The probability of two consecutive non-leap years having 52 Fridays each is $\frac{5}{7}$. How? I took a test on probabilities, and there was this question about finding the probability of two consecutive non-leap years having 52 Fridays each.
I figured it would be $\frac{6}{7} \times \frac{5}{7}$, which is the product of the probability of there being 52 Fridays in a non-leap year and the probability of there being 52 Fridays in the successive year if the previous year had 52 Fridays.
The teacher told me the answer was $\frac{36}{49}$, which is obviously incorrect. I explained my reasoning, but he didn't budge.
I got home and actually analysed years from 1800 to 2000 to find out what the correct answer is. It is actually $\frac{5}{7}$. I cannot seem to figure out how, though, but it must have something to do with the leap years not being taken into consideration.
A year is a leap year if it is divisible by 4. If it is divisible by 100, but not by 400, it is not a leap year.
I changed $\frac{2}{3}$ in my original question to $\frac{5}{7}$ because that is actually the correct answer (which Christian Blatter proves correct in his answer). The $\frac{2}{3}$ came about due to faulty coding.
 A: Everything repeats after 28 years. During such a period there are $14$ pairs of consecutive non-leap years. In $10$ of these pairs none of the two years begins with a Friday. The probability in question therefore comes to ${5\over7}$.
A: The issue of both incorrect derivations (your original one and the teacher's) is that they assume incorrect things about the correlations of the two probabilities.  It's true that if event A has a probability $p_A$ of occurring, and event B has a probability $p_B$ of occurring, and the two events are uncorrelated, then the probability of both A and B occurring is $p_A p_B$.  However, the probability a given year will have 53 Fridays is strongly correlated with whether the previous year had 53 Fridays.  Thus, we need to be more careful.
For the purposes of the derivation below, I'm going to ignore leap years and call a "year" 365 days.  We can see that a year will have 53 Fridays if and only if it starts on a Friday, and that if year $n$ starts on a particular day of the week, then year $n+1$ will start on the next day of the week (since $365 \equiv 1 \mod 7$.)
Your teacher's logic was the following:  "the probability that a given year will not start on a Friday is $\frac{6}{7}$.  Therefore, the probability that neither of two years will start on a Friday is $\left(\frac{6}{7}\right)^2 = \frac{36}{49}$."  The problem is, as stated above, that the start days of two consecutive years is correlated.  Using the same logic, you could prove that the probability that two consecutive years both start on a Friday is $\left(\frac{1}{7}\right)^2 = \frac{1}{49}$.  But obviously the probability should be 0, since two consecutive years never start on the same day.  (Your teacher's answer would be correct if the question was, "What is the probability that two randomly selected years both have 52 Fridays?")
Your initial logic was a bit better, but still off.  It seems something like, "The probability that Year 1 will not start on a Friday is $\frac{6}{7}$.  The next year will start on a different day, so there are only five non-Fridays for Year 2 to start on.  Thus, probability that this other day will also not be a Friday is $\frac{5}{7}$, and the total probability is $\frac{6}{7} \times \frac{5}{7} = \frac{30}{49}$."  The problem here is that there are really only 6 valid choices for the starting day of Year 2, since it will never start on the same day as Year 1.
The correct logic is, I think, as follows:  The probability that Year 1 will not start on a Friday is $\frac{6}{7}$.  Year 2 will never start on the same day as Year 1;  so it must start on one of the other six days of the week.  Out of these remaining six days of the week, five are not Friday.  (We've eliminated the possibility that all six days are non-Fridays, since we're looking at the cases where Year 1 starts on a non-Friday.)  Thus, the probability that Year 2 starts on a non-Friday given that Year 1 starts on a non-Friday is $\frac{5}{6}$.  The overall probability is then $\frac{6}{7} \times \frac{5}{6} = \frac{5}{7}$, as you found "experimentally".
Oh, and there's a much slicker way to solve this problem than by dealing with all these conditional probabilities.  Two consecutive common years will necessarily start on two consecutive days of the week.  Moreover, a randomly chosen common year is equally likely to start on any day of the week.  Thus, your equation is equivalent to, "Given two consecutive days of the week, what is the probability that neither of them is a Friday?"  The answer, is of course, $\frac{5}{7}$, since there are seven possible pairs of consecutive days of the week (Monday-Tuesday, Tuesday-Wednesday, etc.) and only two of them contain Friday.
A: If you use the complete definition of a leap year, then one has 206 pairs of consecutive non-leap years in 400 years. Of these, 147 each has 52 Fridays. That brings the probability to $$\frac{147}{206}$$
This is exactly $\frac{1}{1442}$ less than the $\frac{5}{7}$ in the simplified definition. 
