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  1. 015You roll three fair dice. Is a total of 9 or a total of 10 more likely, given that each can be written as exactly six unordered combinations of three faces?Counting and combinatoricsCoreQuant tradingProp trading firms

    Try it first

    Which total is more likely?

    Show the worked solution

    10 is more likely: 27 ways out of 216 against 25. The equally likely outcomes are the 216 ordered rolls, not the unordered combinations. A combination of three different faces covers six ordered rolls, one with a pair covers three, and a triple covers one. Total 9 includes 3 + 3 + 3, which counts once, and one fewer all-different combination, so it loses two orderings to 10.

    Why are combinations the wrong thing to count?

    Think of dealing two cards and asking whether a pair of kings or a king with a queen is more likely. There is one combination of each, but king and queen can arrive in either order while two kings are just two kings. Probability comes from counting outcomes that are equally likely, and with dice those are the ordered rolls: first die, second die, third die, 6 x 6 x 6 = 216 of them. Unordered combinations bundle different numbers of those outcomes, so counting combinations gives the wrong weights.

    Six combinations each, but the orderings differ: 25 against 27Total of 9orderings1 + 2 + 66 all different1 + 3 + 56 all different1 + 4 + 43 one pair2 + 2 + 53 one pair2 + 3 + 46 all different3 + 3 + 31 tripleTotal25/216 = 11.6%Total of 10orderings1 + 3 + 66 all different1 + 4 + 56 all different2 + 2 + 63 one pair2 + 3 + 56 all different2 + 4 + 43 one pair3 + 3 + 43 one pairTotal27/216 = 12.5%The triple 3 + 3 + 3 counts once; it is what costs 9 the two orderings
    Totals of 9 and 10 each have six unordered combinations, but weighting each combination by its number of orderings gives 27 of 216 rolls for 10 against 25 for 9, mostly because 9 includes the triple 3, 3, 3, which happens only one way.

    How do you count the orderings fast?

    Classify each combination by its repeats. Three different faces give 3! = 6 orders, a pair gives 3 orders, one for each position of the odd die, and a triple gives 1. For 9: 1 2 6, 1 3 5 and 2 3 4 are all different, 18; 1 4 4 and 2 2 5 are pairs, 6; 3 3 3 is a triple, 1; total 25. For 10: 1 3 6, 1 4 5 and 2 3 5 give 18; 2 2 6, 2 4 4 and 3 3 4 give 9; total 27. So 10 comes up 12.5% of the time and 9 only 11.6%.

    The relationship
    P(9)=3⋅6+2⋅3+1216=25216P(10)=3⋅6+3⋅3216=27216P(9) = \frac{3\cdot 6 + 2\cdot 3 + 1}{216} = \frac{25}{216} \qquad P(10) = \frac{3\cdot 6 + 3\cdot 3}{216} = \frac{27}{216}
    6, 3, 1the orderings of an all-different, a pair and a triple combination
    216the ordered outcomes of three dice
    What it says in wordsWeight each combination by its orderings and 10 beats 9 by two rolls in 216.

    Is there a shortcut that avoids listing?

    Yes: symmetry. Replacing each face x by 7 - x maps a total of t to 21 - t, so the distribution of three dice is symmetric about 10.5, and 10 and 11 are the two most likely totals, each 27/216. Anything further from 10.5, including 9, must be less likely or equal; a quick count confirms it is 25. Historically this is the question gamblers put to Galileo, who answered it by counting ordered outcomes, which is still the method.

    Where candidates lose it

    The trap is the question's own framing: six combinations each invites the answer that the totals are equally likely. The interviewer wants you to reject the framing, not accept it.

    The second loss is listing all 216 rolls, or writing out every ordering. Classifying by repeats, six, three or one, gets both totals in under a minute.

    What the interviewer asks next

    • What is the most likely total with four dice, and its probability?
    • What is P(total is 9) with two dice, and why does 9 behave differently?
    • How many ordered outcomes of three dice sum to 7?