Brain Teaser...................

[maverick2112]

You're in a room with 40 people. A friend who does not know any people in the room bets you that at least 2 of them share a birthday(same month and day). Whats your friends chances of winning the bet?

What if the room only has 22?

 

[saint]

The key to this question is that not only will you be asking the 39 other people about their birthdays, but they will all be asking each other. So...the probability is much higher than people would think. To figure it out specifically:

For 40 people:

364/365 x 363/365 x362/365 ? 365-40+1/365

That gives you the chances of no duplicate bdays. To determine the probability of the number of matches, it's 1 minus the above number.


For 22 people:

364/365 x 363/365 x362/365 ? 365-22+1/365

Same deal, subtract from 1.



I'm too lazy to figure out the numbers

 

[taoist]

...read something similar in Amarillo Slim's book. Said something like at 22 people the odds were 50/50, but at 30 people, your chance of 2 of the 30 random people having the same birthday was 70%. :shrug:


...where's Kotysdad? I was wondering how the math worked and just never took the time to figure it out. :shrug:

 

[Waldo]

Taoist, I think you're right. Don't remember how to prove it though. I think it is somewhere around 95% for 40 people.

I remember looking at baseball rosters to see how many had guys with the same birthday. I think well more than half of the 25 man rosters, had them.

 

[maverick2112]

Saint..........let me re-phrase the question......I keep getting a different answer.............

It is you first day of teaching and your class has 40 students. A friend who does not know any children in your class bets you that at least two of them share a birthday (month and day). What are the friend's chances of winning the bet? What if the class only has 22 students?

 

[saint]

The formula I posted above does work. I did some of the math:

For 20 people, there is a around a 41.1% chance

For 22 people, there is around a 47.6% chance

For 40 people, there is around an 89.1% chance.


To get around 50% chance to 2 kids w/ matching bdays, you need 23 kids. To get over 90%, you need 41 kids.

 

[taoist]

...those percentages look right. Thanks, Saint.

 

[saint]

For people who know statistics and want it in permutation form, here it is:


365! / ((365-n)! * 365^n).

 

[KotysDad]

Saint has the right numbers up there. It's the classic "Birthday Problem" from probability courses. In a room with 100 people, if you wanted to bet no two people had the same birthday, you would get approx 1 million to 1 odds. Pretty astounding, and very counter-intuitive IMO.


Saint, here's one for ya. Assume you had a room full of people and you wanted approximately a 50/50 chance of at least 3 people sharing the same birthday. As you shown above, with 2 people sharing, you need 23 people in the room. How many do you need with 3 people sharing the same birthday and 50/50 odds?

 

[maverick2112]

40 people = 365-40+1/365.........326/365= 89.3%

22 people= 365-22+1/365........344/365= 94.0%

Why is me %%% going up and now down like it should with that formula????

 

[saint]

Sorry I didn't explain it better. This is how you would do it for say 4 people:

(364/365)x(363/365)x(362/365)= some number, subtract from 1 to get probability.

For 5 people:

(364/365)x(363/365)x(362/365)x(361/365)=x, then do 1-x

for 6 people:

(364/365)x(363/365)x(362/365)x(361/365)x(360/365)=x, 1-x

So do figure out how far you have to compute it for a certain number of students, you use this formula:

364/365 x 363/365 x362/365 ? 365-n+1/365

So for 5 kids, you go down to 365-n+1, or 365-5+1, so 361/365...which u can see i went to above.

So for 40 kids, you have to keep going down in the fractions all the way to 326/365.

 

[maverick2112]

So for 5 kids......... 361/365= .989

So the probability would be 1-.989=.01 for 5 kids

Is this correct??????

 

[saint]

No, for 5 kids you have to do this whole string of calculations:

(364/365)*(363/365)*(362/365)*(361/365)=x

* is multiplication, / is divide

so for the above, (364/365)x(363/365)x(362/365)x(361/365)=.973, so 1-.973=.273, so there is a 2.73% chance that 5 kids in the room will have 1 pair of overlapping bdays

 

[TORONTO-VIGILANTE]

Sorry I didn't explain it better. This is how you would do it for say 4 people:

(364/365)x(363/365)x(362/365)= some number, subtract from 1 to get probability.

For 5 people:

(364/365)x(363/365)x(362/365)x(361/365)=x, then do 1-x

for 6 people:

(364/365)x(363/365)x(362/365)x(361/365)x(360/365)=x, 1-x

So do figure out how far you have to compute it for a certain number of students, you use this formula:

364/365 x 363/365 x362/365 ? 365-n+1/365

So for 5 kids, you go down to 365-n+1, or 365-5+1, so 361/365...which u can see i went to above.

So for 40 kids, you have to keep going down in the fractions all the way to 326/365.

my eyes are bleeding. :com: :hail