Quant puzzles, solved step by step

1/11, about 9.1%. Any particular sequence with five reds and five blues has probability 5! x 5! / 11!, whatever the order, because the numerators just count up the reds and blues separately. There are 10 choose 5 = 252 such sequences, and 252 x 5! x 5! / 11! = 1/11. The same algebra gives 1/11 for every count from 0 to 10.

Why is this not ten coin flips?

Think of a new café and its first customers. If the first few visitors like it and bring friends, it fills up; if they do not, it stays empty. Early luck compounds. In this urn each red draw adds a red ball, so it raises the chance of the next red: the draws reinforce each other, and runs that start lopsided tend to stay lopsided. That spreads the count of reds far more widely than independent coin flips, which bunch around five.

Why does the order of the draws not matter?

Write out one sequence, say five reds then five blues. The chances are 1/2, 2/3, 3/4, 4/5, 5/6 for the reds, then 1/7, 2/8, 3/9, 4/10, 5/11 for the blues. Shuffle the order and the denominators are still 2 up to 11, while the red numerators still run 1 to 5 and the blue ones 1 to 5, so every arrangement of five reds and five blues has the same probability, 5! x 5! / 11!. Draws whose joint probability ignores order are called exchangeableA sequence of random variables whose joint distribution does not change when you reorder them, even though they need not be independent., and that is the property doing the work.

Is there a picture that makes 1/11 obvious?

Yes. This urn behaves exactly as if nature first picked a hidden red probability p uniformly between 0 and 1, and then flipped ten independent coins with that p. Averaging a binomial over a uniform p gives every count the same weight, 1/(n + 1). The urn is also a small model of momentum and of market share: an early lead makes further gains more likely, so the final split is highly uncertain, even though each step looks like a fair draw.

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.

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.

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.

9 steps. Let E(k) be the expected steps left from state k. Each step costs one and lands on a neighbour with equal chance, so E(k) = 1 + (E(k - 1) + E(k + 1))/2, with E(3) = E(-3) = 0. The solution is E(k) = (3 - k)(3 + k): 9 from 0, 8 from plus or minus 1, 5 from plus or minus 2. A symmetric walk needs about the square of the distance to travel it.

Why is the answer not 3?

Think of someone pacing a narrow corridor, choosing left or right at random with every step. They do reach a wall eventually, but they spend most of their time doubling back. A random walk with no drift covers distance like the square root of time, so covering a distance of 3 takes about 3 squared steps, not 3. The answer 3 would only be right if the walk always moved in one direction. The structure to name in the room is a first-step recursion on the state, which turns the question into seven unknowns with simple links between them.

How do you solve the recursion quickly?

Use symmetry first: E(1) = E(-1) and E(2) = E(-2). Then three equations remain. From 2: E(2) = 1 + (E(1) + 0)/2. From 1: E(1) = 1 + (E(0) + E(2))/2. From 0: E(0) = 1 + E(1), because both neighbours are worth E(1). Substituting gives E(1) = 8 and E(0) = 9, and every value matches the product of the distances to the two walls. The second differences of E are all minus 2, which is why the answer is a parabola: each step must remove exactly one unit of expected time.

How do you check 9 a second way?

Use a martingaleA process whose expected next value, given everything so far, equals its current value: a fair game.. Squaring the position and subtracting the number of steps gives a fair game, because each step adds exactly 1 to the expected square. At the moment the walk stops its square is exactly 9, so the expected number of steps must also be 9. The same argument handles uneven walls: starting between -a and +b, the expected time is a times b. The cousin result is the hitting probability: from +1, the walk reaches +3 before -3 with probability 2/3, the distance to the far wall over the total width.