help with math homework - probabilty

Mehmet · Post by **Mehmet** » Mon Sep 16, 2002 11:41 pm

thanx in advance guys.

people come into a room one at a time right? whenever they come in they call out their birthday.. they keep on doing this until someones birthday matches, then they lock the door... on average how many people will get into the room before this happens?

I need some softa formula or somethign for it.. or some kind of simulation that i can run (with random digits that are 10 digits basically, usually set up like this : .9875462154

Thanx guys.

Immortal · Post by **Immortal** » Mon Sep 16, 2002 11:49 pm

I forgot anything but alll I can remember is that with 30 or more people the chances of two or more person with the same birthday is 80%

Mehmet · Post by **Mehmet** » Mon Sep 16, 2002 11:51 pm

Originally posted by Immortal
I forgot anything but alll I can remember is that with 30 or more people the chances of two or more person with the same birthday is 80%

why?

Immortal · Post by **Immortal** » Mon Sep 16, 2002 11:52 pm

as I've said.. I don't remember why... or how it's done.. man those were some confusing times..

Mehmet · Post by **Mehmet** » Mon Sep 16, 2002 11:55 pm

Originally posted by Immortal
as I've said.. I don't remember why... or how it's done.. man those were some confusing times..

exactly..

I see no point in doing this problem.

Matt615 · Post by **Matt615** » Tue Sep 17, 2002 12:00 am

its late and I am probrably about to make absolutely no sense at all but here goes.

Every time someone walks into the room their chances of having the same birthday as someone else gets greater. The first person has no chance, but the second person that walks in has a 1/31th of a chance of having the same birthday as the other person. When the third person walks in its 2/31ths of a chance, and on an on. I would think that once you get to about a 50% chance or 16/31 or something like that, the door would lock. So if it locked at 50% exactly than it would mean...including the person who just walked in there would be 17 people in the room.

Im sure I have made no sense but it was worth a try. Btw I am half asleep while trying to do this.

Mehmet · Post by **Mehmet** » Tue Sep 17, 2002 12:04 am

Originally posted by Matt615
its late and I am probrably about to make absolutely no sense at all but here goes.

Every time someone walks into the room their chances of having the same birthday as someone else gets greater. The first person has no chance, but the second person that walks in has a 1/31th of a chance of having the same birthday as the other person. When the third person walks in its 2/31ths of a chance, and on an on. I would think that once you get to about a 50% chance or 16/31 or something like that, the door would lock. So if it locked at 50% exactly than it would mean...including the person who just walked in there would be 17 people in the room.

Im sure I have made no sense but it was worth a try. Btw I am half asleep while trying to do this.

yah! thisll work but.. why 31??

Its for the full year btw, the person coming in can have their birthday any day of the year..

although youve got it right..

Matt615 · Post by **Matt615** » Tue Sep 17, 2002 12:23 am

Are they shouting out a day of the month....for example 22 as in the 22nd of whatever month or are they shouting out a number of the year.....281...281st day of the year.

I choose 31 because that it the maximum amount of days in a month. Even though same may have less, you have to use the maximum.

Did I misunderstand how they are shouting out the numbers?

Mehmet · Post by **Mehmet** » Tue Sep 17, 2002 12:35 am

Originally posted by Matt615
Are they shouting out a day of the month....for example 22 as in the 22nd of whatever month or are they shouting out a number of the year.....281...281st day of the year.

I choose 31 because that it the maximum amount of days in a month. Even though same may have less, you have to use the maximum.

Did I misunderstand how they are shouting out the numbers?

no, just their birthday, like "january 30th" and if someone else has the EXACT same birthday.. then the person who last came in sits down, and they lock the door.

Paft · Post by **Paft** » Tue Sep 17, 2002 12:42 am

It would make sense that if you have the target day for one person, then the next who comes in will have a..

1/365 change of having the same birthday (365 days in the year). The next will have a 2/365 (Chance to match one of two), and then 3/365..

I want to say that it increases exponentially, but then again I've never offically taken probability so take what I told you with a grain of salt. Hope it helps point you to an idea though.

Mehmet · Post by **Mehmet** » Tue Sep 17, 2002 12:43 am

Originally posted by Paft
It would make sense that if you have the target day for one person, then the next who comes in will have a..

1/365 change of having the same birthday (365 days in the year). The next will have a 2/365 (Chance to match one of two), and then 3/365..

I want to say that it increases exponentially, but then again I've never offically taken probability so take what I told you with a grain of salt. Hope it helps point you to an idea though.

yah.. im gonna do this man.. best way to go, and seems reasonable.

thanx.. still up for opinions though.

Matt615 · Post by **Matt615** » Tue Sep 17, 2002 12:45 am

than as paft said the second person to walk into the room would have a 1/365 chance of having the same birthday and that would go on for a while.

I am sticking with my original answer that once you get to around a 50% chance than however many people are in the room at the time I think is the average that would get into the room.

I hope we have helped you, but its time for me to go to sleep...Good luck figuring it out

Matt

Chrome · Post by **Chrome** » Tue Sep 17, 2002 2:31 am

I had a high school math teacher who kept track of this informally in all of the classes he taught. Approximately 90% of his classes had two people with the same birthday, throughout his then 15 year career. This, however, had more to do with statistics than probability I believe.