) = k/365^(k+1) * product_of k {from 365-k+1 to 365}
Taking the natural log:
ln(p(k)) = ln(k/365^(k+1)) + sum ln(k) {from 365-k+1 to 365}
The sum of discrete logs can be approximated by an integral of the continuous log function given that the terms in the sum are large (whoever knows of Sterling's factorial approx proof is already familiar) and so we can consider the continuous function q(x):
q(x) = ln(x/365^(x+1)) + integral ln(x) dx {from 365-x+1 to 365}
using parts:
q(x) = ln(x/365^(x+1)) + 365 ln 365 + (366-x) ln(366-x) - x + 1
Differentiating, and equating to zero, we get:
1/x = ln (365 / (366-x)) ... or ... exp(1/x) = 365/(366-x)
This has a unique solution since exp(1/x) is decreasing on [1,365] while 1/(366-x) is increasing on [1,365]. This is computed numerically to give x = 19.3676 so then our k+1 = 20, which is the solution. |
