Probability debate / experiment

[Andrew Fergus]

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

 

[milesUK]

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

 

[erik.van.geit]

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

 

[Oaktree]

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

 

[milesUK]

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?

 

[erik.van.geit]

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

 

[Greg Truby]

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?

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

 

[ExcelChampion]

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

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?

 

[Greg Truby]

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

 

[erik.van.geit]

   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