Question
You need to buy exactly one collectible card to fulfil a customer's order. A card is available immediately at an asking price of $50. Alternatively, you can join a public bid queue at $46 and wait until closing. There are three one-card buy orders ahead of yours and two behind. Before any cards arrive, each order ahead independently cancels with probability 1/2; each order behind independently cancels with probability 3/4. Cancelled orders leave the queue. Incoming cards then execute the remaining buy orders from the front, one card per order.
Before closing, the number N of cards arriving is 1, 2 or 3 with probabilities 1/4, 1/2 and 1/4, independently of cancellations. If your queued order fills, you pay $46; otherwise you must buy the card from a backup supplier at $56 after closing. All routes supply the same card and other costs are zero.
What is the probability your queued bid fills? Which route minimizes expected purchase cost, and why don't cancellations behind you improve your fill probability?
Get hint
Hint 1
Let C be the number of the three orders ahead that cancel. Your bid can fill only if enough arriving cards cover all surviving orders ahead plus yours.
Check your answer
Solution
If C of the three earlier buy orders cancel, 3−C orders remain ahead of yours. With \(N\) incoming sellable cards and price-time priority, you receive a card exactly when \(N\ge(3-C)+1\), equivalently \(N+C\ge4\).
The count \(C\) is \(\operatorname{Binomial}(3,1/2)\), independent of \(N\). Conditional on \(N=1\), filling requires \(C=3\), with probability 1/8. If \(N=2\), filling requires \(C\ge2\), with probability \(\frac{3+1}{8}=\frac12\). If \(N=3\), filling requires \(C\ge1\), with probability \(1-\frac18=\frac78\). Averaging over the specified N distribution,
\[\begin{aligned}P(\text{fill})&=(1/4)(1/8)+(1/2)(1/2)+(1/4)(7/8)\\&=\frac1{32}+\frac14+\frac7{32}=\frac12\end{aligned}\].
The expected cost of queuing plus the guaranteed backup is \(46\cdot P(\text{fill})+56\cdot P(\text{no fill})=46(1/2)+56(1/2)=\$51\). Immediate execution costs $50 with certainty, so crossing the ask is $1 cheaper in expectation. Cancellations behind your order only reduce the number of later orders; they never remove any order standing between yours and incoming supply.
As an independent state calculation, after processing each of the three orders ahead, let a be the number surviving; each survives with probability 1/2, so \(P(a=0,1,2,3)=\frac{(1,3,3,1)}8\). The arriving-volume tail \(P(N\ge a+1)\) equals 1, 3/4, 1/4, 0 respectively. Their weighted sum is \(\frac18\cdot1+(3/8)(3/4)+(3/8)(1/4)+\frac18\cdot0=\frac12\), confirming the fill calculation.
Assumptions
The queue is strictly first-in-first-out; arriving units cannot skip surviving earlier orders.
The three ahead cancellations, two behind cancellations, and arrival quantity N are mutually independent.
An order either cancels before the incoming volume or remains until closing; yours does not cancel.
A failure to fill requires a certain backup purchase at $56, so no unpriced lost-sale penalty is involved.
Final Answer
Queue fill probability 1/2. Waiting costs $51 on average versus $50 to buy immediately, so buy at the $50 ask. Later orders' cancellations cannot advance your place.