# Comic relief probability

Let $$X_0,X_1,...$$ be i.i.d and let $$N = \inf \{ n\ge 1:X_n >X_0 \}$$.

Prove that $$P(N>n) \ge \frac{1}{n+1}$$.

My attempt,for $$E = \{N>n\} = \{N = n+1 ,n+2,n+3,...\}$$ and we know for $$\{N = n+k\}$$ means for the first $$n+k+1$$ element the first one i.e. $$X_0$$ rank exactly $$n+k$$, and $$X_{n+k}$$ rank exactly $$n+k+1$$, which means it has probability $$\frac{1}{(n+k+1)(n+k)}$$ by symmetric reasoning.Hence the sum for all $$N>n$$ is $$\frac{1}{n+1}$$.

What I don't understand is why it may greater than $$\frac{1}{n+1}$$.

This is the example from Durrett probability page 70.Comic relief

Think of the case where all the $$X_i$$'s are 0. What is $$N$$? That should hopefully be enlightening to tell you where the symmetry argument fails, and that it is indeed exact if $$P(X_i = X_j) = 0$$