Guess Two-Thirds of the Average
A group of $n$ players each simultaneously write down an integer between 1 and 100 (inclusive). The winner is the person whose number is closest to $\frac{2}{3}$ of the average of all submitted numbers. Ties are broken randomly.
- What is the unique Nash equilibrium of this game?
- Walk through the iterated reasoning that leads to it.
- In practice, when this game is played with real people, what do you actually observe? Why does the outcome differ from the Nash equilibrium?
Hints
- The target is always strictly less than the average because the multiplier $\frac{2}{3}$ is less than 1. What does that tell you about numbers near the top of the range?
- If no rational player picks above some threshold $M$, the target drops to at most $\frac{2}{3}M$. Apply this reasoning repeatedly.
- After $k$ rounds of elimination, the upper bound is 00 \times (2/3)^k$. Since we need integers and this sequence converges to 0, the process terminates at 1.
Worked Solution
How to Think About It: This is the classic "beauty contest" game, and it is one of the cleanest illustrations of iterated dominance in all of game theory. The key insight is that you are not trying to guess what $\frac{2}{3}$ of the average is -- you are trying to guess what everyone else thinks $\frac{2}{3}$ of the average is. And they are trying to guess what you think. This recursion is the whole game. Before doing any math, notice the fraction $\frac{2}{3} < 1$ -- this means the target is always strictly below the average. That contraction is what drives everything to collapse.
Quick Estimate: If players pick uniformly at random, the average is around 50, so the target is about 33. But if everyone targets 33, the average drops to 33, and the new target is $\frac{2}{3} \times 33 \approx 22$. If everyone targets 22, the target becomes about 15. You can feel this spiraling downward -- each round of "they'll figure it out" pulls the number lower by a factor of $\frac{2}{3}$. After 10 rounds of reasoning:
00 \times (2/3)^{10} \approx 1.7$. So even with a rough feel, you know the equilibrium is at the bottom of the range.Approach: Iterated elimination of strictly dominated strategies.
Formal Solution:
Define the game: each player $i$ chooses $x_i \in \{1, 2, \ldots, 100\}$. The target is $T = \frac{2}{3} \cdot \frac{1}{n}\sum_{j=1}^n x_j$. The player whose choice is closest to $T$ wins.
Round 1: Since every $x_j \leq 100$, the average is at most 100, so $T \leq \frac{2}{3} \times 100 \approx 66.7$. Any number above 67 is strictly dominated by 67 -- it can never be closer to the target than 67 is. Eliminate $\{68, 69, \ldots, 100\}$.
Round 2: Now all rational players know the effective maximum is 67. So $T \leq \frac{2}{3} \times 67 \approx 44.7$. Any number above 45 is now dominated. Eliminate $\{46, \ldots, 67\}$.
Round $k$: After $k$ rounds of elimination, the upper bound is
00 \times (2/3)^k$. The sequence of upper bounds is:$100 \to 67 \to 45 \to 30 \to 20 \to 13 \to 9 \to 6 \to 4 \to 3 \to 2 \to 1$
Since we are restricted to integers and the multiplier $\frac{2}{3} < 1$, this process converges to 1 in finitely many steps.
At the fixed point: If everyone chooses 1, the average is 1, the target is $\frac{2}{3}$, and 1 is the closest integer. No player can profitably deviate -- choosing anything higher moves you further from $\frac{2}{3}$. This is a Nash equilibrium.
Uniqueness: The iterated elimination is strict at every round (any number above the bound is strictly worse than the bound itself, regardless of what others do). Since strict dominance elimination has a unique outcome, the Nash equilibrium is unique.
Part 3 -- What happens in practice:
In experiments (Nagel 1995, and many replications), first-time players typically choose numbers in the range 20-35. This corresponds to roughly 1-2 levels of strategic reasoning: - Level 0: Pick randomly, average $\approx 50$ - Level 1: Best-respond to Level 0 -- pick $\frac{2}{3} \times 50 \approx 33$ - Level 2: Best-respond to Level 1 -- pick $\frac{2}{3} \times 33 \approx 22$
Most people stop at Level 1 or 2. The Nash equilibrium requires infinite levels of mutual reasoning ("I know that you know that I know..."), which is unrealistic. When the game is played repeatedly with the same group, choices do converge toward 1 over many rounds.
Answer: The unique Nash equilibrium is for all players to choose $\mathbf{1}$. It is reached by iterated elimination of dominated strategies: each round of reasoning reduces the upper bound by a factor of $\frac{2}{3}$, which converges to 1. In practice, real players typically choose 20-35, reflecting only 1-2 levels of iterated reasoning.
Intuition
This game is the purest classroom example of iterated dominance and common knowledge of rationality. The logic at each step is simple -- if the target is always below the average, why pick a high number? But the full Nash equilibrium requires not just that you are rational, but that you know everyone else is rational, and that they know you know, and so on to infinity. Each additional "level" of mutual knowledge shaves the upper bound by another factor of $\frac{2}{3}$. The equilibrium only obtains when this tower of knowledge is infinitely tall, which is why it almost never appears in practice.
For quant interviews, this problem tests whether you understand the difference between a Nash equilibrium as a mathematical fixed point and actual human behavior. It also connects to deeper ideas in market microstructure: markets with rational agents should not have trades (no-trade theorem), yet we observe enormous volumes. The gap between the theory and the data is the same gap you see in this game -- bounded rationality, heterogeneous beliefs, and finite depth of reasoning. A good candidate mentions the level-$k$ model and notes that the "correct" answer in a room full of traders is not 1 but something like 15-25, because you are best-responding to real people, not to the Nash equilibrium.