Game theory nash equilibrium question

Question

Suppose $N = 3$ shepherds all graze their sheep on the same land (of size N). Each day, they can choose to have their sheep either

• Graze sustainably (‘Graze’), or

• Gorge on as much grass as possible (‘Gorge’).

A shepherd’s sheep gain 1 pound of meat if they Graze, or N pounds if they Gorge. Additionally, for each other flock that gorges, a a shepherd’s sheep lose an extra 2 pounds regardless of whether they graze or gorge themselves; there is no loss if other flocks graze. Shepherds each attempt to maximize the weight of their own sheep

What NEs exist if the game is played only once?

The answer is:

If the game is played only once, for any number x of other shepherds Gorging, a shepherd gets $1−2 · x$ for Grazing or $N −2 · x $ for Gorging. As $N = 3$, Gorging is always better, and each shepherd wants to Gorge. Thus, Gorge, Gorge, Gorge is the unique NE.

My question is: Why exactly do the shepherds get $N-2x$ or $1-2x$. Since if the $x$ flocks that are on the land are grazing, they can gain $1$ or $N$ ? So why exactly is there subtraction?

Answer 1

The question states that "for each other flock that gorges, a a shepherd’s sheep lose an extra 2 pounds regardless of whether they graze or gorge themselves." So if $x$ is a number of flocks that gorges then sheep will lose $2x$ pounds regardless of its choice.

Now,

• the payoffs to shepherd, if no other flock gorges, is $1$ if his herd grazes and $N$ if his heard gorges
• If one other herd gorges, he gets $1-2$ if his herd grazes, and $N-2$ if his herd gorges.
• More generally, if $x\geq 0 $ other herds gorges, he gets $1-2x$ if his herd grazes and $N-2x$ if his herd gorges.

So clearly, gorging is always better.