Assume that $\sigma\in S\_n$ has the cycle type $(p,.,p,1,..,1)$ where $p>2$ is a prime and the numbers of $1$ maybe $0$. If $\sigma\_1$ and $\sigma\_2$ are chosen uniformly in the conjugacy class of $\sigma$. Assume the cycle type of $\sigma\_1 \sigma\_2=(k\_1,k\_2,.,k\_l)$ . Is there a lower bound $B$ such that $$\mathrm{Pr}\big(\exists j\in \{1,2,..,l\}, \text{ s.t. } k\_j \bmod 2=0\big)\geq B \geq 1/T(n),$$ where $T$ is a polynomial.



A probability problem in the conjugacy classes of symmetric group
Let $kp$ be the size of the support of $\sigma$. Let $1,2,3,4$ be four points of the ground set. The probability that $\sigma\_1$ maps $1 \mapsto 2$ is $kp/n(n-1)$, because there is a $kp/n$ chance that $1$ is in the support and conditionally a $1/(n-1)$ chance that $1 \mapsto 2$. Conditional on that, the probability that $3$ is in the support is $(kp-2)/(n-2)$ and then a $1/(n-2)$ chance that $3 \mapsto 4$ (because $3$ can map to anything apart from $2$ and $3$). Therefore the probability that $\sigma\_1$ maps $1\mapsto2$ and $3 \mapsto4$ is $kp(kp-2)/n(n-1)(n-2)^2 \ge 1/n^4$. Likewise the probability that $\sigma\_2$ maps $2 \mapsto 3$ and $4 \mapsto 1$ is also at least $1/n^4$. The probability that both these happen is at least $1/n^8$ and in this case $\sigma\_1\sigma\_2$ has a $2$-cycle.


