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  1. 028You roll two fair dice and are paid the larger of the two faces in rupees. What is the expected payout?Expected value and optimal stoppingCoreJane StreetNew York · 2026

    Try it first

    Pick the expected payout before you count.

    Show the worked solution

    161/36, about Rs 4.47. The larger face equals k in 2k minus 1 of the 36 equally likely outcomes: 1, 3, 5, 7, 9 and 11 cells for k from 1 to 6. Multiply each value by its count and add: 1 + 6 + 15 + 28 + 45 + 66 = 161. Divided by 36, that is 4.47, almost a full point above a single die's 3.5.

    Why is the answer well above 3.5?

    When two friends each suggest a restaurant and you always go with the better rated one, your average dinner beats either friend's average. Taking the larger of two draws pulls the result toward the top, because a low result survives only if both draws are low. A payout of 1 needs both dice on 1, one cell in 36. A payout of 6 needs just one six, and 11 cells in 36 contain at least one.

    The larger face is k in 2k - 1 of the 36 cells: L-shaped bands123456223456333456444456555556666666112233445566Die 1Die 2Larger faceCellsFace x cellsk = 111k = 236k = 3515k = 4728k = 5945k = 61166Total36161161 / 36 = 4.47one die alone: 3.50
    The larger face is 1 in one cell, 2 in three cells and so on up to 6 in eleven cells; face times count sums to 161, so the expected payout is 161/36, about 4.47, against 3.50 for one die.

    How do you count the cells without listing all 36?

    Count the outcomes where the larger face is at most k: both dice must be at most k, which is k squared cells. The cells where the larger face is exactly k are k squared minus (k - 1) squared, which is 2k - 1. That is the L-shaped band in the grid: a new row and a new column, sharing one corner cell. The bands are 1, 3, 5, 7, 9 and 11, and they add to 36, which is the check that nothing was double counted.

    The relationship
    E[max⁡]=∑k=16k 2k−136=16136≈4.47E[\max] = \sum_{k=1}^{6} k\,\frac{2k-1}{36} = \frac{161}{36} \approx 4.47
    kthe value of the larger face
    2k - 1the number of the 36 outcomes where the larger face is exactly k
    What it says in wordsWeight each possible payout by how many of the 36 outcomes produce it, then divide by 36.

    A second route helps when the interviewer changes the dice. Add up the chance that the payout reaches each level: the payout is at least k unless both dice are below k, so the sum of 1 minus (k - 1) squared over 36, for k from 1 to 6, is 6 minus 55/36, which is 161/36 again. Two methods landing on the same fraction is the check worth saying out loud. By symmetry the smaller face averages 7 minus 4.47, about 2.53, and with three dice the larger face rises to 4.96.

    Where candidates lose it

    The common slip is to treat the six payouts as equally likely and answer 3.5, or to say a bit more than 3.5 without a number. The grid shows how uneven the counts are: eleven ways to be paid 6 against one way to be paid 1.

    The second slip is counting 12 cells for a payout of 6, which counts the double six twice. The row of sixes and the column of sixes share one cell.

    What the interviewer asks next

    • What is the expected value of the smaller face?
    • What is the expected larger face with three dice?
    • I pay you the larger face minus the smaller. What is that worth?

    Asked at Jane Street, Investment Operations, New York, 2026 (Wall Street Oasis): First interview was testing simple math brainteasers (e.g. expected value of dice throws, etc.)