Gambler's ruin: The fair game is the longest Consider a gambler who, in every trial of a game, wins or loses a dollar with
probability $p\in\left(  0,1\right)  $ and $q=1-p$, respectively. Let his
initial capital be $z>0$ and let him play against an adversary with the same
capital $z>0$. The game continues until one of the players is ruined.
In the language of random variables, this amounts to consider a sequence
$X_{1}^{\left(  p\right)  },X_{2}^{\left(  p\right)  },...\ $of random
variables (on a sample space $\Omega$) taking on two values $+1$ and $-1$ with
probabilities $\Pr\left[  X_{n}^{\left(  p\right)  }=+1\right]  =p$ and
$\Pr\left[  X_{n}^{\left(  p\right)  }=-1\right]  =q$. In particular,
$X_{n}^{\left(  p\right)  }$ describes the gambler's gain on the $n$th trial
and
$$
S_{n}^{\left(  p\right)  }\left(  \omega\right)  =X_{1}^{\left(  p\right)
}\left(  \omega\right)  +\cdots+X_{n}^{\left(  p\right)  }\left(
\omega\right)  ,\quad S_{0}^{\left(  p\right)  }\left(  \omega\right)
\equiv0\qquad\forall \,\omega\in\Omega
$$
describes his net cumulated gain after $n$ trials. The duration of the
game is the random number of trials before he is either ruined or wins the
game (and his adversary wins or is ruined)
$$
D^{\left(  p\right)  }\left(  \omega\right)  =\min\left\{  t\in\mathbb{N}%
:\left\vert S_{t}^{\left(  p\right)  }\left(  \omega\right)  \right\vert
=z\right\}  \qquad\forall \,\omega\in\Omega.
$$

Q. I need a reference or a simple proof of the folk result according to which,
  for each $p\in\left(  0,1\right)  $,
  $$
\Pr\left[  D^{\left(  p\right)  }\leq s\,\right]  \geq\Pr\left[  D^{\left(
1/2\right)  }\leq s\,\right]  \qquad\forall\, s=1,2,...
$$

 A: Let us show a bit more, that $F_p(s):=P(D^{(p)}\le s)$ is nondecreasing in $|p-1/2|$, for any real $s$. That is, take any $p$ and $p_1$ in $(0,1)$ such that $|p_1-1/2|>|p-1/2|$, and let, for brevity, $F:=F_p$ and $G:=F_{p_1}$. We shall show that then $F\le G$. 
Indeed, by the formula (10), referenced by Carlo Beenakker, 
\begin{equation*}
 f_p(n):=P(D^{(p)}=n)=a_{n,z} b_{p,z}(pq)^{n/2} 
\end{equation*}
for natural $n$, where $a_{n,z}$ depends only on $n$ and $z$; $b_{p,z}:=(p/q)^{z/2}+(q/p)^{z/2}$; and $q:=1-p$. Since $pq$ decreases in $|p-1/2|$, the family $(f_p)$ of pmf's has the monotone likelihood ratio (MLR) property, meaning that 
\begin{equation*}
 r(n):=\frac{f(n)}{g(n)}
\end{equation*}
increases in $n\in N$, where $N:=N_z:=\{n\colon a_{n,z}\ne0\}$, $f:=f_p$, $g:=f_{p_1}$, 
and $p$ and $p_1$ are as before. 
Therefore, for any given real $s$, 
\begin{align*}
 F(s)-G(s)&=F(s)[1-G(s)]-G(s)[1-F(s)] \\ 
 &=\sum_{m,n\colon\, N\ni m\le s<n\in N}[f(m)g(n)-g(m)f(n)] \\ 
 &=\sum_{m,n\colon\, N\ni m\le s<n\in N}[r(m)-r(n)]g(n)g(m)\le0, 
\end{align*}
since $r$ is increasing. Thus, indeed $F\le G$. 
