+0  
 
+5
404
9
avatar+18 

how many people in a room does it take for there to be a 50/50 chance of 2 people sharing a birthday? show and explane work

yixz33  Jun 2, 2014

Best Answer 

 #9
avatar+92806 
+5

How rude, how rude!

Still, 2 can play at thine game.

Check the points again dear Sir.

And the Lady still hath her head!

Have thou found the Roman zero amounst Zeus's offerings yet?

Thou waste time here annoying the Queen when thou should be attending thine own most urgent tasks.

Be off with you Sir Headless CPhill!

You are but a knat upon Her Royal steed!!

Melody  Jun 2, 2014
 #1
avatar+92806 
+5

23 people and the reasoning (ignoging leap years) goes :- Start with one person in a room. Add a second person and the probability that he has a different birthday is 364/365. Add a third and the probability that he has a different birthday to the first two is 363/365. The probability that there are now no duplicate birthdays is (364/365)*(363/365). Add a fourth and the probability that he has a different birthday to the first three is 362/365. The probability that there are now no duplicate birthdays is (364/365)*(363/365)*(362/365). etc. The probability that there are no duplicate birthdays falls below 50% when the 23rd person arrives.

Melody  Jun 2, 2014
 #2
avatar+92806 
+5

Look what Sir CPhill just sent me!

How dare he talk to the Lady of the Land like that!

"CHEAT!!! 
CHEAT!!! 
CHEAT!!!! 
.....This "procedure" violates "forum protocol"...... 
I'll "out" you on the forum, if you're not careful!! 
At least give me some credit for remembering that it WAS 23 !!!"

Guards! Guards!

Off with his head!

 

Queen Guinevere.

                                 

Melody  Jun 2, 2014
 #3
avatar+87333 
+5

....And the probability that Melody got this answer off the internet  =

[(10000)(1/2)] %

CPhill  Jun 2, 2014
 #4
avatar+92806 
+5

ROF LOL !!!

Melody  Jun 2, 2014
 #5
avatar+87333 
+5

Let us calculate another "probability"........

Question: What is the probability that, of N forum members, one steals another's idea and makes it his/her own ???

 

P.S. ......it's more than 50%....and we don't even need 23 people, either ......!!!

 

CPhill  Jun 2, 2014
 #6
avatar+92806 
+5

It is the quick or the headless Sir CPhill.

I'm quick and you're headless!

Melody  Jun 2, 2014
 #7
avatar+92806 
+5

This cheeky anonymous person who gave themselves 5 points just for asking a question.  

Should shower us both in points for out supreme efforts here!

You can have yours posthumously!

Melody  Jun 2, 2014
 #8
avatar+87333 
+5

Checketh thou point totals on the question under scrutiny, again, Fair Lady  !!!

 

It appearest that MINE hath increased without bound.........while THINE have suffered a most frightful decline....

 

CPhill  Jun 2, 2014
 #9
avatar+92806 
+5
Best Answer

How rude, how rude!

Still, 2 can play at thine game.

Check the points again dear Sir.

And the Lady still hath her head!

Have thou found the Roman zero amounst Zeus's offerings yet?

Thou waste time here annoying the Queen when thou should be attending thine own most urgent tasks.

Be off with you Sir Headless CPhill!

You are but a knat upon Her Royal steed!!

Melody  Jun 2, 2014

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