Quant puzzles, solved step by step

86/99, about 86.9%. Track only the current streak: none, me on 1 or 2, you on 1 or 2. Each game I win with probability 1/3, you win with 1/6, and a draw, 1/2, resets the streak. Write my chance of taking the match from each state in terms of the others, solve the five equations, and the start state comes out at 86/99.

What is the state, and why is the running score not it?

A door lock that opens after three correct codes in a row does not care how many wrong codes came before the last mistake. The only thing that matters for the rest of this match is the current streak, so the states are: no streak, my streak of 1 or 2, and your streak of 1 or 2. Games won earlier, draws played, games elapsed: all irrelevant once the streak is known. That is the Markov propertyThe future depends on the past only through the present state., and spotting it turns an infinite tree of game sequences into five numbers. Per game, I win with probability 1/2 x 2/3 = 1/3, you win with 1/2 x 1/3 = 1/6, and the rest are draws.

How do the equations go, and how do you solve them quickly?

Let x be my chance from the start, a1 and a2 from my streaks, b1 and b2 from yours. Every equation reads the same way: play one more game, three things can happen. A draw always returns to the start, a win for me always moves to a1 or one step up, and a win for you always moves to b1 or one step up. From a2 my next win ends the match in my favour; from b2 your next win ends it against me.

Solve by substituting: b2 is in terms of a1 and x, which gives b1 in the same terms; feed that into a2 and a1, and the start equation leaves one unknown. The answers: x = 86/99, a1 = 29/33, a2 = 10/11, b1 = 28/33, b2 = 8/11. Check the ordering: the further I am ahead, the higher the value, and every value sits between 0 and 1. Your chance is 13/99. A quick cross-check: three wins in a row is (1/3) cubed for me and (1/6) cubed for you, a ratio of 8 to 1, which would suggest about 89%; landing within two points of that rough race is a sign no term was dropped.

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.

Two thirds of days are calm. In the long run the fraction of days moving from calm to stressed must equal the fraction moving back, so calm share x 0.2 = stressed share x 0.4. Calm days are therefore twice as common as stressed days, 2/3 against 1/3. The same answer comes from spell lengths: calm spells average 5 days and stressed spells 2.5, and 5 / 7.5 = 2/3.

Why can you balance flows instead of solving equations?

Think of two rooms at a party. Every few minutes, one in five people in the kitchen wanders to the lounge, and two in five people in the lounge wander back. Once the crowd settles, the numbers crossing each way must match, or one room would keep filling up. In a two-state chain the long-run shares are fixed by one equation: the share of days leaving calm must equal the share of days leaving stressed. Calm leaves at rate 0.2 and stressed at rate 0.4, so calm must hold twice as many days.

How do you check it a second way?

Use spell lengths. A calm spell ends each day with probability 0.2, so it lasts 1/0.2 = 5 days on average; a stressed spell ends with probability 0.4, so it lasts 2.5 days. The chain alternates calm spell, stressed spell, calm spell, so the calm share is 5 out of every 7.5 days, which is 2/3. Two methods, one answer, in under a minute.

How fast does the chain forget where it started?

The transition matrix has a second eigenvalue of 0.8 + 0.6 - 1 = 0.4, and any gap between today's probabilities and the long-run split shrinks by that factor each day. After five days the starting state explains only 0.4^5, about 1%, of the gap, so the answer does not depend on how the week began. A trader would add the limitation: real regimes are not memoryless, and a stress spell that has already lasted a month is not as likely to end tomorrow as one that started yesterday.