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  1. 042A queue holds between 0 and 3 orders. Each tick at most one thing happens: with probability 0.3 a new order arrives (if there is room), with probability 0.5 one order is filled (if the queue is not empty), and otherwise nothing changes. In the long run, what fraction of ticks is the queue full?Markov chains and random walksCoreDRWNew York · 2026

    Try it first

    Roughly what share of ticks is the queue full?

    Show the worked solution

    27/272, about 9.9% of ticks. In a birth-death chain the long-run flow up across each cut equals the flow down, so share(k) x 0.3 = share(k + 1) x 0.5. Each state's share is 0.6 times the one below: weights 1, 0.6, 0.36 and 0.216, summing to 2.176. The full state gets 0.216/2.176, about 9.9%, and the queue is empty about 46% of the time.

    Why can you skip solving the full set of equations?

    Stand at a doorway between two rooms at a party that has settled down. Over an evening, the number of people walking through one way must match the number walking back, or one room would keep filling. In a chain that only steps up or down by one, the long-run flow across the boundary between neighbouring states must balance, which gives one simple equation per cut. Here flow up from state k is its share times 0.3, and flow down from state k + 1 is its share times 0.5.

    Birth-death chain: across each cut, flow up equals flow down0 ordersempty1 order2 orders3 ordersfull0.30.50.30.50.30.5Cut balance: share(k) x 0.3 = share(k + 1) x 0.5, so each share is 0.6 times the last46.0%weight 127.6%weight 0.616.5%weight 0.369.9%weight 0.216
    Arrivals push the queue up with probability 0.3 and fills pull it down with probability 0.5, so each state's long-run share is 0.6 times the one below: 46.0% empty, 27.6% with one order, 16.5% with two and 9.9% full.

    How do the cut equations give the answer?

    Write each share relative to the empty state. Each cut gives share(k + 1) = share(k) x 0.3/0.5 = 0.6 x share(k), so the weights are 1, 0.6, 0.36 and 0.216. They sum to 2.176, so the full queue holds 0.216/2.176 = 27/272 of the time, about 9.9%. The staying probabilities, 0.2 in the middle states and 0.5 when full, never enter; a chain that pauses on a state does not change the balance across cuts.

    The relationship
    πk+1=πk⋅0.30.5π3=0.631+0.6+0.62+0.63=27272≈9.9%\pi_{k+1} = \pi_k\cdot\frac{0.3}{0.5} \qquad \pi_3 = \frac{0.6^3}{1 + 0.6 + 0.6^2 + 0.6^3} = \frac{27}{272} \approx 9.9\%
    pi_klong-run share of ticks with k orders in the queue
    0.3chance of an arrival when there is room
    0.5chance of a fill when the queue is not empty
    What it says in wordsEach state is visited 0.6 times as often as the one below it; normalise the four weights to add to one.

    Check with conservation. Orders accepted per tick are 0.3 x (1 - 0.099) = 0.2702, and orders filled per tick are 0.5 x (1 - 0.460) = 0.2702: the same, as they must be. That gives a useful business number: arrivals turned away because the queue is full run at 0.3 x 0.099, about 0.030 per tick, or one arrival in ten. The model assumes one event per tick; if an arrival and a fill could happen in the same tick, the chain changes and so do the numbers.

    Where candidates lose it

    The common loss is assuming the four states are equally likely, or writing out all four balance equations with the self-loops and solving a 4 by 4 system under time pressure. The cut method needs three one-line ratios.

    The second is inverting the ratio, using 0.5/0.3, which makes the full state the most common. Fills are faster than arrivals, so the queue must lean toward empty; check the direction before you normalise.

    What the interviewer asks next

    • What is the average queue length?
    • What arrival probability would make the queue full 25% of the time?
    • How does the answer change if the queue can hold unlimited orders?

    Asked at DRW, Quantitative Research, New York, 2026 (Wall Street Oasis): There was a problem on Chi-squared distributions which was difficult and also one on birth death chains.