Expected survival time in Russian Roulette not monotone?

Question

Let $a, n$ be positive integers with $a < n$. A revolver with $n$ chambers is loaded with $a$ bullets, where the distribution is uniform among all $\binom{n}{a}$ possible choices of $a$ objects filling $n$ slots.

A lone player is playing Russian Roulette using the loaded gun. In this game the gun is spun, which we conceptualise as a chamber being picked uniformly at random. Subsequently, the player fires at his own head without spinning the gun again - which we conceptualise as going through the chambers deterministically in sequence.

In this game, the player intends to shoot himself until he dies. We assume that there is no chance of surviving a headshot - the player dies the moment a non-empty chamber is encountered.

We denote by $X^{(a, n)}_t$ the survival process of a player using the loaded gun, where at each time $t$, $X^{(a, n)}_t$ takes value $1$ if the player is still alive after the $t$’th round, and $0$ otherwise.

Write $S^{(a, n)} := \sum_{t = 1}^{\infty} X^{(a, n)}_t$ for the survival time of the player.

Question: Is it true that the mean survival time $\mathbb E[S]$ is not monotone decreasing in the ratio $\frac{a}{n}$ of bullets to chambers?

That is, does there exist $(a, n), (a^{*}, n^{*})$ with $\frac{a^{*}}{n^{*}} > \frac{a}{n}$, but

$$\mathbb E[S^{(a^*, n^*)}] > \mathbb E[S^{(a, n)}]?$$

Bonus points if it can be shown that $n$ can be taken arbitrarily large.

Answer 1

Let $S$ be the survival time. Then $$P(S\ge s)=\binom{n-s}a\Big/\binom na$$ for $s=0,1,\dots$. So, $$ES=-1+\sum_{s=0}^\infty P(S\ge s)=\frac{n+1}{a+1}-1.$$

So, $n=3$, $a=1$, $n^*=8$, $a^*=3$ will do.

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More generally, any integers $n$, $a$, $m:=n^*$, $b:=a^*$ such that $n\ge2$, $1\le a<n$, $m>n$, and $$\frac{ma}n<b<\frac{ma+m+a-n}{n+1} \tag{10}\label{10} $$ will do. Note that $$\frac{ma+m+a-n}{n+1}-\frac{ma}n=\frac{(m-n) (n-a)}{n (n+1)}>1$$ if $m$ is large enough, so that then there is an integer $b$ satisfying \eqref{10}. So, indeed, $n,a,m,b$ can be taken arbitrarily large.