Probability debate / experiment

[milesUK]

Mortgageman, Yes your perfectly right. I realised with Eric's post that I was getting mixed up.

I then demonstrated to myself that probability of any 2 people having the same birthday does indeed approach certainty (1, 100%,..) as the sample number approaches infinity. I guess its exponential.

I bow to the better education / memories of my peers. :wink:

A very interesting discussion though; any more out there?

 

[steve case]

This is one is older than me. The probability is very high, so I said 75%. It's an old parlor trick and's been around for a long time.

Here's a link:

http://mathforum.org/library/drmath/view/56511.html

Looks like I guessed high )-:

 

[steve case]

I skimmed through the replies, and there seems to be a fundamental error in quite a few responses. The questions wasn't what are the odds of someone having the the same birthday as yours. but rather what are the odds of two people in the room having the same birthday? Those are two very different questions.

 

[milesUK]

StACase, That was the very mistake I made; the question asked for the probability NOT the odds.

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?

 

[steve case]

Hmmmm, I'll have to look up the definition of "odds" and "probability"

 

[steve case]

Main Entry: odds

(1): the probability that one thing is so or will happen rather than another

Main Entry: prob·a·bil·i·ty

(2): the chance that a given event will occur

 

[milesUK]

There's a very good chance that i'm probably even more confused now!

 

[steve case]

That's odd (-:

 

[Lewiy]

I think I have it….and used Excel to get there!!

I created a 25 x 25 grid and in each cell entered “=1/365” being the probability that each person will share a birthday with a specific other person. (obviously cells which refer to a person sharing with themselves is left blank).

Summing each row will give you 24/365 which is the probability of one person sharing a birthday with anyone else in the room.

Summing the whole 25 x 25 grid gives you 25(24/365) or 600/365 which needs to be divided by 2 to remove duplicates (i.e. person 2 sharing with person 3 and person 3 sharing with person 2).

So the final answer is:
In a room of 25 people, the probability that any 2 of them share the same birthday is 600/730 (or 60/73), which equates to approximately 0.82192

 

[Richard Schollar]

I must admit I calculate it differently as a 57% chance that at least 2 are born on the same day derived from Erik's approach - I think - that 43% chance of them all being born on different days (different Day/Month combinations to be more precise).