Probability Practice: Auction Theory Problems Based on the 1517 Renaissance Portrait Auction
educationstatisticsart

Probability Practice: Auction Theory Problems Based on the 1517 Renaissance Portrait Auction

ccalculation
2026-02-13
11 min read
Advertisement

Classroom-ready auction theory probability problems using a 1517 portrait case: expected value, Bayesian updates, winner's curse, and spreadsheet simulations.

Hook: Stop wasting class time on messy algebra — teach auction theory with a real postcard-sized 1517 Hans Baldung Grien work and spreadsheet-ready problems

Statistics instructors and students often struggle with abstract probability exercises that don't connect to real decisions. If you want classroom-ready probability problems that teach expected value, estimation, and the winner's curse using a single high-profile auction scenario, this set of practice problems is ready to drop into your next lab or exam.

Inspired by the discovery reported in Artnet News (a postcard-sized 1517 Hans Baldung Grien work that could fetch up to $3.5M), these exercises use that auction as a motivating data point. Use them to teach probability, Bayesian updating, and Monte Carlo estimation — all with spreadsheet-ready problems.

The 2026 context: why these problems matter now

By 2026 the art-auction world has become progressively more data-driven. Late 2025 and early 2026 saw wider adoption of AI-powered valuation models, expanded online bidding platforms, and blockchain-based provenance records that provide richer signals for bidders and appraisers. That means statistics students need practice with:

  • Translating appraisal signals into probabilistic estimates
  • Modeling strategic bidding under uncertainty (private vs. common value settings)
  • Using Monte Carlo and spreadsheet tools to compute expected revenue and bidding risk

How to use this worksheet pack

This article gives: (A) five classroom-ready problems, each with step-by-step solutions or solution paths; (B) spreadsheet formulas for Excel/Google Sheets; and (C) a teacher's rubric and time estimate. Everything is designed for a 50–90 minute lab or a take-home assignment. If you want a download-ready .xlsx template, follow the call-to-action at the end.

Learning objectives

  • Compute probabilities for order statistics and use them to find event probabilities (e.g., “sale exceeds reserve”).
  • Calculate expected revenue under auction rules (second-price / Vickrey) with a reserve price.
  • Use Bayesian updating for appraisal signals.
  • Simulate auctions using RAND()/NORM.INV() and summarize results with averages, histograms, and confidence intervals.

Problem 1 — Private-values second-price auction (Uniform model)

Scenario

A small, rare 1517 portrait is estimated publicly with a high-end value of $3.5 million (we treat this as an upper technical bound for theoretical practice). Imagine 4 independent bidders. Each bidder's private valuation is drawn i.i.d. from Uniform(0, A) with A = $3.5M. The auction is a sealed-bid second-price (Vickrey) auction with a reserve price r = $500k. Bidders bid their valuation.

Questions

  1. What is the probability the painting sells (i.e., the highest bid clears the reserve)?
  2. What is the probability the final sale price exceeds the reserve (i.e., second-highest > r)?
  3. Compute the seller's expected revenue (in millions) under truthful bidding.

Solution outline (teacher-ready)

Key facts for Uniform(0,A): if X_(j) is the j-th order statistic out of n draws then X_(j)/A ~ Beta(j, n+1-j). For n=4, the second-highest is X_(3) (the 3rd order statistic).

Step 1 — probability item sells
The item sells if at least one bidder's valuation > r. Probability that a single valuation ≤ r is (r/A). So P(no one > r) = (r/A)^4. Therefore P(sells) = 1 - (r/A)^4.

Numeric: r/A = 0.5/3.5 = 1/7 ≈ 0.142857. (r/A)^4 ≈ 1/2401 ≈ 0.0004165. So P(sells) ≈ 0.9995835 (≈99.96%).

Step 2 — probability final price exceeds reserve
For the sale price in a Vickrey auction to exceed r, we need at least three valuations > r (because the second-highest must exceed r). Let p = P(valuation > r) = (A - r)/A = 6/7 ≈ 0.8571429. For n=4, P(at least 3 > r) = C(4,3) p^3 (1-p)^1 + p^4 = p^3 (4(1-p) + p). Numerically this equals ≈0.8996 (≈89.96%).

Step 3 — expected seller revenue
Revenue rules in Vickrey with reserve r:

  • If at least two bids exceed r, price = second-highest X_(3).
  • If exactly one bid exceeds r, winner pays r.
  • If zero exceed r, sale fails (revenue = 0).
So E[revenue] = E[X_(3) * I{X_(3) > r}] + r * P{exactly 1 valuation > r}.

Compute E[X_(3) * I{X_(3) > r}] by integrating the Beta-scaled density. For n=4, j=3, the density gives a closed-form integral. Using A = 3.5 and r = 0.5 (in millions), this evaluates to ≈ 2.096125 (millions). P{exactly 1 > r} ≈ 24/2401 ≈ 0.009996, so r * P{=1} ≈ 0.004998M. Thus expected revenue ≈ $2.1011M.

Teacher note: show the spreadsheet implementation — use Excel: A2:A5 = RAND()*3.5 to simulate 4 valuations (in millions), B1 = LARGE(A2:A5,2) for second largest, compute revenue rule with IF logic. Repeat with 50,000 rows and summary averages to verify theory.

Problem 2 — Common-value signals and the winner's curse (simulation)

Scenario

The community believes the true market value V of the painting is centered near $2.2M but uncertain. Model V ~ Normal(μ=2.2, σ_v=0.4) (millions). Each of 5 bidders receives an independent signal S_i = V + e_i where e_i ~ Normal(0, σ_e = 0.5). Bidders are risk neutral and do not want to fall victim to the winner's curse.

Questions

  1. Explain how a bidder should form a truthful estimate of the painting's value given her private signal s (i.e., find E[V | S_i = s]).
  2. Show why naive bidding (bidding E[V | S_i = s]) can lead to the winner's curse in the auction with 5 bidders.
  3. Use simulation (spreadsheet) to estimate the expected overpayment when bidders bid the naive posterior mean in a second-price auction.

Solution steps

Analytic update: Because V and the errors are Gaussian, the posterior mean given signal s is
posterior_mean = w * s + (1 - w) * μ, where w = σ_v^2 / (σ_v^2 + σ_e^2).
With σ_v^2 = 0.16, σ_e^2 = 0.25, we get w ≈ 0.39024. So posterior_mean ≈ 0.39024 * s + 0.60976 * 2.2.

Winner's curse explanation: The highest signal among bidders is biased upward relative to V. If every bidder bids their posterior_mean based only on their signal (ignoring the information contained in being the winner), the winner tends to overpay: the conditional expectation of V given that their signal is the highest is below the unconditional posterior based only on their own signal.

Simulation (spreadsheet recipe, Excel/Google Sheets) — 50k trials recommended:

  1. Column A: draw V = NORM.INV(RAND(), 2.2, 0.4)
  2. Columns B:F: draw errors e_j = NORM.INV(RAND(), 0, 0.5); signals S_j = V + e_j
  3. Columns G:K: compute posterior_mean_j = w * S_j + (1-w)*2.2 (w = 0.39024)
  4. Determine winner: idx = MATCH(MAX(G:K), G:K, 0). Sale price (second-price) = second-highest of G:K.
  5. Compute actual V and compute overpayment = (winning bid - V); record negative if underpaying.
  6. Aggregate mean overpayment conditional on sale and distribution of overpayment quantiles.
This outputs an empirical estimate of the winner's curse when bidders use the naive posterior.

Interpretation guidance: Ask students to compute:

  • Average overpayment (mean) and its standard error
  • Proportion of trials where the winner overpays by > $200k
  • How the overpayment changes if n increases from 3 to 10

Problem 3 — Bayesian appraisal update (closed-form)

Scenario

An institution's prior on the portrait's value is N(μ0 = $2.0M, σ0 = $0.6M). An external appraisal produces an independent estimate x = $3.2M with measurement SD σ = $0.4M. Use conjugate Normal updating.

Questions

  1. Compute the posterior mean and posterior SD.
  2. Interpret the result: how much did the appraisal move the institution's belief?

Solution

Posterior mean formula: posterior = (σ0^2 * x + σ^2 * μ0) / (σ0^2 + σ^2). Posterior variance = (σ0^2 * σ^2) / (σ0^2 + σ^2).

Plug numbers: σ0^2 = 0.36, σ^2 = 0.16. Posterior mean = (0.36*3.2 + 0.16*2.0) / 0.52 = (1.152 + 0.32) / 0.52 ≈ 2.8308M. Posterior SD = sqrt(0.0576 / 0.52) ≈ 0.333M.

Teaching point: The appraisal moves the prior mean from $2.0M to $2.83M and reduces uncertainty — a realistic exercise to show how expert signals (and signal precision) change estimates in 2026’s data-driven market.

Problem 4 — Choosing a reserve price by expected revenue maximization

Scenario

Return to Problem 1: same Uniform(0, 3.5M) and n=4. Suppose the seller can choose reserve r to maximize expected revenue in a second-price auction. Use a spreadsheet to search for the optimal r.

Instructions

  1. Use the closed-form revenue formula (numerical integration shown earlier) or simulate with 50k trials where each trial you: draw four Uniform(0,A) valuations, compute revenue under given r.
  2. Evaluate expected revenue for a grid of r values from 0 to A in increments of $50k (or $10k for precision).
  3. Find r that maximizes average revenue.

Teaching observation: In many uniform/independent private-value settings, there is a non-zero optimal reserve > 0. Use the spreadsheet to show how the optimal r depends on A and n. If you want to build reproducible class materials and host them on an LMS, consult field guides on reproducible notebooks and hybrid edge workflows to help students run larger Monte Carlo experiments.

Problem 5 — Risk aversion and entry costs (extension)

Scenario and tasks

Extend Problem 1 by assuming bidders pay an entry cost c to participate and are risk-averse with utility u(w) = sqrt(w) (square-root utility). Ask students to:

  • Compute the ex-ante expected utility of entry for a bidder of mean type.
  • Find the participation threshold (minimum private valuation where expected utility of entry ≥ 0).
  • Simulate how participation changes seller revenue and optimal reserve.

Why this matters: In 2026, many online auctions attract more bidders but also impose bidding fees or verification costs; modelling costs and risk attitudes yields more realistic classroom results. You can also link the classroom pack to practical notes on due diligence and how provenance or domain-level metadata gets verified in modern marketplaces.

Answer key summary (quick numbers and spreadsheet checks)

  • Problem 1 (n=4, A=3.5, r=0.5): P(sells) ≈ 0.99958; P(final price > r) ≈ 0.8996; E[revenue] ≈ $2.1011M.
  • Problem 2: Posterior weight w ≈ 0.39024; posterior mean = 0.39024*s + 0.60976*2.2. Expect a positive mean overpayment when bidders bid naïvely; quantify via Monte Carlo (students will produce the numeric estimate).
  • Problem 3: Posterior mean ≈ $2.8308M, posterior SD ≈ $0.333M.

Spreadsheet formulas & shortcuts (practical)

  • Uniform draw in millions: =RAND()*3.5
  • Normal draw with mean μ and SD σ: =NORM.INV(RAND(), μ, σ)
  • Second-highest of 4: =LARGE(range, 2)
  • Posterior mean (Problem 2 style): =w * signal + (1 - w) * mu_prior
  • Repeat 50,000 trials: set up table of trial rows, then use AVERAGE() and STDEV() as summary stats; use Data Table or drag formulas—Excel quick-fill handles 50k rows fine.

Assessment rubric & classroom timing

  • Short lab (50 minutes): Problems 1 and 3 (closed-form + short simulation). Grade for correctness of calculation and correct spreadsheet outputs.
  • Long lab or take-home (90 minutes): Add Problem 2 simulation + reserve optimization. Grade also for interpretation and correct formulation of Bayesian logic.
  • Rubric: 50% computations, 30% spreadsheet simulation accuracy (reproducible results), 20% interpretation & write-up.

Classroom variations and advanced extensions

  • Change distribution families: use lognormal valuations to model asymmetric heavy-tailed art-market prices.
  • Model correlated signals (provenance blockchain info that all bidders see but interpret differently) to explore information cascades and common-value extremes — tying into practical guides for automating provenance and metadata extraction.
  • Introduce strategic bidding rules for first-price auctions and derive equilibrium numerically by solving for best responses in the spreadsheet.
  • Use Python or R notebooks for larger Monte Carlo experiments — many university labs in 2026 prefer reproducible notebooks connected to LMSes.

References and further reading (2024–2026 context)

Inspiration: a 2024–2026 wave of art-market analysis increased the use of probabilistic and machine-learning valuation models. The Artnet News piece reporting the 1517 discovery provided the motivating headline estimate ("could fetch up to $3.5M"). For deeper auction-theory treatment see standard references (Milgrom, Klemperer) and recent 2025–2026 reviews on AI appraisal integration in auction design. Also see practical notes on data storage and provenance cost tradeoffs when archiving high-resolution provenance and appraisal records.

Takeaways — actionable classroom checklist

  • Download the spreadsheet template (see CTA) containing these five problems, pre-built simulations, and answer sheets.
  • Run the uniform example live in class: show how changing reserve price affects expected revenue.
  • Assign Problem 2 as a take-home where students submit simulation code or spreadsheet and a 1-page interpretation of the winner's curse magnitude.

Call to action

Ready-made worksheet and spreadsheet template: get the editable Excel/Google Sheets pack with all 5 problems, pre-set simulations (50k trials), detailed answer keys, and a teacher rubric. Use it to save class prep time and deliver reproducible, exam-grade problems that match 2026 auction-market realities. Click to download the free student worksheet or upgrade for an instructor bundle with autograded Google Forms and an LMS-ready notebook. Also consider how sustainable packaging and physical provenance practices affect how prints and small-format works circulate among collectors.

Advertisement

Related Topics

#education#statistics#art
c

calculation

Contributor

Senior editor and content strategist. Writing about technology, design, and the future of digital media. Follow along for deep dives into the industry's moving parts.

Advertisement
2026-02-13T02:16:35.482Z