Consider Pascal's triangle with $30$ rows (the top $1$ is the $0$th row). The centre number is the number in the middle of row $30\times \frac23=20$, which is $\binom{20}{10}=184756$. The proportion of the numbers in the triangle that are less than this centre number is $\frac{364}{496}\approx0.7339$.
In the following graph, the horizontal axis is $n$, the number of rows in the triangle ($n$ is a multiple of $3$). The vertical axis is $P$, the proportion of numbers that are less than the centre number.
It looks like $P$ is approaching some limit. The rightmost data point is $n=138$ and $P=\frac{7078}{9730}\approx0.72744$. I put $0.72744$ into Wolfram, and it suggested $e^{-1/\pi}\approx 0.72738$.
Is the following conjecture true or false:
Conjecture: In Pascal's triangle with $n$ rows, where $n$ is a multiple of $3$, the proportion of numbers less than the centre number approaches $e^{-1/\pi}$ as $n\to\infty$.
I would not be too surprised if this is in fact the limiting proportion, since $e$ is related to Pascal's triangle and so is $\pi$.
My attempt: The numbers in the top $\frac23$ of the triangle are all less than the centre number. So we just need to determine the proportion of numbers in the bottom $\frac13$ that are less than the centre number. I considered the row that is $m$ rows below the centre number: I tried to work out the proportion of numbers in this row that are less than the centre number, to obtain an asympototic boundary curve, but I didn't succeed.
Context: I was trying to approximate the median of the numbers in the first $n$ rows of Pascal's triangle, and in my investigation I stumbled upon this possible result.
1-sum(outer(0:999,0:999,"choose")>=choose(666,333))/choose(1001,2)
using R gives $0.72689$ for $n=999$, in the same way as1-sum(outer(0:30,0:30,"choose")>=choose(20,10))/choose(32,2)
gives $0.73387$ for $n=30$ $\endgroup$