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> **Question:** Is there a sequence $(\delta\_n)\_n$ of real numbers with $\delta\_n \to 0$ as $n \to \infty$, such that the following holds:
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> Let $F$ be a free group on two generators, let $F \curvearrowright X$ be a transitive action on an infinite set, and let $x \in X$. Then, the probability that a random word of length $n$ in $F$ fixes $x \in X$ is smaller than $\delta\_n$.
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I would like to consider random unreduced words (so that there are $4^n$ such words of length $n$ and each is equally likely), but probably this does not matter much. It corresponds to the nearest neighbor random walk on the Schreier graph corresponding to the action of $F$ on $X$.


It is clear that for each individual action $x \in X$, the return probability decays; and it seems plausible that this happens uniformly over all actions.



Return probabilities for random walks on infinite Schreier graphs
So, if I understand things right, we have a random walk on an undirected connected graph with possibly multiple edges and degree 4 at each vertex that starts at some vertex $x$. 


The usual way to proceed is to consider the vector $P=P\_n$ whose components $P\_n(y)$ are the probabilities to end at $y$ after $n$ steps. The dynamics is $P\_{n+1}=AP\_n$ where $A$ is one quarter of the incidence matrix. The key is that the $\ell^2$ norm of $P\_n$ drops at each step by some constant multiple of the "gradient energy" $E=\sum\_y \sum\_{z,w\in N(y)}|P\_n(z)-P\_n(w)|^2$ where $N(y)$ is the set of neighbors of $y$. This allows to control the $L^\infty$ norm of $P\_n$ (that also can go only down) very easily. If we made $n$ steps, the energy at some step was at most $C/n$. Now, take any vertex $v$ and join it with a faraway vertex by a shortest path $v=v\_0--v\_1--v\_2--v\_3--\dots$. Then the energy is at least $\sum\_{k=0}^\infty |P(v\_{2k})-P(v\_{2k+2})|^2$. Now it remains to note that the $\ell^1$ norm of $P$ is always $1$, so we can estimate $P(v)$ by $1/M+\sqrt{M/n}$ with any integer $M$ we want (among first $M$ vertices with even indices, we have at least one with the probability $1/M$ or less and the sum of differences on the return path to $v$ from there is bounded by $\sqrt{M/n}$ by Cauchy-Schwarz. Taking $M=n^{1/3}$ gives $\delta\_n=O(n^{-1/3})$. This is by no means optimal but it gives you the general idea of how such proofs go. 


I leave it to the professional probabilists to refer you to the relevant literature.


