Trading Against a Better-Informed Opponent

A classic interview game: the interviewer picks a number secretly (or knows something you don't — a card, a die roll, a piece of news) and asks you to name a price at which you'll buy or sell against them. They will only trade when it benefits them. You cannot ask what they know. What price do you name?

Why "your best guess" is the wrong instinct

The instinctive move is to quote your best unconditional estimate of the value — but this ignores that you only get traded with when you're wrong in your counterparty's favor. This is the same structural problem as the classic "used car" (lemons) market: a seller who knows the car's true condition will only sell at a price that's good for them, so any buyer naively offering the average car's worth loses money, because only below-average cars get sold to them. The fix isn't guessing harder — it's recognizing that being selected to trade is itself informative and pricing around that selection, not around the naive average.

Worked example

The interviewer will reveal a card drawn uniformly from 2 through 10 (all equally likely) and ask you to buy it from them at a price you name, sight unseen, with the rule that they'll only sell if your price is at or above the card's value. If you price at the unconditional average, $(2+3+\cdots+10)/9 = 6$, they'll sell you every card worth 6 or less (2 through 6) and keep every card worth more than 6 (7 through 10) — you only ever buy the cheap half. Your actual average purchase price paid is 6, but the average value of what you receive is $(2+3+4+5+6)/5 = 4$: you lose 2 on average, every single trade, despite pricing at the "correct" unconditional mean. The correct approach is to solve for the price $p$ where the average value of cards at or below $p$ equals $p$ itself — the point where you stop being adversely selected. Checking $p = 4$: cards 2, 3, 4 would sell, averaging 3, which is below 4 — still losing. Checking progressively lower values shows that the only price where you break even is $p = 2$, the very lowest possible card value: at any price above the minimum, the seller only ever hands you cards worth less than what you paid, because they always keep anything worth more than your bid. Bidding exactly 2 means you only ever buy the one card that's worth exactly what you paid.

What this means in practice

This is the mathematics behind why market makers widen sharply, or refuse to quote at all, when they suspect a counterparty is systematically better informed — the "average value" of what you'd trade is not the average value of the underlying, it's the average value conditional on someone choosing to trade with you at that price, which is always worse. Real desks manage this with information about counterparty identity, order size, and venue rather than assuming every trade is adversarial to this degree, but the underlying selection effect never fully disappears.