Miscellaneous · Q5: A leap year with 53 Tuesdays, $P = \frac27$

Basics you need

Solution

1. Weeks and extra days

   A leap year has $366$ days, and

   $$366 = 7 \times 52 + 2.$$

   So it is 52 complete weeks followed by 2 more days.

   Days of the week repeat every 7 days, so we group the year into blocks of 7. What matters is how many days are left over.
   Division with remainder (earlier classes).

2. The full weeks always give 52 Tuesdays

   Each block of 7 consecutive days contains each day of the week exactly once, so the first $7 \times 52 = 364$ days contain exactly 52 Tuesdays. Every leap year has at least 52 Tuesdays, and it has 53 exactly when one of the 2 extra days is a Tuesday.

   This turns a question about 366 days into a question about just 2 days.
   An event described in words becomes a set of outcomes. Basics: events as sets

3. The 7 possible pairs of extra days

   The 2 extra days are the 365th and 366th days. Day 365 falls on the same day of the week as 1 January (364 days later is exactly 52 weeks later), and day 366 on the next day. So the pair is fixed by the day on which the year starts:

   (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun).

   These are the outcomes of the sample space $S$⁠, so $n(S) = 7$⁠, and they are equally likely.

   The year is equally likely to start on any day of the week (the "at random" in the question), so each of these 7 pairs has probability $\frac17$⁠. Note that (Sat, Sun) is a pair too: the week wraps round from Saturday to Sunday.
   List every outcome of the sample space. Basics: events as sets

4. Count the favourable pairs and divide

   Let $E$⁠: the leap year has 53 Tuesdays. The pairs containing a Tuesday are (Mon, Tue) and (Tue, Wed), so $n(E) = 2$ and

   $$P(E) = \frac{n(E)}{n(S)} = \frac27.$$
   Tuesday can be the first extra day (pair (Tue, Wed)) or the second (pair (Mon, Tue)); it cannot be both, so there are exactly 2 favourable pairs.
   $P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes. Basics: classical probability

5. Compare with an ordinary year

   An ordinary year has $365 = 7 \times 52 + 1$ days: one extra day, so P(53 Tuesdays) $= \frac17$ there. The second extra day of a leap year doubles the chance, to $\frac27$⁠.

   In the real calendar the seven starting days are not quite equally common: in the 400-year cycle 2001 to 2400 there are 97 leap years, and 27 of them have 53 Tuesdays ($\frac{27}{97} \approx 0.278$⁠, close to $\frac27 \approx 0.286$⁠). The question's "at random" means the textbook model, in which all 7 pairs are equally likely.

   Each extra day gives one chance in seven of being a Tuesday, and two different extra days can never both be Tuesdays, so the chances add: $\frac17 + \frac17 = \frac27$⁠.
   Mutually exclusive events add. Basics: rules of probability

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