# Probability debate / experiment



## Andrew Fergus (Feb 26, 2007)

I'm having a debate with a friend about the probablity of something and would appreciate your opinion on this subject.  I also want to get peoples guesses as to where they think the answer will be - so take a punt and have a guess before you look for / work out the answer.

If there were 25 people in a room, what is the probability that at least any 2 people in that room have the same birthday anniversary?

Approximations are ok - in other words I don't require this to be expressed as a percentage to 10 decimal places - at this point in time my friend and I are leagues apart with our 'guesses' so approximations will be ok.

Any thoughts and opinions and are welcome, proofs are optional but I intend to rely on a proof to win the argument.

What is your initial guess?  If you want to stay anonymous I have provided a poll with a range of options.  I purposely haven't looked on Google (and don't expect you to) because I want to get a feel for what people think based on intuition - please take a guess before you seek the answer elsewhere!

Cheers, Andrew


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## milesUK (Feb 26, 2007)

My first guess is (25/365)x(24/365). So less than 10%.


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## erik.van.geit (Feb 26, 2007)

Hi,

wouldn't it be easier to ask
what's the chance we have NOT the same birthday ?

2 persons: I assume 364/365
3 persons: the above x 363/365
continue calculation

the SAME birthday would be complementary
2 persons = 1 - 364/365
3 persons = 1 - (364/365 * 363/365)

I think this would be a start if no error made

will check it out using Excel   

kind regards,
Erik

EDIT: Andrew, I'll PM you a l"table-it" with my results


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## Oaktree (Feb 26, 2007)

My guess was somewhere around 60%, given previous knowledge of the significance of 23 people to this puzzle.


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## milesUK (Feb 26, 2007)

Given 2 people in a room then there is a 1:365 chance that one has the same anniversary as the other. With 25 in the room my second guess (after discussion with a collegue) is 24:365 or 6.6% chance that one of those 24 will celebrate on the same date as the 25th.

I have not voted a second time! Put us out of misery who is right?


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## erik.van.geit (Feb 26, 2007)

> Put us out of misery who is right?


it's a bit counter-intuitive (if that word exists) but following the approach I posted above would lead you to a certain result

what's the chance you get 2 same dice-results ?
what's the chance you get 2 different results ?
both together are equal to 100%

2 different: 5/6
2 same 1/6 = 17%

3 dices
different 5/6*4/6
same 1-5/6*4/6 = 44%

4 dices
different 5/6*4/6*3/6
same 1-5/6*4/6*3/6 = 72%

5 = 95%

as you can see, the multiplication of 5/6*4/6*etcetera makes the chance of a difference "increasingly smaller" (hmm, my english ?) resulting in a rapid increase of the SAME-chance

364/365*363/365*362/365*... you get the picture ?

so the result is
1 - number in brown = little bit more than 50%

o, o, I hope this is clear


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## Greg Truby (Feb 26, 2007)

If I'm #1, then I have 24 chances to match somebody else: 24/365 = .065
If I'm #2, then I have 23 chances to match everybody else except #1: 23/365 = .063
...
If I'm #24, then I have 1 chance to match person #25: 1/365 = .0027

The odds that #1 matched against #2-25 = .065
The odds that #2 matched against #3-25 = .063
The odds that #1 OR #2 matched against somebody = .0128 ? Correct?

So, add 'em all up for around 0.82 probability?

<sup>edit</sup> This must be fundamentally flawed though... because soon you get > 1.00 probability and that can't be right... <sub>/edit</sub>


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## ExcelChampion (Feb 26, 2007)

> as you can see, the multiplication of 5/6*4/6*etcetera makes the chance of a difference "increasingly smaller" (hmm, my english ?) resulting in a rapid increase of the SAME-chance
> 
> 364/365*363/365*362/365*... you get the picture ?
> 
> ...



I think erik is on to something.  Each time your birthday does not match another person's birthday, your chance of matching the next person's birthday increases.  Right?


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## Greg Truby (Feb 26, 2007)

Yeah, I too think Erik's on the right trail.  With that approach you get the exponential curve you'd expect.


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## erik.van.geit (Feb 26, 2007)

*A*   *B*       *C*      
 *1*     DIFF    SAME   
 *2* 365 100,00% 0,00%  
 *3* 364 99,73%  0,27%  
 *4* 363 99,18%  0,82%  
 *5* 362 98,36%  1,64%  
 *6* 361 97,29%  2,71%  
 *7* 360 95,95%  4,05%  
 *8* 359 94,38%  5,62%  
 *9* 358 92,57%  7,43%  
*10* 357 90,54%  9,46%  
*11* 356 88,31%  11,69% 
*12* 355 85,89%  14,11% 
*13* 354 83,30%  16,70% 
*14* 353 80,56%  19,44% 
*15* 352 77,69%  22,31% 
*16* 351 74,71%  25,29% 
*17* 350 71,64%  28,36% 
*18* 349 68,50%  31,50% 
*19* 348 65,31%  34,69% 
*20* 347 62,09%  37,91% 
*21* 346 58,86%  41,14% 
*22* 345 55,63%  44,37% 
*23* 344 52,43%  47,57% 
*24* 343 49,27%  50,73% 
*25* 342 46,17%  53,83% 
*26* 341 43,13%  56,87% 

test
[Table-It] version 06 by Erik Van Geit

```
RANGE   FORMULA (1st cell)
B2:B26  =PRODUCT($A$2:$A2)/365^(ROW()-1)
C2:C26  =1-B2

[Table-It] version 06 by Erik Van Geit
```


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## PaddyD (Feb 26, 2007)

"I think erik is on to something"

Indeed:

*** SPOILER ALERT ***

http://en.wikipedia.org/wiki/Birthday_paradox


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## NateO (Feb 26, 2007)

Hello,


> If there were 25 people in a room, what is the probability that at least any 2 people in that room have the same birthday anniversary?


If my sister I are are in the room, the probability is 100%!  

I'm finding 50% a little hard to believe... I've been in plenty of rooms of 25 people or more and I've only met three other people, in my entire life, with my birthday and one of them is my twin (not so random).

Granted we don't always compare notes on birth-dates... But, that number sounds way too high to me...



> Each time your birthday does not match another person's birthday, your chance of matching the next person's birthday increases. Right?


Not sure why... These should be independent random variables, no?

http://en.wikipedia.org/wiki/Bernoulli_process



> Independence of Bernoulli trials implies memorylessness property: past trials do not provide any information regarding future outcomes. From any given time, future trials is also a Bernoulli process independent of the past (fresh-start property).
> {snip}



http://en.wikipedia.org/wiki/Statistical_independence


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## Andrew Fergus (Feb 26, 2007)

Hi Nate

A fair point but consider the question (i.e. any two people), such that if the other 24 people in the room do not share your birthday, then person number 2 may in fact have the same birthday as someone else in the room.  But for one individual / one birthday you are correct - the probability is very low.

The methodology outlined by Greg does result in absurdly high probabilities (as Greg guessed) - consider a room of 365 people.  Under Greg's scenario the probability is 364/365 + 363/365 etc which yields a number > 100% so intuitively it must be wrong, given we know there is at least one combination of birthdays (i.e. 365 unique) in which we can't find 2 matches so the probability must be < 100% (is the probability of 365 unique birthdays  1/Factorial(365)?)

But I think Erik hit on the answer which is to turn the question around and determine the cumulative probability that you don't share a birthday, and Paddy's link outlines the proof very nicely.

Thanks for the input!  I think what this also shows is that, in a significant number of cases based on the votes so far, human intuition is not a good match for probabilities.  Take for example lotteries - I have been asked on a number of occasions why there is often 2 consecutive numbers in the 6 or 7 balls drawn from a population of 40 - I suspect the proof will be the same as the 'birthday paradox'.

Cheers
Andrew

P.S.  If my brother and I are in the same room then we too have a 100% probability!



> I'm finding 50% a little hard to believe... I've been in plenty of rooms of 25 people or more and I've only met three other people, in my entire life, with my birthday and one of them is my twin (not so random).
> 
> Granted we don't always compare notes on birth-dates... But, that number sounds way too high to me...


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## PaddyD (Feb 26, 2007)

"... human intuition is not a good match for probabilities"

Not simply not a good match, but notoriously poor.  

Here's another:

You're in a quiz.  There are 3 doors.  Quiz master says there's a prize behind one of the doors & invites you to choose one.  You do.  Quiz master then, for free, opens a door (other than the one you have chosen) & shows you that it wasn't the door with the prize.  Quiz master then offers you the option of changing the door you've picked before they reveal whether you've got the prize-winning door.

Q: Should you change your pick? Why?


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## PA HS Teacher (Feb 26, 2007)

Book2ABCDEFGHIJKLMNOPQRSTUVWXYZ1427No Matches21000Total Simulations357.30%% of the time there was a birthday Match in the room.452418525135725528370161277140243136241155101182351425816673102283319186966241852771673592632463531022602651648181441004329619110210429719394194166Sheet1


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## Oaktree (Feb 26, 2007)

Fess up, Nate... your birthday is Feb 29th, isn't it? 

As to Paddy's puzzler, the statisticians say you should switch, since the probabity of your door winning is 1/3 (the only way you win is if your original choice was right) and the probability of the other winning is 2/3 (the only way you lose is if your original door was the winner).  Of course, they're the same ones who say the expected value of playing the lottery is something other than ($1.00).

Personally, if 2 was my lucky number, I'd pick door number 2 and stick to my guns.


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## Andrew Fergus (Feb 26, 2007)

Given intuition is no match for stats I suspect my guess at Paddy's question will be flawed in some way, but here goes : given you now know 1 of the doors that doesn't contain the prize, are you not back to a 50:50 chance with whichever door you choose?  So IMO there is no expected benefit of changing your pick, or not changing your pick.

Although I think I can see where the statisticians may be coming from : you have a 2/3 chance of getting it wrong but the host had a 1/2 chance of getting it wrong (assuming you also got it wrong) so given he too got it wrong (albeit with better odds), it stands to (*illogical*) reason the most likely door is the last?!?!?  It sounds illogical in a logical kind of way.....

Andrew

P.S.  I think I won the birthday debate - although I'm still waiting for a concession of defeat.......


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## PaddyD (Feb 26, 2007)

If the quiz master knew what was behind the doors, always switch.  

Full story:

http://en.wikipedia.org/wiki/Monty_Hall_problem


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## RichardS (Feb 27, 2007)

> If my sister I are are in the room, the probability is 100%!



Is this birth date, or birthday? If my brother and I are in the same room, it's 100% birthday, but not date. 2 years apart. What are the odds of brothers being born on the same date in different years.

Richard


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## Lewiy (Feb 27, 2007)

I'm sure this is simple:

You would have to "assume" that 1 in 365 people share the same birthday as you.  Therefore in a room with 24 other people, there is a 24/365 chance that one of then shares your birthday.  This equates to 0.06575.


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## mortgageman (Feb 27, 2007)

> Hello,
> 
> 
> > If there were 25 people in a room, what is the probability that at least any 2 people in that room have the same birthday anniversary?
> ...




Nateo - The question is not will at least two people share YOUR birthday - that indeed is a low probability event.  The question was - what is the probability of ANY TWO PEOPLE sharing the same birthday.  That is a horse of a different species.


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## NateO (Feb 27, 2007)

> > If my sister I are are in the room, the probability is 100%!
> 
> 
> 
> Is this birth date, or birthday? If my brother and I are in the same room, it's 100% birthday, but not date. 2 years apart. What are the odds of brothers being born on the same date in different years.


Both in my case... The probability of your case has to be even smaller, Richard?

Although footnote 2 in Paddy's link is an interesting one in terms of birthdays being skewed...

I definitely didn't get the question right, which makes it really hard to get the answer right... So pardon the interruption, and please continue. 

And, no, I'm one of the masses, born in August. While I'm sure there are a lot of people with my birthday, e.g., Bill Clinton, I've only met three others, that I can recall...


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## Richard Schollar (Feb 27, 2007)

> And, no, I'm one of the masses, born in August. While I'm sure there are a lot of people with my birthday, e.g., Bill Clinton, I've only met three others, that I can recall...



Nate's a Taurean??  I've always said he was full of Bull...

  

Just kidding Nate


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## NateO (Feb 27, 2007)

Heh...  



> Nate's a Taurean??


Negative. I've dated a couple, though... Scary stuff! 

http://en.wikipedia.org/wiki/Taurus_(astrology)



> Under the tropical zodiac, Taurus is occupied by the Sun from April 20 to May 20
> {snip}


My birthday makes me a Leo:

http://en.wikipedia.org/wiki/Leo_(astrology)



> Under the tropical zodiac, Leo is occupied by the Sun from July 23 to August 22, and under the sidereal zodiac, it is currently from August 17 to September 16.
> {snip}


Hear me roar!


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## Richard Schollar (Feb 27, 2007)

Bugger - I knew one of my parents was a Taurus (one born in April, other in August) - just got it round the wrong way...  ****, makes me look bloody stupid.   

Still, Kristy's frequently telling us what a Pu55y Cat you are - case proved!  


EDIT: the correct spelling for pu55y gets asterisked out!!!  Same doesn't happen to Doggy though...


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## NateO (Feb 27, 2007)

No worries, Richard, not being an expert in Astrology isn't a real short-coming, in my estimation... 

That whole Leo/Taurus combo is a scary one! Been there, done that, twice! If I ask a woman what her sign is, it's simply to make sure that it's not a Taurus!!


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## Von Pookie (Feb 27, 2007)

> Still, Kristy's frequently telling us what a Pu55y Cat you are - case proved!


::looks around:: I do what, now? 



> EDIT: the correct spelling for pu55y gets asterisked out!!!


Hooray for the filters! 
However, if you had used it as all one word, I believe it would have gone through: pussycat.


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## milesUK (Mar 1, 2007)

Greg wrote:





> This must be fundamentally flawed though... because soon you get > 1.00 probability and that can't be right...


But it can! If you are in a room of more than 365 people then the probability of one person having the same birthday as you becomes better than 1:1. It therefore follows that a point occurs (a group of 28 people) where the likelihood of ANY 2 having the same birthday is better than evens.


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## erik.van.geit (Mar 1, 2007)

> But it can!


from a mathematic point of view probability is "0" to "1"; "impossible" to "sure"
you can never be more sure than 100%


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## mortgageman (Mar 1, 2007)

> Greg wrote:
> 
> 
> 
> ...



I'm no bookie, but you are mixing up odds with certainty.  A 1 (100%) probabiltiy is equivaltent to certainty.  Greg's question was how can you be > 1 probability.  The answer is (and I didn't look at his math) you can't.  I sure, that a mistake was made in the math somewhere.  I suspect as well that either you are guessing about the odds here or you think that better than 1:1 means >1 probability.  If the latter, then clearly you must agree that it is not certain that you and someone must share the same birthday if there are 365 people in the room! If the former, than I would guess that the odds are better than 1:1 with far  less than 365 people.


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## milesUK (Mar 1, 2007)

Mortgageman, Yes your perfectly right. I realised with Eric's post that I was getting mixed up.  

I then demonstrated to myself that probability of any 2 people having the same birthday does indeed approach certainty (1, 100%,..) as the sample number approaches infinity. I guess its exponential.

I bow to the better education / memories of my peers.  :wink:

A very interesting discussion though; any more out there?


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## steve case (Mar 2, 2007)

This is one is older than me.  The probability is very high, so I said 75%.  It's an old parlor trick and's been around for a long time.   

Here's a link:

http://mathforum.org/library/drmath/view/56511.html

Looks like I guessed high )-:


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## steve case (Mar 2, 2007)

I skimmed through the replies, and there seems to be a fundamental error in quite a few responses.  The questions wasn't what are the odds of someone having the the same birthday as yours. but rather what are the odds of two people in the room having the same birthday?  Those are two very different questions.


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## milesUK (Mar 2, 2007)

StACase, That was the very mistake I made; the question asked for the probability NOT the odds.



> If there were 25 people in a room, what is the probability that at least any 2 people in that room have the same birthday anniversary?


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## steve case (Mar 2, 2007)

Hmmmm, I'll have to look up the definition of "odds" and "probability"


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## steve case (Mar 2, 2007)

Main Entry: odds 

(1): the probability that one thing is so or will happen rather than another

Main Entry: prob·a·bil·i·ty

(2): the chance that a given event will occur


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## milesUK (Mar 2, 2007)

There's a very good _chance _that i'm _probably_ _even _more confused now!


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## steve case (Mar 2, 2007)

That's odd (-:


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## Lewiy (Mar 2, 2007)

I think I have it….and used Excel to get there!!

I created a 25 x 25 grid and in each cell entered “=1/365” being the probability that each person will share a birthday with a specific other person. (obviously cells which refer to a person sharing with themselves is left blank).

Summing each row will give you 24/365 which is the probability of one person sharing a birthday with anyone else in the room.

Summing the whole 25 x 25 grid gives you 25(24/365) or 600/365 which needs to be divided by 2 to remove duplicates (i.e. person 2 sharing with person 3 and person 3 sharing with person 2).

So the final answer is:
In a room of 25 people, the probability that any 2 of them share the same birthday is 600/730 (or 60/73), which equates to approximately 0.82192


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## Richard Schollar (Mar 2, 2007)

I must admit I calculate it differently as a 57% chance that at least 2 are born on the same day derived from Erik's approach - I think - that 43% chance of them all being born on different days (different Day/Month combinations to be more precise).


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## mortgageman (Mar 2, 2007)

> I think I have it….and used Excel to get there!!
> 
> I created a 25 x 25 grid and in each cell entered “=1/365” being the probability that each person will share a birthday with a specific other person. (obviously cells which refer to a person sharing with themselves is left blank).
> 
> ...




I didn't try to follow your post, but as I hope you have seen from the links provided in the past few days, your answer of 82% for n=25 is incorrect.


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## erik.van.geit (Mar 2, 2007)

> I think I have it….and used Excel to get there!!
> 
> Summing each row will give you 24/365 which is the probability of one person sharing a birthday with anyone else in the room.
> 
> ...


I cannot agree with this explanation
there are at least two posts in this thread which have links to complete explanations
if you apply your trick to 50 persons, what would be the result ?

EDIT: for some reason missed Gens post: Hi, Gene


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## Lewiy (Mar 2, 2007)

I see the error of my ways.  I think what's happened is I have failed to take into account that each individual themselves has a 1/365 chance of having a birthday on any given day.  Will have to re-think this.


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## Lewiy (Mar 2, 2007)

Ok, taking a different angle now.

Probability equals the number of ways an event can occur divided by the number of possible outcomes.

The number of possible outcomes in this example is given by 365^25 as each person could have a birthday on one of 365 days (if we ignore leap years).

The number of ways that 365 people could all have *different *birthdays is given by 365*364*363*362……etc.  Or 365! (factorial 365).  Thus the number of ways 25 people could all have *different *birthdays is 365!-340!

Consequently the number of ways in which 25 people *do not *have the same birthday is (365^25)-(365!-340!) which is the number of possible outcomes less the combinations where all birthdays are different.

Hence, I put it to you that the probability of at least 2 people in a room of 25 have the same birthday is:
((365^25)-(365!-340!))/(365^25)

I have yet to evaluate this as the numbers are huge.  However, you can test the maths with a simpler example:

You have 3 regular dice (with numbers 1 to 6 on each) the probability of any 2 dice having the same number when rolled (using the above logic) would be:
((6^3)-(6!-3!))/(6^3) which is 0.697478992.


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## EE2006 (Mar 2, 2007)

> Ok, taking a different angle now.
> 
> Probability equals the number of ways an event can occur divided by the number of possible outcomes.
> 
> ...




I don't really remember a whole lot from the statistics courses I've done in the past, but something about your method seems familiar. However, I'm pretty sure there is a problem with the factorial terms you are using. There are a couple of issues that I can see with your response right away.


(1) As Eric already detailed on page 1 of this thread, the probability of any 2 dice having the same number when 3 are rolled is 0.44444.

(2) The equation you have posted doesn't actually work out to 0.697.... It calculates to -2.30556.


If you do figure out the correct formula for your method, be sure to post it. I've been tryinig to figure it out but haven't been able to.


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## Lewiy (Mar 2, 2007)

Yeah you're right, I should be dividing the factorials rather than taking them away:

Corrected formula:

((365^25)-(365!/340!))/(365^25)

And for the dice example:

((6^3)-(6!/3!))/(6^3) which is now 0.4444444


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## Lewiy (Mar 2, 2007)

Additional:

Have now managed to evaluate using the following:

365! / (340!(365^25)) gives the probability that no one has the same birthday.  This equals 0.4313

Therefore 1-0.4313 = 0.5687 so the answer is 56.87%


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## mortgageman (Mar 2, 2007)

> Big Snip of material addressed elsewhere anyway
> 
> EDIT: for some reason missed Gens post: Hi, Gene



Hi Eric - I see that you have turned you face to the right.  Was that a political move?


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## erik.van.geit (Mar 3, 2007)

> > Big Snip of material addressed elsewhere anyway
> >
> > EDIT: for some reason missed Gens post: Hi, Gene
> 
> ...


to stay on topic
the probability to turn it to the left is equal the right: no statistical conclusions can be retreived from this "choice"
also the entire picture explains why I'm looking to the *LEFT *(you missed the correct angle !!   )

best regards,
Erik


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