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RACINGPOST1
20 Aug 11 21:32
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Date Joined: 07 Mar 08
| Topic/replies: 69 | Blogger: RACINGPOST1's blog
IS IT TRUE THAT IF THERES 52 PEOPLE INA ROOM 2 HAVE THE SAME BIRTHDAY..THKS
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Report Eddie the eagle August 20, 2011 9:35 PM BST
Not necessarily, but very likely.
Report Betfair-Donk August 20, 2011 9:57 PM BST
no it isnt true

you can have 350 people in a room and not have 2 people sharing the same birthday
Report Banks. August 20, 2011 10:04 PM BST
If you have 23 it is slight odds on that 2 will have the same birthday.

If you have 52 then it is something like 1/33 that 2 people will share a birthday.
Report TheInvestor2 August 20, 2011 10:10 PM BST
More like 1/44 I think
Report TheInvestor2 August 20, 2011 10:13 PM BST
P (no one shares birthday in a group of 23) = 0.493
P (no one shares birthday in a group of 52) = 0.022
Report TheInvestor2 August 20, 2011 10:14 PM BST
Taking it as a given that no one is born on 29 Feb.
Report Banks. August 20, 2011 10:24 PM BST
More like 1/44 I think

I only quote in traditional prices!
Report catfloppo August 20, 2011 10:36 PM BST
What if the room is hosting an annual get together of local sagittarians?
Report bf_fananatic August 20, 2011 10:54 PM BST
If the 52 are babies in a maternity hospital they may all share the same birthday, what a ridiculous
questionConfused
Report weatherman2004 August 20, 2011 11:51 PM BST
It's a good question to anyone interested in probability

But not of much interest to those unable to work out the answer.
Report Coachbuster August 20, 2011 11:54 PM BST
he meant 365 babies
Report bf_fananatic August 21, 2011 12:06 AM BST
lol, It wouldn't be difficult to work out the probability but it would be pointless
A more useful statistic is that persons using bookmakers are born one every minuteCool
Report weatherman2004 August 21, 2011 12:29 AM BST
You do know how to work it though, don't you?

Just post how. Give you 5 minutes, so you don't go googling.
Report bf_fananatic August 21, 2011 12:41 AM BST
average chance of being born on any date
=52/365=0.142465753

average chance of 2 being born on same day is
(52/365) *(51/365) = 0.019906174 or 1 in 50.23567
Report bf_fananatic August 21, 2011 12:45 AM BST
Your question contains a grammar mistake it contains a statement with past/present tense and future tense

In same line "you do not know" is future if questioned or present if declared and "dont you" is past tense.
Report bf_fananatic August 21, 2011 12:47 AM BST
Its ok though, its accepted and widely used bad English!
Report weatherman2004 August 21, 2011 12:48 AM BST
there they're their

but took a little long

let me think of some others
Report bf_fananatic August 21, 2011 12:59 AM BST
To be honest, I didn't get any english education beyond 11 years of age and I taught myself
everything I know about science, computers, physics, statistics and not when the Internet was available
Even einstien didn't start talking a word till he was 4 and isaac newton was a failure as a child.

I Have been told about the difference between where were there there they're their but I cant
really be bothered with English as it doesn't pay me any reward, perhaps if I was a writer
then I would clean up my english!
Report TheInvestor2 August 21, 2011 1:03 AM BST
weatherman, bff's response is incorrect anyway. Do you know how to work it out yourself? Laugh

Or am I missing something Confused
Report bf_fananatic August 21, 2011 1:09 AM BST
Well of course i gave the average chance of any date having 2 birthdays linked so to arrive
at the chance per year you times by 365

0.019906174 * 365 = 7.265753425 days will have 2 birthdays on average
Report TheInvestor2 August 21, 2011 1:13 AM BST
I don't really understand the question you are answering.
P (no one shares birthday in a group of 52) = 0.022 (rounded)

Are you answering a different question?
Report bf_fananatic August 21, 2011 1:14 AM BST
Perhaps I am wrong investor can you enlighten me as weatherman has vanished!
Report bf_fananatic August 21, 2011 1:16 AM BST
What is the question lol
Report weatherman2004 August 21, 2011 1:18 AM BST
Weatherman bets on things

I can only post when Zvon and Pova change ends

Investor - I didn't understand his answer. I assumed he'd used an approximation and wouldn't be dumb enough to post the wrong answer on the forum
Report bf_fananatic August 21, 2011 1:21 AM BST
Ok I see you are busy, what is the question to this thread, its open to interpretation ?
Report TheInvestor2 August 21, 2011 1:37 AM BST
It's quite fast to calculate in excel, but requires a lot of input.
The way I did this is by assigning a number to each person in the group.
Then I work out the chance of 1 and 2 not sharing a birthday.
Then the probability of 3 not sharing a birthday with 1 or with 2.
Then the probability of 4 not sharing a birthday with 1 or with 2 or with 3 and so on.

Finally I take the product of these which gives the probability that all these statements are true and solves the problem.


I did it like this:

in cell a1 put the value 365.
in a2 put =a1-1
drag this down until you fill a number of cells equal to size of group minus 1. It's minus 1 because person 1 does not share a birthday with anyone with a lower number.
in cell b1 put =1-1/a1
in cell c1 put =PRODUCT($B$1:B1)

Drag b1 and c1 down to complete the table. For 52 people this should go from A1 to C51. The answer =PRODUCT(C1:C51).
Report weatherman2004 August 21, 2011 3:11 AM BST
I think it's more straightforward to put numbers 1 to 51 in column A, and then 1-A1/365 in column C.

Intuitively I find this easier - I can see that 1/365 is the chance of the second person having the same birthday as the first, so 1-1/365 must be the chance of him not.

But I'm struggling to see why {(1-1/365)*(1-1/365)} * {(1-1/365)*(1-1/364)} * {(1-1/365)*(1-1/363)}..., gives the same result as (1-1/365) * (1-2/365) * (1-3/365)...
Report weatherman2004 August 21, 2011 3:24 AM BST
Sorry, first factor should just be (1-1/365)

So I'm just trying to work out why (1-1/365)*(1-1/364) equals 1-2/365.

I know no-one cares. But it's annoying me now.
Report weatherman2004 August 21, 2011 4:15 AM BST
Ok. (1-1/365)*(1-1/364) = 364/365 * (1 - 1/364) = 364/365 - 1/365 = 363/365.

Hah!
Report waynebrayn August 21, 2011 7:14 AM BST
Have a look here:

.http://www.probabilitytheory.info/content/item/4-converse-complementary-probabilities
Report Eddie the eagle August 21, 2011 9:08 AM BST
TheInvestor2's way of doing it is the correct way.

And bff, you really aren't in a position to have a go at others spelling or grammar.
Report Billposter August 21, 2011 11:09 AM BST
Will it help if we knew the size of the room?
Report bf_fananatic August 21, 2011 1:41 PM BST
How can I be wrong when I break it down like this to make it understandable;

1....The odds of a date in a year being linked to a birthday of a group of people
      is 365 days divided by the group size = 365/52 = 7.019 to 1

2....The odds of a date in a year being linked to a 2 birthdays of a group of people
      is the first argument product(7.019) multiplied by 365/51(1 less as this is in the first
      argument condition)7.156862745 therefore the result is 50.23567119

3....so if each day of the year has a 1 in 50.23567119 of 2 birthdays then divide the number
     of days in a year by the chance to arrive at the
     average amount in the room that will share a birthday = 365/50.23567119 = 7.265753425

This is indeed the same figure I posted before and is based on chance , if you cant explain it properly as a formula with all the paremeters present and conditions then you will not convince anyone you know how to calculate probability Blush
Report bf_fananatic August 21, 2011 1:47 PM BST
I currently use excel in an extended fashion with multiple apps running to look at over 1200
races say at one track over 6 runner over 100 factors based on
over 20 million pieces of form data
and I don't really have time for simple probability questions like this but I am glad to help[;)]
Report Eddie the eagle August 21, 2011 1:55 PM BST
bff, you have just shons us you don't know how to calculate probabilities.............either
Report bf_fananatic August 21, 2011 1:59 PM BST
sorry should of noted that  7.265753425 is the figure of days in a year that will on average coincide with 2 birthdays so in fairness just over 14 persons will share a birthday with another
Report bf_fananatic August 21, 2011 2:01 PM BST
lets stick it on yahoo questions and see what comes back?
Report Eddie the eagle August 21, 2011 2:04 PM BST
You can stick it wherever you want.
  The correct way to calculate it is given earlier on this thread by TheInvestor2.
Report bf_fananatic August 21, 2011 2:27 PM BST
I am quite sure my formula is correct and if I programed it with random into a program that ran it
many thousands of times I know It would give me the same figure eventually

investor is giving some product that gives the answer for not being a birthday match which isn't
even the answer that is wanted?
Report _kawasaki_ August 27, 2011 8:30 PM BST
maybe, if pay attention to only DAYS of month, forget about 12 months Laugh
Report Contrarian August 27, 2011 9:52 PM BST
Proof, if proof were needed that bf_fananatic knows f*ck all about probability.
Report Contrarian August 27, 2011 9:59 PM BST
1....The odds of a date in a year being linked to a birthday of a group of people
      is 365 days divided by the group size = 365/52 = 7.019 to 1



If by 'a date being linked to a birthday of a group of people', you mean 'one of the group having that date as their birthday', then by your logic, if there were 365 people in the room, it would be certain that one of them was born on that date!
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