Probability about center of mass of 2n+1 random variables based on the 4 vertices

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There are four points in the coordinate plane as following.

A=(1,1), B=(1,-1), C=(-1,1), D=(-1,-1)

We will choose $2n+1$ points and each point is one of these 4 points, but the probability of choice is different.

P(point is A)=$p$, $\quad$ P(point is B)=$\frac{1}{2}-p$

P(point is C)=$\frac{1}{2}-p$,$\quad$ P(point is D)=$p$

This probability is for choice of one point and all $2n+1$ choice are independent.

Let $x$ be the center of mass of these $2n+1$ points.

and $f(n,p)$=probability that $x$ is on the first quadrant.

For example, $f(0,\frac{1}{2})=\frac{1}{2}$, $f(1,\frac{1}{2})=\frac{1}{2}$, $f(0,\frac{1}{3})=\frac{1}{3}$, $f(1,\frac{1}{3})=\frac{17}{54}$.

Prove: If $\frac{1}{4}<p<\frac{1}{2}$, $f(n,p)$ is decreasing at n.

If you tell me references for this proof, it would be great thanks.

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This is my answer to my question itself. At least for this problem, I didn't use deep probability theory other than elementary method.

Let block-P be a square which has four vertices ((0,0),(0,1),(1,1),(1,0)) and let block-Q be a square of ((0,0),(-1,0),(-1,-1),(0,1)).

Let's decompose this block-P and block-Q into each $(n+1)^2$ cells. For example, if n=2, it will be divided as 9 small cells for each block like following figure.

figure 1

Let Core-P(n) be the small square cell of $((0,0),(0,\frac{1}{n+1}), (\frac{1}{n+1},\frac{1}{n+1} ) , (\frac{1}{n+1},0) )$ and Core-Q(n) is defined similarly. You can see it at following figure.

enter image description here

Let $f(P,n)$ be the probability that x is on the block-P with n. Let $f(C_p,n)$ be the probability that x is on the Core-P(n). $f(Q,n)$ and $f(C_q,n)$ is defined similarly.

Lemma 1. $f(C_q,n)<f(C_p,n)$ if $p>q$: Trivially true.

Lemma 2. $f(P,n)-f(P,n+1)=2pq(f(C_p,n)-f(C_q,n))$

I will skip the proof here. It is not difficult. Hint: Use symmetry of p and q.

The problem was that $f(P,n)-f(P,n+1)$ is positive. Actually it is because $f(C_p,n)-f(C_q,n)$ is positive. Q.E.D.

Although this proof doesn't seem to have much to do with coupling argument, I will give bounty to @mathworker21 because he is only one who answered to this question and the reference would be helpful for my future research. Thank you.

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A coupling argument should do the trick. It seems you just want references for a proof, so here ya go: http://websites.math.leidenuniv.nl/probability/lecturenotes/CouplingLectures.pdf