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| dj_pulse |
| quote: | Originally posted by nervous
I admit to not knowing a real lot about breaks, but the two times i did venture up there i was greeted with The Knack - My Sharona(original) and the second time Nirvana - Smells like Teen Spirit(Original).
Had to check i hadn't walked out of bohem into a pub:conf: |
hehe.. dont know about my sharona, but you probably heard Nirvana vs Adam Freeland - Smells Like Freeland .. |
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| bragi |
| quote: | Originally posted by Hyperdimension
Yes that is bizarre, that we were born on exactly the same date, and that we found out right there and then on the dancefloor!
I think it's very rare to have met someone that shares the same birth date as you!
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<smartarse>
Statistically speaking, in a room with 23 people, there's a 50/50 chance of 2 people having the same birthday (not counting the year)... and with 41 people in the room, there's a approx 90% chance of the same :)
</smartarse> |
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| djway |
| quote: | Originally posted by bragi
<smartarse>
Statistically speaking, in a room with 23 people, there's a 50/50 chance of 2 people having the same birthday (not counting the year)... and with 41 people in the room, there's a approx 90% chance of the same :)
</smartarse> |
Go the B'day Paradox :)
| quote: |
You've probably heard that in a group of about two dozen people there is a high chance that at least two of the people have the same birthday. If you've ever wondered why this is so, keep reading.
The calculation of probabilities where multiple events are involved can be a daunting task. So, this particular poser has been reduced to a relatively simple calculation and is based on the following relationships:
the total probability of all possible events occurring is equal to one (1)
it is possible to calculate the probability of a particular event (e. g., no occurrence)
subtracting the calculated probability from one gives the probability of at least one occurrence
So, in order to calculate the probability that, among a group of people, at least two of them have the same birthday, all we have to do is calculate the probability that none of them have the same birthday and subtract that from one (1). Start with one person in the room, we'll call her Amanda. The probability that her birthday is unique among all the people in the room is one; after all, Amanda is the only person in the room!
Now, let Brenda enter the room. The probability that Brenda's birthday is different from Amanda's is 364/365 (we'll ignore leap years in all of this) since there are 364 available days other than Amanda's birthday. With just two people in the room, the probability that they do not share a birthday is 1 x (364/365) or 0.997 and the probability that they share a birthday is 1 - 0.997 or 0.003.
In walks Carla. In order for Carla's birthday to be different from either Amanda or Brenda, it must occur on one of the remaining 363 days. The probability that Carla does not share a birthday with either Brenda or Amanda is 1 x (364/365) x (363/365) or 0.992. And, the probability that at least two of the three people share a birthday is 1 - 0.992 or 0.008.
We can continue this calculation as each new person enters the room and this is shown graphically below. Shown in red is the probability that no two people share the same birthday {1x(364/365)x(363/365)x ...} for 2 to 50 people. The green line is the probability that at least two people share a birthday. You can see that with 23 people in the room, the green curve has reached a probability of about 0.5 (0.507). This means there is a 50% chance (even odds) among a group of 23 people that two of them have the same birthday
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For those who are interested :)
http://www.people.virginia.edu/~rjh9u/birthday.html
--djway |
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| Pointy |
| quote: | Originally posted by bragi
<smartarse>
Statistically speaking, in a room with 23 people, there's a 50/50 chance of 2 people having the same birthday (not counting the year)... and with 41 people in the room, there's a approx 90% chance of the same :)
</smartarse> |
Dude, you're wrong....i thought you were smart?
:p |
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| bragi |
| quote: | Originally posted by Pointy
Dude, you're wrong....i thought you were smart?
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So did I, but it turned out to just be my butt :) |
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