I would like to show that $\mathbb P(\operatorname{Binomial}(n,p) = \operatorname{Binomial}(n,q))$ decreases when $n$ increases for a fixed pair $(p,q)$. This can be reformulated as $\mathbb P(X_n=0)$ decreases where $X_n=\sum_{i=1}^n S_i$ is a lazy random walk where $S_i=-1,0,1$ with probability $p(1-q),pq+(1-p)(1-q),(1-p)q$. For $p=q$ this can be done by characteristic functions. Any ideas for general $(p,q)$? I feel like this should be well known.

P.S. This question is crossposted from mathstackexchange, where it receives no answer even after I put a bounty.

Edit: To be clear the actual question is to show for any pair of $(p,q)$, we have $P(Bin(n,p)=Bin(n,q))\le P(Bin(n-1,p)=Bin(n-1,q))$ for any $n$, and everything is independent of each other.

$\newcommand{\Bin}{\operatorname{Bin}}$I would like to show that $\mathbb P(\operatorname{Binomial}(n,p) = \operatorname{Binomial}(n,q))$ decreases when $n$ increases for a fixed pair $(p,q)$. This can be reformulated as $\mathbb P(X_n=0)$ decreases where $X_n=\sum_{i=1}^n S_i$ is a lazy random walk where $S_i=-1,0,1$ with probability $p(1-q),pq+(1-p)(1-q),(1-p)q$. For $p=q$ this can be done by characteristic functions. Any ideas for general $(p,q)$? I feel like this should be well known.

P.S. This question is crossposted from mathstackexchange, where it receives no answer even after I put a bounty.

Edit: To be clear the actual question is to show for any pair of $(p,q)$, we have $P(\Bin(n,p)=\Bin(n,q))\le P(\Bin(n-1,p)=\Bin(n-1,q))$ for any $n$, and everything is independent of each other.

I would like to show that $\mathbb P(\operatorname{Binomial}(n,p) = \operatorname{Binomial}(n,q))$ decreases when $n$ increases for a fixed pair $(p,q)$. This can be reformulated as $\mathbb P(X_n=0)$ decreases where $X_n=\sum_{i=1}^n S_i$ is a lazy random walk where $S_i=-1,0,1$ with probability $p(1-q),pq+(1-p)(1-q),(1-p)q$. For $p=q$ this can be done by characteristic functions. Any ideas for general $(p,q)$? I feel like this should be well known.

P.S. This question is crossposted from mathstackexchange, where it receives no answer even after I put a bounty.

I would like to show that $\mathbb P(\operatorname{Binomial}(n,p) = \operatorname{Binomial}(n,q))$ decreases when $n$ increases for a fixed pair $(p,q)$. This can be reformulated as $\mathbb P(X_n=0)$ decreases where $X_n=\sum_{i=1}^n S_i$ is a lazy random walk where $S_i=-1,0,1$ with probability $p(1-q),pq+(1-p)(1-q),(1-p)q$. For $p=q$ this can be done by characteristic functions. Any ideas for general $(p,q)$? I feel like this should be well known.

P.S. This question is crossposted from mathstackexchange, where it receives no answer even after I put a bounty.

Edit: To be clear the actual question is to show for any pair of $(p,q)$, we have $P(Bin(n,p)=Bin(n,q))\le P(Bin(n-1,p)=Bin(n-1,q))$ for any $n$, and everything is independent of each other.

I would like to show that $\mathbb P(Binomial(n,p)=Binomial(n,q))$ decreases when $n$ increases for a fixed pair $(p,q)$. This can be reformulated as $\mathbb P(X_n=0)$ decreases where $X_n=\sum_{i=1}^n S_i$ is a lazy random walk where $S_i=-1,0,1$ with probability $p(1-q),pq+(1-p)(1-q),(1-p)q$. For $p=q$ this can be done by characteristic functions. Any ideas for general $(p,q)$? I feel like this should be well known.

P.S. This question is crossposted from mathstackexchange, where it receives no answer even after I put a bounty.

I would like to show that $\mathbb P(\operatorname{Binomial}(n,p) = \operatorname{Binomial}(n,q))$ decreases when $n$ increases for a fixed pair $(p,q)$. This can be reformulated as $\mathbb P(X_n=0)$ decreases where $X_n=\sum_{i=1}^n S_i$ is a lazy random walk where $S_i=-1,0,1$ with probability $p(1-q),pq+(1-p)(1-q),(1-p)q$. For $p=q$ this can be done by characteristic functions. Any ideas for general $(p,q)$? I feel like this should be well known.

P.S. This question is crossposted from mathstackexchange, where it receives no answer even after I put a bounty.