Probability of day-date problems
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How to solve -Leap years have 366 days; other years have 365 days. Year n > 0 is a leap year iff (1) 4 divides n, but 100 does not divide n, or (2) 400 divides n. n is chosen at random from the natural numbers. Show that the probability that December 25 in year n is a Wednesday is not 1/7.
combinatorics
asked Jan 24 at 11:08
Lalla95Lalla95
564
$endgroup$
add a comment |
$begingroup$
How to solve -Leap years have 366 days; other years have 365 days. Year n > 0 is a leap year iff (1) 4 divides n, but 100 does not divide n, or (2) 400 divides n. n is chosen at random from the natural numbers. Show that the probability that December 25 in year n is a Wednesday is not 1/7.
combinatorics
asked Jan 24 at 11:08
Lalla95Lalla95
564
$endgroup$
$begingroup$
Well, I think that the problem definition already gives a quite big hint by talking about leap years. You can take two different cases: Leap years and non-leap years.
$endgroup$
– Matti P.
Jan 24 at 11:10
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"$n$ is chosen at random from the natural numbers". On base of what distribution? A canonical/natural/uniform one does not exist. Actually this problem has no answer.
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– drhab
Jan 24 at 11:38
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You need to show that there is an integral number of weeks in $400$ years, so that the Gregorian calendar repeats every $400$ years.
$endgroup$
– saulspatz
Jan 24 at 13:06
add a comment |
$begingroup$
How to solve -Leap years have 366 days; other years have 365 days. Year n > 0 is a leap year iff (1) 4 divides n, but 100 does not divide n, or (2) 400 divides n. n is chosen at random from the natural numbers. Show that the probability that December 25 in year n is a Wednesday is not 1/7.
combinatorics
asked Jan 24 at 11:08
Lalla95Lalla95
564
$endgroup$
How to solve -Leap years have 366 days; other years have 365 days. Year n > 0 is a leap year iff (1) 4 divides n, but 100 does not divide n, or (2) 400 divides n. n is chosen at random from the natural numbers. Show that the probability that December 25 in year n is a Wednesday is not 1/7.
combinatorics
combinatorics
asked Jan 24 at 11:08
Lalla95Lalla95
564
asked Jan 24 at 11:08
Lalla95Lalla95
564
asked Jan 24 at 11:08
Lalla95Lalla95
564
asked Jan 24 at 11:08
Lalla95Lalla95
564
asked Jan 24 at 11:08
Lalla95Lalla95
564
564
$begingroup$
Well, I think that the problem definition already gives a quite big hint by talking about leap years. You can take two different cases: Leap years and non-leap years.
$endgroup$
– Matti P.
Jan 24 at 11:10
$begingroup$
"$n$ is chosen at random from the natural numbers". On base of what distribution? A canonical/natural/uniform one does not exist. Actually this problem has no answer.
$endgroup$
– drhab
Jan 24 at 11:38
$begingroup$
You need to show that there is an integral number of weeks in $400$ years, so that the Gregorian calendar repeats every $400$ years.
$endgroup$
– saulspatz
Jan 24 at 13:06
add a comment |
$begingroup$
Well, I think that the problem definition already gives a quite big hint by talking about leap years. You can take two different cases: Leap years and non-leap years.
$endgroup$
– Matti P.
Jan 24 at 11:10
$begingroup$
"$n$ is chosen at random from the natural numbers". On base of what distribution? A canonical/natural/uniform one does not exist. Actually this problem has no answer.
$endgroup$
– drhab
Jan 24 at 11:38
$begingroup$
You need to show that there is an integral number of weeks in $400$ years, so that the Gregorian calendar repeats every $400$ years.
$endgroup$
– saulspatz
Jan 24 at 13:06
$begingroup$
Well, I think that the problem definition already gives a quite big hint by talking about leap years. You can take two different cases: Leap years and non-leap years.
$endgroup$
– Matti P.
Jan 24 at 11:10
$begingroup$
Well, I think that the problem definition already gives a quite big hint by talking about leap years. You can take two different cases: Leap years and non-leap years.
$endgroup$
– Matti P.
Jan 24 at 11:10
$begingroup$
"$n$ is chosen at random from the natural numbers". On base of what distribution? A canonical/natural/uniform one does not exist. Actually this problem has no answer.
$endgroup$
– drhab
Jan 24 at 11:38
$begingroup$
"$n$ is chosen at random from the natural numbers". On base of what distribution? A canonical/natural/uniform one does not exist. Actually this problem has no answer.
$endgroup$
– drhab
Jan 24 at 11:38
$begingroup$
You need to show that there is an integral number of weeks in $400$ years, so that the Gregorian calendar repeats every $400$ years.
$endgroup$
– saulspatz
Jan 24 at 13:06
$begingroup$
You need to show that there is an integral number of weeks in $400$ years, so that the Gregorian calendar repeats every $400$ years.
$endgroup$
– saulspatz
Jan 24 at 13:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider a period of 400 years: four of them are divisible by 100 and only one by 400. So the number of leap years is $(400/4)-3=97$. The total number of days in a 400-year period is therefore $400*365+97=146097$ days and that number is divisible by 7 - so you have a whole number of weeks there. In other words, history repeats itself every 400 years. So if December 25th is Wednesday in year X, it will be Wednesday again in year $X+400$, $X+800$... So you don't have to consider a period longer than 400 years.
But 400 is not divisible by seven so it's impossible that December 25th (or any other particular date) will have a chance to hit a particular day of week with probability which is exactly $1/7$.
If you dig deeper and use a bit of modular artihmetics you will find out that in a single century (starting with year divisible by 100), December 25th hits five days of week 14 times and two days of week 15 times. If you consider 400-year long period (starting with year divisible by 400), December 25th hits two days of week 56 times, two days of week 57 times and three days of week 58 times. So the probability that December 25th will hit any particular day in a week is either $56/400$, $57/400$, $58/400$, fairly close to $1/7$ but not equal to.
answered Jan 24 at 16:02
OldboyOldboy
8,62711036
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add a comment |
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1 Answer
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$begingroup$
Consider a period of 400 years: four of them are divisible by 100 and only one by 400. So the number of leap years is $(400/4)-3=97$. The total number of days in a 400-year period is therefore $400*365+97=146097$ days and that number is divisible by 7 - so you have a whole number of weeks there. In other words, history repeats itself every 400 years. So if December 25th is Wednesday in year X, it will be Wednesday again in year $X+400$, $X+800$... So you don't have to consider a period longer than 400 years.
But 400 is not divisible by seven so it's impossible that December 25th (or any other particular date) will have a chance to hit a particular day of week with probability which is exactly $1/7$.
If you dig deeper and use a bit of modular artihmetics you will find out that in a single century (starting with year divisible by 100), December 25th hits five days of week 14 times and two days of week 15 times. If you consider 400-year long period (starting with year divisible by 400), December 25th hits two days of week 56 times, two days of week 57 times and three days of week 58 times. So the probability that December 25th will hit any particular day in a week is either $56/400$, $57/400$, $58/400$, fairly close to $1/7$ but not equal to.
answered Jan 24 at 16:02
OldboyOldboy
8,62711036
$endgroup$
add a comment |
$begingroup$
Consider a period of 400 years: four of them are divisible by 100 and only one by 400. So the number of leap years is $(400/4)-3=97$. The total number of days in a 400-year period is therefore $400*365+97=146097$ days and that number is divisible by 7 - so you have a whole number of weeks there. In other words, history repeats itself every 400 years. So if December 25th is Wednesday in year X, it will be Wednesday again in year $X+400$, $X+800$... So you don't have to consider a period longer than 400 years.
But 400 is not divisible by seven so it's impossible that December 25th (or any other particular date) will have a chance to hit a particular day of week with probability which is exactly $1/7$.
If you dig deeper and use a bit of modular artihmetics you will find out that in a single century (starting with year divisible by 100), December 25th hits five days of week 14 times and two days of week 15 times. If you consider 400-year long period (starting with year divisible by 400), December 25th hits two days of week 56 times, two days of week 57 times and three days of week 58 times. So the probability that December 25th will hit any particular day in a week is either $56/400$, $57/400$, $58/400$, fairly close to $1/7$ but not equal to.
answered Jan 24 at 16:02
OldboyOldboy
8,62711036
$endgroup$
add a comment |
$begingroup$
Consider a period of 400 years: four of them are divisible by 100 and only one by 400. So the number of leap years is $(400/4)-3=97$. The total number of days in a 400-year period is therefore $400*365+97=146097$ days and that number is divisible by 7 - so you have a whole number of weeks there. In other words, history repeats itself every 400 years. So if December 25th is Wednesday in year X, it will be Wednesday again in year $X+400$, $X+800$... So you don't have to consider a period longer than 400 years.
But 400 is not divisible by seven so it's impossible that December 25th (or any other particular date) will have a chance to hit a particular day of week with probability which is exactly $1/7$.
If you dig deeper and use a bit of modular artihmetics you will find out that in a single century (starting with year divisible by 100), December 25th hits five days of week 14 times and two days of week 15 times. If you consider 400-year long period (starting with year divisible by 400), December 25th hits two days of week 56 times, two days of week 57 times and three days of week 58 times. So the probability that December 25th will hit any particular day in a week is either $56/400$, $57/400$, $58/400$, fairly close to $1/7$ but not equal to.
answered Jan 24 at 16:02
OldboyOldboy
8,62711036
$endgroup$
Consider a period of 400 years: four of them are divisible by 100 and only one by 400. So the number of leap years is $(400/4)-3=97$. The total number of days in a 400-year period is therefore $400*365+97=146097$ days and that number is divisible by 7 - so you have a whole number of weeks there. In other words, history repeats itself every 400 years. So if December 25th is Wednesday in year X, it will be Wednesday again in year $X+400$, $X+800$... So you don't have to consider a period longer than 400 years.
But 400 is not divisible by seven so it's impossible that December 25th (or any other particular date) will have a chance to hit a particular day of week with probability which is exactly $1/7$.
If you dig deeper and use a bit of modular artihmetics you will find out that in a single century (starting with year divisible by 100), December 25th hits five days of week 14 times and two days of week 15 times. If you consider 400-year long period (starting with year divisible by 400), December 25th hits two days of week 56 times, two days of week 57 times and three days of week 58 times. So the probability that December 25th will hit any particular day in a week is either $56/400$, $57/400$, $58/400$, fairly close to $1/7$ but not equal to.
answered Jan 24 at 16:02
OldboyOldboy
8,62711036
answered Jan 24 at 16:02
OldboyOldboy
8,62711036
answered Jan 24 at 16:02
OldboyOldboy
8,62711036
answered Jan 24 at 16:02
OldboyOldboy
8,62711036
8,62711036
add a comment |
add a comment |
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$begingroup$
Well, I think that the problem definition already gives a quite big hint by talking about leap years. You can take two different cases: Leap years and non-leap years.
$endgroup$
– Matti P.
Jan 24 at 11:10
$begingroup$
"$n$ is chosen at random from the natural numbers". On base of what distribution? A canonical/natural/uniform one does not exist. Actually this problem has no answer.
$endgroup$
– drhab
Jan 24 at 11:38
$begingroup$
You need to show that there is an integral number of weeks in $400$ years, so that the Gregorian calendar repeats every $400$ years.
$endgroup$
– saulspatz
Jan 24 at 13:06