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  1. 047The index has fallen five days in a row and a client says it is now due to rise. Suppose each day is independent and a down day has a 48% chance. How many losing runs of five or more days should you expect in a 250-day year, and what does the streak say about tomorrow?Behavioural trapsHardWealth management

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    After five down days in a row, what is the chance tomorrow is an up day, under these assumptions?

    Show the worked solution

    About 3.3 losing runs of five or more days a year, and the streak says nothing about tomorrow. A run of five falls has probability 0.48 to the power 5, about 2.55%, and it can start on almost any of the 250 days, provided the day before was up. That gives roughly 3.3 runs a year, and a 97% chance of at least one. With independent days, tomorrow's chance of rising is still 52%.

    How often should a five-day losing run turn up?

    Toss a slightly unfair coin 250 times and look for five tails in a row. Any one stretch of five is unlikely, 2.55%, but there are about 245 stretches to look at. A run of five down days needs an up day, or the start of the year, followed by five falls, so the expected count is about 0.0255 x (1 + 245 x 0.52), roughly 3.3 runs a year. Across a year, the chance of seeing at least one is about 97%.

    The relationship
    E[runs]=p5[1+(n−5)(1−p)]=0.485 [1+245×0.52]≈3.3E[\text{runs}] = p^{5}\left[1 + (n-5)(1-p)\right] = 0.48^{5}\,[1 + 245 \times 0.52] \approx 3.3
    pthe chance of a down day, 48%
    ntrading days in the year, 250
    (1-p)the up day that must come just before a run for it to be a new run
    What it says in wordsCount every day a new five-day losing run could start, and multiply by the chance it does.
    One simulated year of coin-flip days: long losing runs appear anyway1511011512017 down in a row5 down in a row5 down in a rowdown dayup daypart of a run of 5 or moreExpected runs of 5+ down days a year: 3.3. Chance of at least one: 97%. Chance day 6 is up: still 52%.
    In one simulated year of 250 independent days with a 48% chance of a fall, three losing runs of five or more days appear, of 7, 5, 5 days. Independence alone predicts about 3.3 such runs a year, and none of them tells you anything about the next day.

    Why does the client feel the market is due?

    Because people expect short sequences to look like long-run averages, so a run of losses feels like a debt the market must repay. If the days are independent, the market keeps no ledger: the chance of a rise after five falls is the same 52% as after five rises. The adviser's job is to take the streak out of the decision and bring the conversation back to the client's plan and time horizon.

    The limit is the independence assumption. Real markets show some short-term momentum and some mean reversion at different horizons, and volatility clusters, so streaks are a little more common than a coin predicts. None of that makes five falls a reliable signal to buy.

    Where candidates lose it

    One trap is agreeing with the client that the market is due, which is the gambler's fallacy with a market label. The other is calling a five-day run rare because 0.48 to the fifth is small, forgetting how many days it has to appear on.

    Give the expected count, about three a year, and the 52% for tomorrow. Then say how you would steer the client back to his plan, because that is what the desk actually does with the maths.

    What the interviewer asks next

    • How many runs of ten or more down days would you expect in a year?
    • What would you look for in the data before believing streaks carry information?
    • The client wants to add money after every three-day fall. How do you respond?