Wednesday puzzle

[sasguru]

That is not something I here very often!!!



Hmmm - let me look again

2 people out of 3 share a birthday:

I thought it was 3 x (1/365 + 1/365) = 6/365

2 out of 3 share a birthday would be 1/365 + 1/365 as the first person could either be the same as the second or third.

Oooh I think I get it.

The first person could be the same as the second or third - 2/365
The second could be the same as the third 1/365

Hence would it be 1/365 + 2/365 = 3/365?

It is a hard one - took me a while to work it out too, I don't have the answers given, just the question:

here's my thinking:

P(2 have the same birthday) =3 x [ P(any 2 have the same birthday) X P(1 doesn't)]

= 3 x (1/365 x 364/365).

You were very close, just forgot to take into account that 1 doesn't share a birthday

[eek]

1965(!) How fookin old are you!!!!!!!!

1965 entrance paper marks him out as born 1947-8.

Which makes him 65 or 66 years old.

Last years retirement wasn't fake or early he had to retire because Sas has hit retirement age.

It does explain the old man "sports" car.

[sasguru]

That's embarrassing (for you)

At 11:03 you thought the answer was wrong
At 11:10 you realised the answer was correct.

Show some consistency Muppet and actually read and think before trying to look clever.

I don't have the answers and have to work it out for myself, dumbkopf.
I notice you couldn't have a go, which is what I expext from you.

[sasguru]

1965 entrance paper marks him out as born 1947-8.

Which makes him 65 or 66 years old.

Last years retirement wasn't fake or early he had to retire because Sas has hit retirement age.

It does explain the old man "sports" car.

Try reading the rest of the thread. If you can read.

[eek]

I don't have the answers and have to work it out for myself, dumbkopf.
I notice you couldn't have a go, which is what I expext from you.

Had work to do so I'm only back here for a few minutes.

I know you've all screwed up as the figures should all be based on 365.25 (to cover leap years).

1 is 364.25/365.25*363.25/365.25 (1 day has to be excluded for person 2, 2 days for person 3)
2 odds for 2 people with the same birthday is 1/365.25
odds for the other person to not have the same birthday is 364.25/365.25
and because there are 3 possible variations
its 3 x (1/365.25*364.25/365.25)
3 is 1/365.25*1/365.25

[sasguru]

Probably wrong...

Correct.

[Robinho]

Probably wrong...

All 3) 1/365^2
2) (364+364)/365^2
None) 364*363/365^2

I'm amending this...

2) (364+364)/365*364
None) 364*363/365*364

Probably still wrong.

[sasguru]

Had work to do so I'm only back here for a few minutes.

I know you've all screwed up as the figures should all be based on 365.25 (to cover leap years).

1 is 364.25/365.25*363.25/365.25
2 is 2 x (1/365.25*364.25/365.25)
3 is 1/365.25*1/365.25

So your contribution is to take someone else's probabilistic thinking and pedantically add 0.25 to the answer.?
I can tell you were sort of middling at school.
As it happens you're completely wrong.

[sasguru]

End of term report:

Top of the class: MyUserName

OK attempt: Bunk, Robby

Destined for a life of drudgery in boring low-level IT: DP, Doogie, eek.

Sayonara: got some models to look at now.

[eek]

So your contribution is to take someone else's probabilistic thinking and pedantically add 0.25 to the answer.?
I can tell you were sort of middling at school.

I will admit to making an assumption.

odds of being born on Feb 29th (if age unknown) is 1 in 365.25*

This of course is only the case if the entire population was born after 1900 and before 2100 (when there won't be a leap day). Its a valid assumption now but wasn't 10 years ago.