The eternal vertex colouring game, recently introduced byKlostermeyer and Mendoza (2018), is a version of the vertex colouring game, a well studied graph game. The game is played by two players, Alice and Bob on a graph $G$ with a set of colours $\{1,\dots ,k\}$. The game consists of rounds, such that in each round, every vertex is coloured exactly once. In the first round, players alternate turns with Alice playing first. During their turn, each player first picks yet uncoloured vertex and then colours it by any colour so that the resulting partial colouring of the graph is proper (if at least one such colour exists). During all further rounds, players keep choosing vertices alternately. After choosing a vertex, the player assigns a colour to the vertex which is distinct from its current colour such that the resulting colouring is still proper. Each vertex retains its colour between rounds until it is recoloured. Bob wins if at any point the chosen vertex does not have a legal recolouring, while Alice wins if the game is continued indefinitely. The eternal game chromatic number ${\chi }_g^{\infty }\left(G\right)$ is the smallest number $k$ such that Alice has a winning strategy.

In this note, we study the eternal game chromatic number of random graphs. We show that with high probability ${\chi }_g^{\infty }\left(G_{n,p}\right)=(\frac{p}{2}+o\left(1\right))n$ for odd $n$, and also for even $n$ when $p=\frac{1}{k}$ for some $k\in \mathbb{N}$. The upper bound applies for even $n$ and any other value of $p$ as well, but we conjecture in this case this upper bound is not sharp. Finally, we answer a question posed by Klostermeyer and Mendoza.

永恒顶点着色游戏是由Klostermeyer和Mendoza（2018）最近推出的，是顶点着色游戏的一个版本，是一个经过深入研究的图形游戏。该游戏由两个玩家Alice和Bob玩在一个图表上$G$ 有一套颜色 $\{\mathrm{1个}，\dots ，ķ\}$。该游戏由多个回合组成，因此在每个回合中，每个顶点都被精确着色一次。在第一轮中，玩家交替轮流，而爱丽丝则先行。在回合中，每个玩家首先选择尚未着色的顶点，然后以任何颜色对其进行着色，以使图形产生的部分着色是正确的（如果存在至少一种这样的颜色）。在接下来的所有回合中，玩家继续交替选择顶点。选择一个顶点后，播放器将一种颜色分配给该顶点，该颜色与其当前颜色不同，因此所得的颜色仍然是正确的。每个顶点在回合之间保留其颜色，直到重新着色为止。如果所选顶点在任何时候都没有合法的重新着色，则Bob获胜；而如果游戏无限期继续，则Alice获胜。永恒游戏的色数${\chi }_G^{\infty }（G）$ 是最小的数字 $ķ$ 这样爱丽丝就有了制胜法宝。

在本文中，我们研究了随机图的永恒游戏色数。我们很有可能证明${\chi }_G^{\infty }（G_{ñ，p}）=（\frac{p}{\mathrm{2个}}+Ø（\mathrm{1个}））ñ$ 奇数 $ñ$，甚至 $ñ$ 什么时候 $p=\frac{\mathrm{1个}}{ķ}$ 对于一些 $ķ\in \mathbb{ñ}$。上限适用于偶数$ñ$ 以及任何其他值 $p$同样，但是在这种情况下，我们猜想这个上限并不尖锐。最后，我们回答克洛斯泰米尔和门多萨提出的问题。