Game Of Chance!

$9$ can be obtained through the combinations: $(6,1,2)$ , $(5,3,1)$ , $(5,2,2)$ , $(4,3,2)$ , $(4,1,4)$ , $(3,3,3)$

$10$ can be obtained through the combinations: $(6,1,3)$ , $(6,2,2)$ , $(5,1,4)$ , $(5,3,2)$ , $(4,3,3)$ , $(4,2,4)$

However, just because there are 6 different combinations to obtain each number, doesn't mean they have the same probability.

If ${ r }_{ 1 }$ represents the number that is rolled on the first die, and ${ r }_{ 2 }$ and ${ r }_{ 3 }$ for the second and third dice, respectively, then, for example, the combination $(6,1,2)$ can be obtained by triplets $({ r }_{ 1 }, { r }_{ 2 }, { r }_{ 3 })$ :

$(6,1,2)$

$(6,2,1)$

$(1,6,2)$

$(1,2,6)$

$(2,1,6)$

$(2,6,1)$

Any of the combinations above that have 3 distinct numbers have 6 permutations, while those with 1 repeated number have 3 permutations, and those that are comprised of 3 of the same number only have 1 permutation.

Both 9 and 10 have 3 combinations with 3 distinct numbers, but 9 has 2 combinations with 1 repeated number and 1 combination made of 3 of the same number, while 10 has 3 combinations 1 repeated number. Therefore, 10 has 27 permutations and a greater probability of being obtained than 9, with 25 permutations.

Thus, $P(9) < P(10)$ .

The expected value when you roll one dice is 3.5 so when you roll three die its 10.5 which is closer to 10 then 9.

Yee-Lynn has written a nice explanation for this problem. Here, I will use the lame method by listing out all possibilities.

The sample space of throwing 3 dice, $S =\{(1,1,1),(1,1,2),\ldots,(1,6,6),(2,1,1),\ldots,(2,6,6),\ldots,(6,6,6)\}$

$n(S) = 6^3 = 216$

Define event $A$ as getting a $9$ and event $B$ as getting a $10$ .

$A = \{(1,2,6),(1,3,5),(1,4,4),(1,5,3),(1,6,2),\\ (2,1,6),(2,2,5),(2,3,4),(2,4,3),(2,5,2),(2,6,1),\\ (3,1,5),(3,2,4),(3,3,3),(3,4,2),(3,5,1),\\ (4,1,4),(4,2,3),(4,3,2),(4,4,1),\\ (5,1,3),(5,2,2),(5,3,1),\\ (6,1,2),(6,2,1)\}$

$n(A)=25$

Therefore, $P(9)=P(A)=\dfrac{n(A)}{n(S)}=\dfrac{25}{216}$

$B=\{(1,3,6),(1,4,5),(1,5,4),(1,6,3),\\ (2,2,6),(2,3,5),(2,4,4),(2,5,3),(2,6,2),\\ (3,1,6),(3,2,5),(3,3,4),(3,4,3),(3,5,2),(3,6,1),\\ (4,1,5),(4,2,4),(4,3,3),(4,4,2),(4,5,1),\\ (5,1,4),(5,2,3),(5,3,2),(5,4,1),\\ (6,1,3),(6,2,2),(6,3,1)\}$

$n(B)=27$

Therefore, $P(10)=P(B)=\dfrac{n(B)}{n(S)}=\dfrac{27}{216}$

From here, we know that $\boxed{P(9)<P(10)}$

Relative probabilities of different outcomes are encoded by coefficients of the the polynomial $(x + x^{2} + ... + x^{6})^{6}$ (cf. the Generating functions wiki). One can compute them directly or notice that their values increase monotonically up to the median power on the interval [6, 36] and then decrease. 10 is closer to 21 than 9, so necessarily its coefficient will be greater.

The sum of the dice follows a concave symmetrical distribution over the values 3, 4, ..., 18. Since 10 is the midpoint of such range, P(10) is the highest, thus P(10) > P(9).