A List of Birthdays
Sep 19, 2003 | 5:31 pm
#32
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SXJ,
Same back to you!
Craig
_________________
Interesting sidelight about 25 July. My office has 80 people in it, and 5 have our same birthdays. Odds of that would be pretty high, I'd guess.
Same back to you!
Craig
_________________
Interesting sidelight about 25 July. My office has 80 people in it, and 5 have our same birthdays. Odds of that would be pretty high, I'd guess.
Sep 20, 2003 | 6:55 am
#33
Original Member
Join Date: May 1998
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bourne on the foreth of july
arturo mak misteak on hes birfdey, et es march sekond.
edioted bye arturo too korect birfdey
------------------
Loving, Caring,
Honest , Intelligent, Empathetic, Creative, and Giving.
[This message has been edited by arturo (edited 09-21-2003).]
arturo mak misteak on hes birfdey, et es march sekond.
edioted bye arturo too korect birfdey
------------------
Loving, Caring,
Honest , Intelligent, Empathetic, Creative, and Giving.[This message has been edited by arturo (edited 09-21-2003).]
Sep 20, 2003 | 5:16 pm
#35
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Interesting fact: if you have 23 people together in a room, the probability that two or more people share a common birthday is greater than 50%. This is true, even taking leap years into account.
FewMiles..
FewMiles..
Sep 20, 2003 | 5:25 pm
#36
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FewMiles,
Can you explain the math behind the 23 people, odds are greater then 50%? Also, how do you calculate the odds on 5 out of 80?
Craig
Can you explain the math behind the 23 people, odds are greater then 50%? Also, how do you calculate the odds on 5 out of 80?
Craig
Sep 21, 2003 | 12:46 am
#39
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Craig6z, here's how to calculate it.
The probability that 23 people in a room all have distinct birthdays is calculated this way:
I hope that made sense.
FewMiles..
[This message has been edited by FewMiles (edited 09-21-2003).]
The probability that 23 people in a room all have distinct birthdays is calculated this way:
- The first person has some birthday X.
- The probability that the second person doesn't have the same birthday is 365/366.
- The probability that the third person has a birthday different than the first two is 364/366.
- Keep going until you reach the 23rd person.
- Multiply all these terms together and you get something just over 0.49 or 49%.
- Remember that this is the probability that all 23 people have distinct birthdays.
- The opposite case, that there is at least one occurence of a common birthday, would be 100% minus the above probability which gives an answer of just over 50%.
I hope that made sense.
FewMiles..
[This message has been edited by FewMiles (edited 09-21-2003).]
Sep 21, 2003 | 5:53 am
#41
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<font face="Verdana, Arial, Helvetica, sans-serif" size="2">Originally posted by skofarrell:
14-Sept for me.
And Blair didn't send a card
</font>
14-Sept for me.
And Blair didn't send a card
</font>
Happy belated, sir!
Sep 21, 2003 | 8:27 am
#42
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I think we should have one standing Happy Birthday thread for everyone, this is plainly going to get out of control!
As for a commie like me, May Day is an approriate birthday.
As for a commie like me, May Day is an approriate birthday.
Sep 21, 2003 | 2:19 pm
#43
Join Date: Feb 2001
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<font face="Verdana, Arial, Helvetica, sans-serif" size="2">Originally posted by FewMiles:
Should blairvanhorn have multiple entries on your list?</font>
Should blairvanhorn have multiple entries on your list?</font>
Sep 21, 2003 | 2:22 pm
#44
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<font face="Verdana, Arial, Helvetica, sans-serif" size="2">Originally posted by pynchonesque:
... well into his thirties!</font>
... well into his thirties!</font>