The Gregorian calendar repeats every 400 years (because there are 97 leap years every 400 years, and $400(365)+97$ is a multiple of $7$), so it suffices to count the number of leap years in the period from $1601$ to $2000$ that contain 53 Tuesdays. In the first 84 years of each century, there will be 21 leap years (xx04, xx08, ..., xx84). Because these leap years occur at regular intervals, 3 of them will start on each day of the week. So 6 will start on a Monday or a Tuesday each century; this gives $24$ leap years that start on a Monday or a Tuesday in the first $84$ years of each century in the overall time frame. It remains to count how many leap years in the last $16$ years of each century start on a Monday or a Tuesday. Note that if year $x$ is a leap year, year $x+4$ will always start on the weekday two days earlier than year $x$, because $4(365)+1 \equiv -2 \pmod{7}$. Also, if year $x$ is not in the final century of a Gregorian cycle, year $x+100$ will also always start on the weekday two days earlier than year $x$, because $100(365) + 24 \equiv -2 \pmod{7}$. Finally, it can be seen that $1688$ starts on a Thursday in the Gregorian calendar. So:
Adding these to the $24$ from before gives a total of $27$ leap years that start on a Monday or Tuesday in a single calendar period. As there are $97$ leap years total in that period, the probability that a given one starts on a Monday or a Tuesday is $\dfrac{27}{97}$. What is the probability of getting 53 thursdays in a leap year?
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What is the probability that a leap year selected at random has 53 Thursday?
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