the birthday problem

[new fish]

hi guys
some of you will be familiar with the birthday problem

"how many ppl do you need in a room before the probability of two sharing a birthday becomes at least 50%"

the answer to this is 23 people

you can reason it by considering the prob that all have different birthdays:
consider a group of 3 ppl
the number of possible combinations of birtdays for the 3 is (366)(366)(366)
now the first person can have a bday on any of 366 days
the second person can have a bday on any of 365 days
and the third person can have a bday on any of 364 days

so the prob that they all have different bdays is
(366*365*364)/(366*366*366)

so for them to share is 1-(the above)

for a group of 23 ppl this yields just over 50%..



NOW
shrink the problem to the chances that in a group of 3 people, 2 were born on the same day of the week, according to the above the probability is
1-(7*6*5/7^3) which is about 38.7%. i have even written a vb simulation to verify all of this and it works out!!
BUT
i cant help thinking
>all 3 choose a day of the week
>person 1 stands up and says his day
the chances that one of the other 2 picked this day are 2/7
>if neither did choose the same day, person 1 leaves the room along with
the day that he choose
>person two stands up and says his day
the chances the third person picked that day are 1/6

so that it seems the chances should be 2/7+1/6 =45.2%
or even at a stretch 2/7 for a first time hit, 5/7 for a fist time miss*1/6 for a second time hit
2/7+(5/7*1/6)

i know this is WRONG but i am trying to think of it i terms of "what are the chances" not "what are the chances of not"

i just thought from my frustration that i'd share it!!!
and see if anyone has way of explaining it through positive outcomes!!

[Logophobic]

i cant help thinking
>all 3 choose a day of the week
>person 1 stands up and says his day
the chances that one of the other 2 picked this day are 2/7

This is incorrect. The probability that neither of the other two picked the same day as the first is 6/7 * 6/7, which is 36/49. As such, the probability that at least one of them picked the same day is 13/49. I'm guessing you calculated 1/7 + 1/7 instead of 1/7 + 6/7 * 1/7 for this.

The approach you are looking for is:

(1) Probability that person A picked same day as person B is 1/7 = 7/49. In this case, person C's pick is not relevant.
(2) If A and B picked different days (6/7), then the probability that C picked one of those is 2/7. Probability of this scenario is 6/7 * 2/7 = 12/49.
(3) Sum of (1) and (2) is 19/49, or about 38.8%

[new fish]

hello logophobic
yup 19/49 is exactly the right answer
i like your (1)(2)(3) reasoning




but lets say john is person A and he chooses monday!:

i play the role of person B and C and i'm asked to choose two random days:

there are 49 combinations i can choose, monday appears in 13 of them.

the chances i pick a combination with monday is 13/49?

the same can be said of tuesday, it appears 13 times
as does every day.
so no matter what day john picks it seems my chances should be 13/49 and not 19/49, even though i KNOW the prob is 19/49.

i just cant see the illegal move in my reasoning, which is very worrying!!


i supoose i'm wondering if the quesiton was reworded as:
pick two random days,
what are the chances that one of the days is saturday?
would 13/49 be the answer
or:
all combinations of two days are laid on a floor and you throw a stone which lands on one combination> what are the chances this combination contains friday?

i'll have a eureka moment one of these days

[Logophobic]

OK, John chose Monday. You randomly choose two days, possibly the same day twice. As you stated, at least one of your days will be Monday 13 times out of 49. This is the probability that either B or C, or both, is the same as A. The 19/49 result includes the 6 possible ways B and C can be the same, but different from A.

[new fish]

ahhh of course!! i have it now!! how the hell did i miss that!!
thanks logo!
although i know and of course accept the method whereby one first calculates the probability of all being different and then does 1-p, i prefer to do it the other way around!

[Code Doc]

The matching birthday problem is an application of the hypergeometric distribution--sampling without replacement. Take a look at this link:
http://en.wikipedia.org/wiki/Hyperge...c_distribution

If I recall, the odds reach 50:50 when about 30 people are in a room. However, birthdays are not uniformly distributed throughout the year. There are lots of June weddings.

[new fish]

assuming birthdays are uniformly distributed you only need 23 people, i've even written a little vb prog to simulate the problem!

also if you have a group of 12 people ask them all to choose a number between 1 and 100, 50% of the time 2 will choose the same!

[metalmidget]

Actually, it's probably a lot higher than 50%, because the chances of people picking 7 or 56 are pretty good. There's no mathematical basis for that, it's just what people tend to do.
People are crappy random number generators

[new fish]

lol 7 and 56!
i always go for 24