Two-Sided Market Making With Overround

Worked Solution

How to Think About It: You are a market maker quoting odds on a yes/no event. Your edge comes from the overround -- the implied probabilities you quote sum to more than 1, so on average you collect more than you pay out. But you are not trading against a single counterparty; customers arrive with random beliefs and only trade when they think they have an edge. The question is: how does this adverse selection eat into your vig?

The key observation is that customers with extreme beliefs (very high $p_c$ or very low $p_c$) are the ones who trade. Customers in the middle see no edge on either side and walk away. Your profit depends on the fraction of customers who trade and how much you make (or lose) on each trade on average.

Quick Estimate: Take $p = 0.5$ and symmetric odds $O_{\text{yes}} = O_{\text{no}} = 1.8$ (implied overround $2/1.8 \approx 1.11$). The breakeven belief for "yes" is $p_c > 1/1.8 \approx 0.556$, and for "no" is $p_c < 1 - 1/1.8 \approx 0.444$. About $44.4\%$ of customers bet each side, and $11.1\%$ walk away. On each trade, your expected profit is $1 - 0.5 \times 1.8 = 0.10$. Total: $2 \times 0.444 \times 0.10 \approx 0.089$ per customer. That is a healthy vig.

Approach: Define the breakeven beliefs, split customers into regions, and compute the expected profit in each region. Then optimize over $(O_{\text{yes}}, O_{\text{no}})$.

Formal Solution:

Define the breakeven beliefs:

$$a = \frac{1}{O_{\text{yes}}}, \qquad b = 1 - \frac{1}{O_{\text{no}}}$$

A customer bets "yes" when $p_c > a$ (EV of yes is positive) and bets "no" when $p_c < b$ (EV of no is positive). For a viable market with overround, we need $O_{\text{yes}} > 1$, $O_{\text{no}} > 1$, and $1/O_{\text{yes}} + 1/O_{\text{no}} > 1$, which gives $a > b$. In this regime, customers with $p_c \in [b, a]$ see no positive EV on either side and do not trade.

Since $p_c \sim \text{Uniform}[0,1]$:

• Customer bets "no" (probability $b$): market maker's expected profit per trade is $1 - (1-p) O_{\text{no}}$.
• Customer bets "yes" (probability $1 - a$): market maker's expected profit per trade is $1 - p \, O_{\text{yes}}$.
• Customer walks away (probability $a - b$): profit is $0$.

Total expected profit:

$$E[\pi] = b \bigl(1 - (1-p)\, O_{\text{no}}\bigr) + (1 - a)\bigl(1 - p\, O_{\text{yes}}\bigr)$$

Substituting $a = 1/O_{\text{yes}}$ and $b = 1 - 1/O_{\text{no}}$:

$$E[\pi] = \left(1 - \frac{1}{O_{\text{no}}}\right)\!\left(1 - (1-p)\, O_{\text{no}}\right) + \left(1 - \frac{1}{O_{\text{yes}}}\right)\!\left(1 - p\, O_{\text{yes}}\right)$$

Expanding and simplifying:

$$E[\pi] = 3 - p\, O_{\text{yes}} - (1-p)\, O_{\text{no}} - \frac{1}{O_{\text{yes}}} - \frac{1}{O_{\text{no}}}$$

This is the answer to part 1.

Optimization (Part 2): To maximize over $(O_{\text{yes}}, O_{\text{no}})$, take partial derivatives:

$$\frac{\partial E[\pi]}{\partial O_{\text{yes}}} = -p + \frac{1}{O_{\text{yes}}^2} = 0 \quad \Longrightarrow \quad O_{\text{yes}}^{*} = \frac{1}{\sqrt{p}}$$

$$\frac{\partial E[\pi]}{\partial O_{\text{no}}} = -(1-p) + \frac{1}{O_{\text{no}}^2} = 0 \quad \Longrightarrow \quad O_{\text{no}}^{*} = \frac{1}{\sqrt{1-p}}$$

The second derivatives are $-2/O^3 < 0$, confirming this is a maximum. Substituting back:

$$E[\pi]^{*} = 3 - \frac{p}{\sqrt{p}} - \frac{1-p}{\sqrt{1-p}} - \sqrt{p} - \sqrt{1-p} = 3 - 2\sqrt{p} - 2\sqrt{1-p}$$

To verify this is strictly positive for all $p \in (0,1)$: by the Cauchy-Schwarz inequality (or direct calculus), $\sqrt{p} + \sqrt{1-p} \leq \sqrt{2}$, with equality at $p = 1/2$. Therefore:

$$E[\pi]^{*} \geq 3 - 2\sqrt{2} \approx 3 - 2.828 = 0.172 > 0$$

The minimum profit occurs at $p = 1/2$ with symmetric optimal odds $O_{\text{yes}}^{*} = O_{\text{no}}^{*} = \sqrt{2} \approx 1.414$ (implied overround $2\sqrt{2}/2 \cdot 100\% \approx 141\%$).

Sanity Check: The optimal overround is $1/O_{\text{yes}}^{*} + 1/O_{\text{no}}^{*} = \sqrt{p} + \sqrt{1-p}$. This exceeds 1 for all $p \in (0,1)$ (since each term is positive and their sum at $p = 0$ or $p = 1$ equals 1, and it is concave with interior maximum $\sqrt{2} > 1$). The implied probabilities are $\sqrt{p}$ and $\sqrt{1-p}$, which compress probabilities toward $1/2$ -- exactly the "shading" a market maker applies to protect against adverse selection.

Answer:

Expected profit per customer:

$$E[\pi] = 3 - p\, O_{\text{yes}} - (1-p)\, O_{\text{no}} - \frac{1}{O_{\text{yes}}} - \frac{1}{O_{\text{no}}}$$

Optimal odds: $O_{\text{yes}}^{*} = 1/\sqrt{p}$, $O_{\text{no}}^{*} = 1/\sqrt{1-p}$. Maximum profit: $3 - 2\sqrt{p} - 2\sqrt{1-p} > 0$ for all $p \in (0,1)$.