What is the chance that a leap year will have 53 Sundays?

As per my trial, will it be 2/7? Since 366 days in a leap year means 52 weeks and 2 more days, so from the extra two days, the probability of Sunday is 2/7.

P.S : This was a question I found in a basic statistics book.

**Answer**

The Gregorian calendar favors five of the seven weekdays during leap years. Therefore **the chance is not precisely 2/7.**

This was essentially problem B3 in the 1950 Putnam Mathematics Competition:

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.

In the Gregorian Calendar, years that are multiples of 4 are leap years (with 7×52+2=366 days), *but* years that are multiples of 100 are not leap years (and therefore have 7×52+1=365 days), *with the exception* that years that are multiples of 400 are leap years. (Many of us remember the most recent exception in 2000.) This creates a 400 year cycle containing 400/4−400/100+400/400=97 leap years.

What is especially interesting is that the total number of days in this cycle is a whole multiple of seven:

400×(7×52+1)+97×1≡400+97≡71×7≡0mod7.

This shows that the 400 year cycle comprises a whole number of weeks. Consequently, **the pattern of days of the week is exactly the same from one cycle to the next.**

We may therefore interpret the question as asking for the chance of 53 Sundays when sampling randomly and uniformly from any 400-year cycle of leap years. A brute-force calculation (using, say, the fact that January 1, 2001, was a Monday) shows that 28 of the 97 leap years in each cycle have 53 Sundays. Therefore the chance is

Pr(53 Sundays)=2897.

Note that this does *not* equal 28/98 = 2/7: it is slightly greater.

Incidentally, there is the same chance of 53 Wednesdays, Fridays, Saturdays, or Mondays and only a 27/97 chance of 53 Tuesdays or Thursdays.

For those who would like to make more detailed calculations (and might mistrust any mathematical simplifications), here is brute-force code that computes and examines ever day of the week for a given set of years. At the end it displays the number of years with 53 appearances of each day of the week. It is written in `R`

.

Here is its output for the 2001-2400 cycle:

```
Friday Monday Saturday Sunday Thursday Tuesday Wednesday
28 28 28 28 27 27 28
```

Here is the code itself.

```
leapyear <- function(y) {
(y %% 4 == 0 & !(y%% 100 == 0)) | (y %% 400 == 0)
}
leapyears <- seq(2001, length.out=400)
leapyears <- leapyears[leapyear(leapyears)]
results <- sapply(leapyears, function(y) {
table(weekdays(seq.Date(as.Date(paste0(y, "-01-01")), by="1 day", length.out=366)))
})
rowSums(results==53)
```

**Attribution***Source : Link , Question Author : Manali Chatterjee , Answer Author : whuber*