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  1. 067A friend offers a bet on one roll of a fair die: you win Rs 10,000 if it shows a six and you pay Rs 2,000 if it shows anything else. Should you take it?Expected value and gamesWarm upBulge bracket IBConsulting style brainteasers

    Try it first

    What is the expected value of one roll?

    Show the worked solution

    The bet is exactly fair: its expected value is zero. A six comes up one time in six, so the win is worth 1/6 x Rs 10,000 = Rs 1,667. The other five faces each cost Rs 2,000, worth 5/6 x Rs 2,000 = Rs 1,667. They cancel. Whether to take it is then a question about risk: five rolls in six you hand over Rs 2,000, and the swing on one roll is about Rs 4,472 either way.

    How do you weigh a big rare win against a small frequent loss?

    Imagine playing the game six hundred times. You would expect a six about a hundred times, collecting Rs 10 lakh, and the other five hundred rolls would cost Rs 10 lakh. Expected value is each outcome multiplied by its probability, added up, and here the two sides come to Rs 1,667 each, so they cancel exactly. Say the arithmetic as fractions of 6 so the interviewer can follow: 10,000 over 6 against 5 x 2,000 over 6, and 10,000 equals 10,000.

    Weigh each branch by its chance: the two sides come out exactly levelWin: a six123456chance 1/61/6 x Rs 10,000Rs 1,667worth, on averageone roll:+Rs 10,000pocket swingLose: 1 to 5123456chance 5/65/6 x Rs 2,000Rs 1,667worth, on averageone roll:-Rs 2,000pocket swinglevelExpected value Rs 1,667 - Rs 1,667 = 0. Fair bet; the spread of about Rs 4,472 a roll is what you are really deciding on.
    Weighted by its one-in-six chance the Rs 10,000 win is worth Rs 1,667, and weighted by its five-in-six chance the Rs 2,000 loss is worth the same Rs 1,667, so the expected value is zero and the bet is fair.

    If the maths is a tie, what decides it?

    Risk appetite, and the question is really about whether you can say so. A fair bet with an uneven shape is not neutral to everyone: you lose on five rolls in six, and the spread of outcomes, about Rs 4,472 on a single roll, is large next to the stake. Someone for whom Rs 2,000 is lunch money might play for fun; someone for whom it is a week's groceries should not, because the most likely result of a single roll is a loss. Interviewers want the number, then a sentence that separates expected value from the experience of one roll.

    The relationship
    E[X]=16(10,000)−56(2,000)=1,667−1,667=0E[X] = \tfrac{1}{6}(10{,}000) - \tfrac{5}{6}(2{,}000) = 1,667 - 1,667 = 0
    E[X]the expected value of one roll, in rupees
    1/6the chance of a six
    5/6the chance of any other face
    What it says in wordsMultiply each payoff by its chance and add; a result of zero means the bet is fair.

    How would you make the bet worth taking?

    Move one number and watch the sign. Break-even on the win is 5 x Rs 2,000 = Rs 10,000, so any win above Rs 10,000 makes it positive, and at Rs 12,000 the expected value is Rs 333 a roll. Alternatively, ask to play many times: over 600 rolls the expected result is still zero, but the swing shrinks relative to the stake, which is why a casino is happy with a tiny edge and a huge number of hands. Offering that framing shows you understand why desks care about both edge and variance.

    Where candidates lose it

    The fast wrong answer is to take the bet because Rs 10,000 is bigger than Rs 2,000. The interviewer is checking that you weigh each payoff by its chance before you compare.

    The second loss is stopping at zero. A fair bet is still a decision, and saying one sentence about the shape of the outcomes, five losses for every win, is what separates a calculator from a candidate.

    What the interviewer asks next

    • What win on a six would make the expected value Rs 500 a roll?
    • Now you roll twice and win if either roll is a six. Is the bet fair?
    • Why might a trader take a fair bet with a tiny edge thousands of times but refuse it once?