Probability debate / experiment

[PaddyD]

"I think erik is on to something"

Indeed:

*** SPOILER ALERT ***

http://en.wikipedia.org/wiki/Birthday_paradox

 

[NateO]

Hello,

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?

If my sister I are are in the room, the probability is 100%!

I'm finding 50% a little hard to believe... I've been in plenty of rooms of 25 people or more and I've only met three other people, in my entire life, with my birthday and one of them is my twin (not so random).

Granted we don't always compare notes on birth-dates... But, that number sounds way too high to me...

Each time your birthday does not match another person's birthday, your chance of matching the next person's birthday increases. Right?

Not sure why... These should be independent random variables, no?

http://en.wikipedia.org/wiki/Bernoulli_process

Independence of Bernoulli trials implies memorylessness property: past trials do not provide any information regarding future outcomes. From any given time, future trials is also a Bernoulli process independent of the past (fresh-start property).
{snip}

http://en.wikipedia.org/wiki/Statistical_independence

 

[Andrew Fergus]

Hi Nate

A fair point but consider the question (i.e. any two people), such that if the other 24 people in the room do not share your birthday, then person number 2 may in fact have the same birthday as someone else in the room. But for one individual / one birthday you are correct - the probability is very low.

The methodology outlined by Greg does result in absurdly high probabilities (as Greg guessed) - consider a room of 365 people. Under Greg's scenario the probability is 364/365 + 363/365 etc which yields a number > 100% so intuitively it must be wrong, given we know there is at least one combination of birthdays (i.e. 365 unique) in which we can't find 2 matches so the probability must be < 100% (is the probability of 365 unique birthdays 1/Factorial(365)?)

But I think Erik hit on the answer which is to turn the question around and determine the cumulative probability that you don't share a birthday, and Paddy's link outlines the proof very nicely.

Thanks for the input! I think what this also shows is that, in a significant number of cases based on the votes so far, human intuition is not a good match for probabilities. Take for example lotteries - I have been asked on a number of occasions why there is often 2 consecutive numbers in the 6 or 7 balls drawn from a population of 40 - I suspect the proof will be the same as the 'birthday paradox'.

Cheers
Andrew

P.S. If my brother and I are in the same room then we too have a 100% probability!

I'm finding 50% a little hard to believe... I've been in plenty of rooms of 25 people or more and I've only met three other people, in my entire life, with my birthday and one of them is my twin (not so random).

Granted we don't always compare notes on birth-dates... But, that number sounds way too high to me...

 

[PaddyD]

"... human intuition is not a good match for probabilities"

Not simply not a good match, but notoriously poor.

Here's another:

You're in a quiz. There are 3 doors. Quiz master says there's a prize behind one of the doors & invites you to choose one. You do. Quiz master then, for free, opens a door (other than the one you have chosen) & shows you that it wasn't the door with the prize. Quiz master then offers you the option of changing the door you've picked before they reveal whether you've got the prize-winning door.

Q: Should you change your pick? Why?

 

[PA HS Teacher]

Book2
	A	B	C	D	E	F	G	H	I	J	K	L	M	N	O	P	Q	R	S	T	U	V	W	X	Y	Z
1	427	No Matches
2	1000	Total Simulations
3	57.30%	% of the time there was a birthday Match in the room.
4
5	24	185	251	357	255	283	70	161	277	140	243	136	241	155	101	182	351	42	58	166	73	102	283	319	186	96
6	24	185	277	167	359	263	246	353	102	260	265	16	48	18	144	100	43	296	191	102	104	297	193	94	194	166
Sheet1

 

[Oaktree]

Fess up, Nate... your birthday is Feb 29th, isn't it?

As to Paddy's puzzler, the statisticians say you should switch, since the probabity of your door winning is 1/3 (the only way you win is if your original choice was right) and the probability of the other winning is 2/3 (the only way you lose is if your original door was the winner). Of course, they're the same ones who say the expected value of playing the lottery is something other than ($1.00).

Personally, if 2 was my lucky number, I'd pick door number 2 and stick to my guns.

 

[Andrew Fergus]

Given intuition is no match for stats I suspect my guess at Paddy's question will be flawed in some way, but here goes : given you now know 1 of the doors that doesn't contain the prize, are you not back to a 50:50 chance with whichever door you choose? So IMO there is no expected benefit of changing your pick, or not changing your pick.

Although I think I can see where the statisticians may be coming from : you have a 2/3 chance of getting it wrong but the host had a 1/2 chance of getting it wrong (assuming you also got it wrong) so given he too got it wrong (albeit with better odds), it stands to (*illogical*) reason the most likely door is the last?!?!? It sounds illogical in a logical kind of way.....

Andrew

P.S. I think I won the birthday debate - although I'm still waiting for a concession of defeat.......

 

[PaddyD]

If the quiz master knew what was behind the doors, always switch.

Full story:

http://en.wikipedia.org/wiki/Monty_Hall_problem

 

[RichardS]

If my sister I are are in the room, the probability is 100%!

Is this birth date, or birthday? If my brother and I are in the same room, it's 100% birthday, but not date. 2 years apart. What are the odds of brothers being born on the same date in different years.

Richard

 

[Lewiy]

I'm sure this is simple:

You would have to "assume" that 1 in 365 people share the same birthday as you. Therefore in a room with 24 other people, there is a 24/365 chance that one of then shares your birthday. This equates to 0.06575.