# Don't Fail MY Exam [duplicate]

The solution to the 100 Prisoner's problem works here. Give every student a number between $$1$$ and $$n$$, and let the $$i$$th student start with the $$i$$th test from the top. Then, assume this is the $$j$$th person's test, then check the $$j$$th test from the top. Keep on following this; all the students will find their own test iff there is no cycle greater than $$\frac{n}{2}$$ in size. The probability of there being a cycle greater than this in size is at least $$ln(2)$$, giving a lower bound of $$1-ln(2)$$.