Given the numbers ${1,2,3,4,5,6...n}$ ($n >= 4$) I need to find the probability that in a permutation of the number I will have local maximum in place number 2 and in place number 4. I know that the probability for 1 local maximum point is $1/3$ in a specific location, and the total without endpoints is $(n-2)/{3}$.
So for my question, I know that I need for every 5 numbers I am able to arrange them such that there will be local maximum in place 2 and in place 4, but I'm not able to compute how many of such options exist for 5 number since it depends also on the gap between the numbers.
Hope you guys can help thanks!
Let's assume $n\ge5$ to avoid the awkward special case $n=4$. Then without loss of generality we can assume $n=5$. Let $a_k$ denote the elements of the permutation.
We must have $a_2=5$ or $a_4=5$, and the other of $a_2$ and $a_4$ must be either $3$ or $4$ (as the two numbers next to it are smaller). If it's $4$, there are $2!=2$ ways to arrange the numbers at the maxima and $3!=6$ ways to arrange the other numbers, which makes $2\cdot6=12$. If it's $3$, then the $4$ must be at the opposite end, so there are still $2!=2$ ways to arrange the numbers at the maxima but then only $2!=2$ to arrange the $1$ and the $2$, which makes $2\cdot2=4$. Thus, in total there are $12+4=16$ permutations with local maxima at $2$ and $4$, out of a total of $5!=120$, so the desired probability is $\frac{16}{120}=\frac2{15}$.