Probability of Christmas on Wednesday
Original Problem
Proof: In the system of the current Common Era, the probability of Christmas on Wed. is not 1/7.
What exactly the probability is
In 400 years there are 303 non-leap years and 97 leap years. Total number of days:
$$T = 303\cdot 365 + 97\cdot 366\equiv 303 + 97\cdot 2\equiv 0\pmod 7.$$
This means: If the Christmas is on Mon. in year x, then it would be on Mon. in year (x + 400). That’s to say, there’s a difference between a day of the week and another day of the week.
Solve in C
/*
*
=
*
* Filename: LeapYear_Xmas.cpp
*
* Description: A program to count which day of the week does the X-mas falls.
*
* Version: 1.0
* Created: 2021/10/9 20:31:35
* Revision: none
* Compiler: gcc
*
* Author: CuSO4_Deposit, CuSO4D@protonmail.com
*
*
=
*/
#include <stdio.h>
#include <iostream>
#define TRUE 1
#define FALSE 0
#define Status int
#define SUCCESS 1
#define FAILURE 0
#define FEB29 59
#define DEC25_nonleap 358
//Function Declaration
bool ifLeap(int year);
Status RunCalendar(int StartYearDay, int w[], int TrackDay);
int RunAYear(int weekday, int TrackDay);
//Function Implement
bool ifLeap(int year){
/*
*
= FUNCTION
* Name: ifLeap
* Description: if the year is leap, return 1; else return 0.
* Arguments: year: the year to be judged.
*
=
*/
if(!(year % 4)){
if(!(year % 400)) return 1;
if(!(year % 100)) return 0;
return 1;
}
else return 0;
}
Status RunCalendar(int StartYearDay, int w[], int TrackDay){
/*
*
= FUNCTION
* Name: RunCalendar
* Preconditon: TrackDay can only be in a non-leap year(i.e. can't be Feb. 29).
* Description: Run a calendar for 400 years, when the 1st day is StartYearDay, and
* return the day of TrackDay through w[].
* Arguments: StartYearDay: what day of the week is the 1st day.
* w[]: how many times does TrackDay falls in weekday[i].
* TrackDay: the day to track.
*
=
*/
int NextYearStartDay = StartYearDay;
if(TrackDay >= FEB29)
for(int i = 0; i < 400; i++){
w[RunAYear(NextYearStartDay, ifLeap(i) ? TrackDay + 1 : TrackDay)]++;
NextYearStartDay = ifLeap(i) ? (NextYearStartDay + 366) % 7 : (NextYearStartDay + 365) % 7;
}
else
for(int i = 0; i < 400; i++){
w[RunAYear(NextYearStartDay, TrackDay)]++;
NextYearStartDay = ifLeap(i) ? (NextYearStartDay + 366) % 7 : (NextYearStartDay + 365) % 7;
}
return SUCCESS;
}
int RunAYear(int weekday, int TrackDay){
/*
*
= FUNCTION
* Name: RunAYear
* Description: Run a year, return what day of the week the targetday is on.
*
=
*/
return (weekday + TrackDay % 7) % 7;
}
int main(){
int w[7] = {0}; //the day of the week array
for(int i = 0; i < 7; i++){
for(int j = 0; j < 7; j++) //array initialization
w[j] = 0;
RunCalendar(i, w, DEC25_nonleap);
printf("When the 1st day of Year 0 is %d:\n", i);
printf("S\tM\tT\tW\tT\tF\tS\n");
for(int i = 0; i < 7; i++){
printf("%d\t", w[i]);
}
printf("\n");
}
return 0;
}
result:
When the 1st day of Year 0 is 0: S M T W T F S 56 58 56 58 57 57 58 When the 1st day of Year 0 is 1: S M T W T F S 58 56 58 56 58 57 57 When the 1st day of Year 0 is 2: S M T W T F S 57 58 56 58 56 58 57 When the 1st day of Year 0 is 3: S M T W T F S 57 57 58 56 58 56 58 When the 1st day of Year 0 is 4: S M T W T F S 58 57 57 58 56 58 56 When the 1st day of Year 0 is 5: S M T W T F S 56 58 57 57 58 56 58 When the 1st day of Year 0 is 6: S M T W T F S 58 56 58 57 57 58 56 Hit any key to close this window…
As the Jan. 1st is on Sat in 2000, we know it’s Sat. in year 0 too. Then our chronological system loops as what the 7th result shows. X-mas falls on Wednesday 57 times in 400 years. $$P = \dfrac{57}{400}.$$
Solve in Wolfram Mathematica[ref] 试编程求圣诞节在星期三的具体概率？ [/ref]
Tally[DayName /@ Tuples[{Range[0, 399], Range[12, 12], {25}}]]
result:
{{Monday, 56}, {Tuesday, 58}, {Wednesday, 57}, {Thursday, 57}, {Saturday, 56}, {Sunday, 58}, {Friday, 58}}
#mathematica #NumberTheory