The Price is Right Bidding Strategies
Yes. Note that in the paper the authors assume that the action space is continuous, so it is possible to bid in arbitrarily small fractions of a dollar.

Suppose players $i\in\{1,2,3\}$ have bids $b_i$ that, without loss of generality, are ordered $b_1&lt;b_2&lt;b_3$. Write $p$ for the actual price.

If we forget about player four for a moment, then players' probability of winning are

$$\Pr(1\text{ wins})=\Pr(b_1\leq p&lt; b_2)\equiv \pi_1$$
$$\Pr(2\text{ wins})=\Pr(b_2\leq p&lt; b_3)\equiv \pi_2$$
$$\Pr(3\text{ wins})=\Pr(b_3\leq p)\equiv \pi_3.$$

Suppose $\pi_1&gt;\pi_2,\pi_3$. Then the optimal thing for 4 to do is to bid $b_4=b_1+\epsilon$ ($\epsilon$ positive and small). This bis would result in
$$\Pr(1\text{ wins})=0$$
$$\Pr(2\text{ wins})=\Pr(b_2\leq p&lt; b_3)$$
$$\Pr(3\text{ wins})=\Pr(b_3&lt; p)$$
$$\Pr(4\text{ wins})=\Pr(b_1\leq p&lt; b_2).$$

One can easilly check that $\pi_1&gt;\pi_2,\pi_3$ implies that $\Pr(4\text{ wins})$ is bigger than any rival winning probability. Likewise, if we had $\pi_2&gt;\pi_1,\pi_3$ then the optimal thing for $4$ to do would be to bid $b_4=b_2+\epsilon$.



edit: The above reasoning works if we take $\pi_1&gt;\pi_2,\pi_3$ (with strict inequality). But what if the inequality is not strict?

Suppose that $b_1&lt;b_2&lt;b_3$, and that $\pi_1=\pi_2&gt;\pi_3,\Pr(p&lt;b_1)$. Then it seems like $b_4=b_1+\epsilon$ must yield a lower winning probability than that enjoyed by player 2 for any positive $\epsilon$. indeed we would have
$$\Pr(b_1+\epsilon\leq p&lt;b_2)&lt;\Pr(b_1\leq p&lt;b_2)=\Pr(b_2\leq p&lt;b_3).$$
Similarly, $b_4=b_2+\epsilon$ will yield a lower winning probability than that enjoyed by player 1 for any positive $\epsilon$.

I think that, in equilibrium, this won't happen so I guess the result is fine. But it makes the proof seem a bit hand-wavey.