Weak Value Comparison Degrades to 1/n on the Needle Profile
Counterexample. If the sample-then-threshold auction uses weak value comparison $p \le v$ as the acceptance rule (accepting all ties), it achieves expected welfare only $(1/n) \cdot \max v$ on the needle profile — which is not a constant competitive ratio.
This complements the strict-comparison counterexample ([[mechanism_design.auction.online.secretary_strict_comparison_fails]]), which gives welfare $= 0$ outright. Weak comparison avoids that extreme failure but introduces a subtler one: it cannot distinguish a worthless bidder who happens to match a zero threshold from the valuable bidder who genuinely exceeds it.
Construction
Fix $M > 0$ and $n \ge 2$. The needle profile is
$$ v = (M,\, 0,\, 0,\, \ldots,\, 0). $$
One needle bidder has value $M$; the remaining $n - 1$ haystack bidders have value $0$. The sample-then-threshold rule observes $\lfloor n/2 \rfloor$ bidders, then posts the observed maximum as the threshold. Under weak comparison, the acceptance test is $p \le v_i$.
Case analysis over the needle's arrival position
Under a uniformly random permutation, the needle is equally likely to occupy any of the $n$ positions.
Case 1: needle in the observe phase (position $k < \lfloor n/2 \rfloor$). The threshold value becomes $\ge M$. Every phase-2 bidder has value $0$, and $M \le 0$ is false. Nobody clears. Welfare $= 0$.
Case 2: needle in phase 2, but not first (position $k > \lfloor n/2 \rfloor$). All observed bidders have value $0$, so the threshold drops to $p = 0$. The first phase-2 bidder is a haystack bidder with value $0$; the test $0 \le 0$ is true. This haystack bidder is accepted with welfare $= 0$. The needle never gets a chance.
Case 3: needle is first in phase 2 (position $k = \lfloor n/2 \rfloor$). Threshold is $p = 0$. The needle faces $0 \le M$, which is true. Accepted. Welfare $= M$.
Expected welfare is $M/n$
Exactly one of $n$ equally likely positions (Case 3) gives welfare $M$; all others give $0$. Hence
$$ \mathbb{E}[\mathrm{welfare}] = \frac{1}{n} \cdot M = \frac{1}{n} \cdot \max_i v_i. $$
For any constant $c > 0$, choosing $n > 1/c$ gives $\mathbb{E}[\mathrm{welfare}] < c \cdot \max v$.
Why it fails
The root cause is that weak comparison cannot distinguish a haystack bidder matching a zero threshold by coincidence from the needle bidder genuinely exceeding it. When $p = 0$ and a haystack bidder has $v = 0$, the test $0 \le 0$ is vacuously true — the first phase-2 arrival wins regardless of value.
How lexicographic tie-breaking fixes this
With the full lexicographic threshold ([[mechanism_design.auction.online.secretary_quarter_competitive]]), Case 2 plays out differently:
- The threshold has value $p = 0$ and identity component $\bar{b} = b_{\max}$, the largest identity seen in phase 1.
- A haystack bidder $j$ arriving first in phase 2 faces the value tie $p = v_j = 0$, so acceptance falls to the identity test $\bar{b} \le b_j$. But $b_j < b_{\max} = \bar{b}$ (by identity injectivity — the largest-identity haystack bidder was observed in phase 1 on the favourable event). The haystack bidder is rejected.
- This rejection continues for every subsequent haystack bidder, until the needle arrives. The needle clears strictly: $0 < M$. Welfare $= M$.
On the favourable event (probability $\ge 1/4$), the lex rule rejects all haystack bidders and accepts the needle, giving expected welfare $\ge M/4$.
Summary
| Acceptance rule | Needle welfare | Competitive? |
|---|---|---|
| $p < v$ (strict) | $0$ always | No ([[mechanism_design.auction.online.secretary_strict_comparison_fails]]) |
| $p \le v$ (weak) | $M/n$ in expectation | No (this) |
| lex | $\ge M/4$ in expectation | Yes ([[mechanism_design.auction.online.secretary_quarter_competitive]]) |
Remarks
Lean formalization
Weak comparison is modelled by WeakComparison.auction, which sets the
identity component of the threshold to $\bot$ — since $\bot \le b$
holds for all $b$, the acceptance condition degenerates to $p \le v$.
The lemma maxPairFold_fst_zero shows that the threshold value is $0$
when all observed bids have value $0$. The theorem
welfare_eq_zero_needle_last proves welfare $= 0$ for the $n = 3$ case
with the needle at the last position (Case 2).
References
- [Roughgarden 2016, Lecture 2, Problem 2.1(c)] Tim Roughgarden, Twenty Lectures on Algorithmic Game Theory, Cambridge University Press. Analysis of the online auction (stated for distinct values, which excludes the needle profile).