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Topic: Selecting Leap Years at Random (Read 506 times) |
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TenaliRaman
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Selecting Leap Years at Random
« on: Aug 22nd, 2003, 9:03am » |
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What is the chance that a leap year selected at random contains 53 sundays?
//Title modified by Icarus to be more descriptive.
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| « Last Edit: Sep 20th, 2003, 6:48pm by Icarus » |
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Self discovery comes when a man measures himself against an obstacle - Antoine de Saint Exupery
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ThudnBlunder
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Re: Should be damn easy!
« Reply #1 on: Aug 22nd, 2003, 9:56am » |
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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?
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| « Last Edit: Aug 22nd, 2003, 9:58am by ThudnBlunder » |
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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tohuvabohu
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Re: Should be damn easy!
« Reply #2 on: Aug 22nd, 2003, 11:10am » |
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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).
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Sameer
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Re: Should be damn easy!
« Reply #3 on: Aug 22nd, 2003, 11:19am » |
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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
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"Obvious" is the most dangerous word in mathematics.
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Proof is an idol before which the mathematician tortures himself.
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TenaliRaman
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Re: Should be damn easy!
« Reply #4 on: Aug 22nd, 2003, 11:28am » |
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i get some really odd result like 5/28.
My reasoning :
consider 4N years.
N years will be leap years and 3N years will be non-leap years.
so the required probability would be,
P
= limN->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.
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| « Last Edit: Aug 22nd, 2003, 11:30am by TenaliRaman » |
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Sir Col
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Re: Should be damn easy!
« Reply #5 on: Aug 22nd, 2003, 11:37am » |
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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)
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TenaliRaman
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Re: Should be damn easy!
« Reply #6 on: Aug 22nd, 2003, 11:52am » |
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71/400 = .1775
5/28 = .1785
hey i wasn't far off by much 
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Sameer
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Re: Should be damn easy!
« Reply #7 on: Aug 22nd, 2003, 2:33pm » |
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*kicks myself for making silly mistakes*
that's why I never got perfect scores 
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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