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        riddles >> easy >> Selecting Leap Years at Random
        
(Message started by: TenaliRaman on Aug 22^nd, 2003, 9:03am)

Title: Selecting Leap Years at Random
Post by TenaliRaman on Aug 22^nd, 2003, 9:03am

What is the chance that a leap year selected at random contains 53 sundays?

//Title modified by Icarus to be more descriptive.

Title: Re: Should be damn easy!
Post by THUDandBLUNDER on Aug 22^nd, 2003, 9:56am

366 days = 52 weeks + 2 days.
So a leap year will have 53 Sundays if Jan 1st is on Saturday or Sunday.

Therefore chance = 2/7

What is the chance that ANY year chosen at random will have 53 Sundays?

Title: Re: Should be damn easy!
Post by tohuvabohu on Aug 22^nd, 2003, 11:10am

It should be easy, but it's not.
The days of the week are not evenly distributed because of the way 3 out of 4 century years are not leap years.
Every 400 years the calendar repeats itself (400*365+97 leap days is divisible by 7). But since 97 is not divisible by 7, the answer can not possibly be 2/7.
If I did it correctly, the answer is 28/97 on a leap year
and 71/400 for any year (using the gregorian calendar).

Title: Re: Should be damn easy!
Post by Sameer on Aug 22^nd, 2003, 11:19am

I got a different answer:

Considering 1 in 4 years is a leap year...

3*(3/4)*(1/7)+(1/4)*(2/7) = 11/28

Title: Re: Should be damn easy!
Post by TenaliRaman on Aug 22^nd, 2003, 11:28am

i get some really odd result like [hide]5/28[/hide].

My reasoning :
[hide]
consider 4N years.
N years will be leap years and 3N years will be non-leap years.
so the required probability would be,
P
= lim_N->oo[C(N,1)*(2/7)+C(3N,1)*(1/7)]/C(4N,1)
= 5/28

I maybe totally wrong ... if so please do clarify them.
[/hide]

Title: Re: Should be damn easy!
Post by Sir Col on Aug 22^nd, 2003, 11:37am

But as tohuvabohu has already pointed out, a leap year is every 4 years except on a century, unless the century is divisible by 400 (2000 was, 1900 wasn't). So it is not 1/4=100/400, it's 97/400.

As T&B stated, for a leap year (366 days) to have 53 Sundays, 1st Jan must be Sat or Sun. In a normal year, 1st Jan must be Sun
Therefore, P(53 Sundays in a year)=97/400*2/7+303/400*1/7=71/400 (confirming tohuvabohu's answer)

Title: Re: Should be damn easy!
Post by TenaliRaman on Aug 22^nd, 2003, 11:52am

71/400 = .1775
5/28 = .1785

hey i wasn't far off by much ;D

Title: Re: Should be damn easy!
Post by Sameer on Aug 22^nd, 2003, 2:33pm

*kicks myself for making silly mistakes*

that's why I never got perfect scores :-/