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  1. 071You start with Rs 10 and bet Rs 1 at a time on a game you win with probability 0.55, winning or losing Rs 1 each round. What is the chance you reach Rs 20 before you go broke?Random walks and Markov chainsHardTwo SigmaNew York · 2023

    Try it first

    Roughly what is the chance of reaching Rs 20 first?

    Show the worked solution

    About 88%. Let r be the ratio of losing to winning odds, 0.45/0.55 = 9/11. The chance of reaching Rs 20 from Rs 10 is (1 - r to the 10th)/(1 - r to the 20th), which simplifies to 1/(1 + r to the 10th). Since r to the 10th is about 0.134, the answer is 1/1.134, about 88.1%. A fair game would give exactly 50%.

    Why does a small edge per bet become a large edge on the game?

    Think of a tug of war between two teams, one a shade stronger. A single pull is close to a coin toss, but the rope has to travel a long way before either side wins, and every pull leans the same way. Reaching Rs 20 or Rs 0 takes many Rs 1 bets, and the 0.55 edge applies to every one of them, so the chance of winning the whole game rises far above 0.55. Staking all Rs 10 on one bet would give only 55%; betting Rs 1 at a time gives about 88%.

    How do you get the formula?

    Let P(i) be the chance of reaching 20 from a stake of i. One bet later you are at i + 1 with chance p or i - 1 with chance q, so P(i) = p P(i + 1) + q P(i - 1), with P(0) = 0 and P(20) = 1, and the solution is P(i) = (1 - r to the i)/(1 - r to the 20), where r = q/p. For a fair game the formula collapses to a straight line, P(i) = i/20, because a fair game keeps your expected wealth at 10, so 20 times P must equal 10. Checking the fair case is the fastest way to trust the biased one.

    The relationship
    P(10)=1−r101−r20=11+r10,r=0.450.55=911,r10≈0.134P(10) = \frac{1 - r^{10}}{1 - r^{20}} = \frac{1}{1 + r^{10}}, \qquad r = \frac{0.45}{0.55} = \frac{9}{11}, \quad r^{10} \approx 0.134
    P(10)the chance of reaching Rs 20 before Rs 0 from Rs 10
    rthe ratio of the losing chance to the winning chance
    r^20the same ratio over the full distance of Rs 20
    What it says in wordsThe chance of success from the middle is one over one plus the odds ratio raised to the distance to either end.
    A small edge per bet bends the whole curve25%50%75%100%0%Rs 0Rs 5Rs 10Rs 15Rs 20Starting stake, betting Rs 1 a time, target Rs 20p = 0.55: 88.1%p = 0.50: 50%p = 0.45: 11.9%From Rs 10:one Rs 10 betat 0.55 wins 55%Rs 1 betsat 0.55 win 88%edge: bet small
    Betting Rs 1 a time from Rs 10, a fair game reaches Rs 20 first 50% of the time, a 0.55 edge lifts that to 88.1% and a 0.45 disadvantage cuts it to 11.9%, because the small edge on each bet compounds over the many bets the game takes.

    How do you compute r to the tenth in your head, and what is the lesson?

    Square repeatedly: 9/11 squared is 81/121, about 0.669; squared again about 0.448; again about 0.201; times 0.669 gives about 0.134. The desk lesson is about sizing: with an edge, make many small bets so the edge compounds; without one, the same arithmetic works against you, and at 0.45 the Rs 1 strategy reaches Rs 20 only 12% of the time against 45% for one bold bet. This is the classic gambler's ruinThe problem of a gambler betting fixed amounts until reaching a target or losing everything, solved as a random walk with two absorbing ends. result, and it is why a trader with a real but thin edge wants volume, not size.

    Where candidates lose it

    The common wrong answer is 55%, the chance of winning one bet, carried over to the whole game. It ignores that the game takes many bets and the edge applies to each of them.

    The second loss is setting up the recursion and getting lost in the algebra. Write the answer in the form 1/(1 + r to the 10th), check it on the fair case, and compute the power by repeated squaring.

    What the interviewer asks next

    • With a win probability of 0.45, should you bet Rs 1 at a time or everything at once, and why?
    • How many rounds do you expect the game to last at p = 0.5?
    • The casino has unlimited money and you never stop at Rs 20. What is your chance of eventual ruin at p = 0.55?

    Asked at Two Sigma, Research, New York, 2023 (Wall Street Oasis): Biased gamblers ruin problems; Markov Chain problems; sampling uniformly from triangle