Register | Sign In


Understanding through Discussion


EvC Forum active members: 64 (9164 total)
0 online now:
Newest Member: ChatGPT
Post Volume: Total: 916,902 Year: 4,159/9,624 Month: 1,030/974 Week: 357/286 Day: 0/13 Hour: 0/0


Thread  Details

Email This Thread
Newer Topic | Older Topic
  
Author Topic:   The limitations of common sense
cavediver
Member (Idle past 3673 days)
Posts: 4129
From: UK
Joined: 06-16-2005


Message 8 of 66 (302699)
04-09-2006 7:00 PM
Reply to: Message 6 by Percy
04-09-2006 5:09 PM


Birthdays
How many people do you need at a party before the probability of there being a shared birthday reaches 50%?
Common sense answers only

This message is a reply to:
 Message 6 by Percy, posted 04-09-2006 5:09 PM Percy has not replied

Replies to this message:
 Message 10 by RAZD, posted 04-09-2006 8:00 PM cavediver has not replied
 Message 11 by Omnivorous, posted 04-09-2006 8:30 PM cavediver has replied

  
cavediver
Member (Idle past 3673 days)
Posts: 4129
From: UK
Joined: 06-16-2005


Message 17 of 66 (302811)
04-10-2006 3:12 AM
Reply to: Message 13 by RAZD
04-09-2006 9:02 PM


Re: Birthdays
Nice one 50% is exceeded at 23, so that is the correct "answer" though as you say, real world clustering of birthdays, and the odd pair of twins are going to lower this very slightly.
The important point is this is about the size of one of my maths sets. It's a good bet to play with them that two of the students share a birthday...
I would love to see the reasoning that led to 5 though

This message is a reply to:
 Message 13 by RAZD, posted 04-09-2006 9:02 PM RAZD has replied

Replies to this message:
 Message 21 by RAZD, posted 04-10-2006 7:12 AM cavediver has replied

  
cavediver
Member (Idle past 3673 days)
Posts: 4129
From: UK
Joined: 06-16-2005


Message 18 of 66 (302815)
04-10-2006 4:39 AM
Reply to: Message 11 by Omnivorous
04-09-2006 8:30 PM


Re: Birthdays
I love your thinking Omni... I always told my students that poker taught a great stats lesson
For those wanting to know how to calculate, these problems are nearly always much more tractable by reversing them. One shared birthday solves this, but so does two shared birthdays, or three peopel sharing one birthday. There are many many permutations and it is very difficult trying to account for them all. Far easier is to look at there being no shared birthdays:
First person can have any birthday: 365 possibilities
Second person can have any birthday but the first: 364 possibilities
Third person has 363 possible days
Fourth has 362 days to choose from
etc, etc
So probability that 5 people do not share a birthday is:
365   364   363   362   361
--- x --- x --- x --- x --- = .973
365   365   365   365   365
and so prob that there is at least one shared birthday is 1 - 0.973 = 0.027 Fairly low
So just keep adding more people and more terms to the above product and wait for the answer to drop below 50%, which happens with 23 terms, and so 23 people is the answer...
This message has been edited by cavediver, 04-10-2006 04:52 AM

This message is a reply to:
 Message 11 by Omnivorous, posted 04-09-2006 8:30 PM Omnivorous has replied

Replies to this message:
 Message 25 by Omnivorous, posted 04-10-2006 9:33 AM cavediver has not replied
 Message 39 by RAZD, posted 04-11-2006 12:50 AM cavediver has replied

  
cavediver
Member (Idle past 3673 days)
Posts: 4129
From: UK
Joined: 06-16-2005


Message 20 of 66 (302823)
04-10-2006 7:03 AM
Reply to: Message 19 by U can call me Cookie
04-10-2006 5:47 AM


Re: A question...
"perfectly stationary" only can really mean with respect to our local comoving frame in the universe, so you are going to find yourself about 15 million miles from the earth... possibly well on your way to Mars

This message is a reply to:
 Message 19 by U can call me Cookie, posted 04-10-2006 5:47 AM U can call me Cookie has replied

Replies to this message:
 Message 24 by U can call me Cookie, posted 04-10-2006 8:29 AM cavediver has not replied

  
cavediver
Member (Idle past 3673 days)
Posts: 4129
From: UK
Joined: 06-16-2005


Message 23 of 66 (302829)
04-10-2006 7:48 AM
Reply to: Message 21 by RAZD
04-10-2006 7:12 AM


Re: Birthdays
I think there's an entire topic here, in what leads to "common sense" conclusions: false memory, false reasoning, and may own favourite and short-coming: false assumptions...

This message is a reply to:
 Message 21 by RAZD, posted 04-10-2006 7:12 AM RAZD has replied

Replies to this message:
 Message 27 by sidelined, posted 04-10-2006 10:30 AM cavediver has not replied
 Message 32 by RAZD, posted 04-10-2006 6:25 PM cavediver has not replied

  
cavediver
Member (Idle past 3673 days)
Posts: 4129
From: UK
Joined: 06-16-2005


Message 41 of 66 (303104)
04-11-2006 5:01 AM
Reply to: Message 39 by RAZD
04-11-2006 12:50 AM


Re: Birthdays take 2 give or take 1
"Common" sense would say 1/3rd the size (~7)
Ok, this is what we can call the error of assumption of linearity. There must be many examples of this in everyday life...
The actual answer is between 14 and 16, depending on how many of the birthdays actually fall within two days of each other (as two birthdays a day apart only restricts 5 days from the calendar, not 6)

This message is a reply to:
 Message 39 by RAZD, posted 04-11-2006 12:50 AM RAZD has replied

Replies to this message:
 Message 44 by sidelined, posted 04-11-2006 12:06 PM cavediver has not replied
 Message 56 by RAZD, posted 04-12-2006 7:51 PM cavediver has not replied

  
Newer Topic | Older Topic
Jump to:


Copyright 2001-2023 by EvC Forum, All Rights Reserved

™ Version 4.2
Innovative software from Qwixotic © 2024