This question is related to [that](https://mathoverflow.net/questions/37344/orders-of-products-of-permutations) (if $s$ is co-prime with prime $p$ and a permutation in $S\_s$ has order $p$, then it fixes a point). 


Let us fix two (finite) numbers $p\gg 1, n\gg 1$. Say, $p=47, n=18999$. Take a
sequence $s\_1,s\_2,...$ of numbers not divisible by $p$. For each $s$
compute $p(s)$ as follows. Take two permutations $a, b$ in the symmetric group $S\_s$ such that
$\langle a,b\rangle$ is highly (say, $3$-)transitive on $\{1,...,s\}$. Compute the probability $p(s,a,b)$
that for a word $w(x,y)$ of length $n$, $w(a,b)$ has a fixed point. Then $p(s)$ is the maximum of
$p(s,a,b)$ for all (such) $a,b$.


**Question.** Is it true that $$\lim\_{i\to\infty} p(s\_i) = 0?$$


 **Note.**  As Doron Puder explained to me, one cannot replace "highly transitive" by "transitive" because $a,b$ can generate a dihedral group of order $2s\_i$ in which case $p(s\_i,a,b)$ will be at least $1/5$. F. Ladisch explains in his answer to the previous version of the question below that "2-transitive" is not enough either. 


 **Update** Perhaps the "correct" condition instead of "highly transitive" should be "$\langle a,b\rangle$ contains "few" permutations having fixed points, that is the portion of such permutations in $\langle a,b\rangle$ tends to 0 as $i\to \infty$. Examples of Puder and Ladisch contain "many" permutations fixing a point and the arguments were based on that fact. Thus the problem is this: if for many words $w(x,y)$ of length $n$ the permutation $w(a,b)$ has a fixed point, then many elements in $\langle a,b\rangle$ have fixed points, a local-to-global property. 



fixed points of products of permutations
Update: It just occured to me that if you fix the length $n$, then $p(s,a,b)$ is always a fraction with denominator $4\cdot 3^{n-1}$, so only finitely many values are possible. Thus for a sequence $(s\_i,a\_i,b\_i)$ with $\langle a\_i, b\_i\rangle\leq S\_{s\_i}$, the sequence $p(s\_i, a\_i, b\_i)$ can converge only if it is constant eventually. It converges to zero only if $\langle a\_i, b\_i \rangle$ contains no reduced words of length $n$ with fixed points for all but finitely many $i$. So I doubt that only looking at words of length $n$ can tell you much. Also, transitive but nonregular permutation groups always seem to have a large portion of elements with fixed points.


(The following was my original answer to a previous version of the question:)  

Take $C\_{p-1} \ltimes C\_p$ (or somewhat more generally, the affine group of a finite field of order $q=p^k$). This is generated by two elements, has order $(p-1)p$, and all but $p-1$ elements have fixed points. So whithout taking into account the restriction on the length, the probability for fixed points is $\frac{1}{p}$. For $p\gg n$, the probability will be similar, I think. So if you take a sequence of primes tending to infinity for $s\_i$, this shows that $\lim\_{i\to \infty} p(s\_i)$ can be big (probably $1$, in this case). 


EDIT: I justed realized that I misunderstood the question, since you are asking for the fraction of reduced words, and not the fraction of group elements itself. However, if we take $a$ and $b$ to be generators of the cyclic groups of order $p-1$ and $p$ respectively, then the reduced words mapping into the regular subgroup $C\_p= \langle b \rangle$ are those where the numbers of $a$'s and $a^{-1}$'s occurring is equal $\mod p-1$, that is, equal, if $p\gg n$. The number of these words is independent of $p$, and there are other words, so $\lim\_{i\to \infty} p(s\_i)\neq 0$. Hope now it's correct.


