This is a basic probability question.
Persons A and B decide to arrive and meet sometime between 7 and 8 pm. Whoever arrives first will wait for ten minutes for the other person. If the other person doesn't turn up inside ten minutes, then the person waiting will leave. What is the probability that they will meet? I am assuming uniform distribution for arrival time between 7 pm and 8 pm for both of them.
The exact question is given here: Probability of meeting
I am aware of geometric probability and the methods they have used there. The answer seems to be unanimously $9/36$. However, here is where I am confused. Considering A reaches before B, and that he reaches before the first $50$ minutes, the probability of meeting should be $5/6 \cdot 1/6= 5/36$ (since B would have to reach within $10$ minutes of A). By symmetry, this means that if B reaches early and reaches before the first $50$ minutes, the meeting probability is again $5/36$. It seems that even without considering the probability of what happens if the earlier person reaches in the last $10$ minutes, we have a probability of $10/36$ already, greater than the total probability calculated in the post in the link.
Can anyone please point out my logical flaw?