Ways to Choose Dice Sum

Question

Jordan rolls eight fair dice, computing the sum of the numbers on the top of the dice after the first three rolls. What is the probability that Jordan is able to find at least one way to pick exactly three of the latter five dice in such a way that the sum of these three dice is the same sum as that of the first three dice?

Answer 1

In retrospect, since this is a problem that the OP (i.e. original poster) just made up, it is reasonable to suppose that the OP lacks the training in Combinatorics or Probability theory to pursue this problem. Further, the OP's subsequent comment that expresses the computation of $~\dfrac{65}{108},~$ while wrong, does represent showing work.

I don't feel it is reasonable to ask the OP to do more, just to get an answer.

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For $~x \in \{3,4,\cdots,18\},~$ let

• $f(x)~$ denote the probability that the sum of the first three dice will be $~x.~$

• $g(x)~$ denote the probability that you can select $~3~$ of the remaining $~5~$ dice so that the sum of the $~3~$ selected dice is $~x.$

Then, the overall probability that you can select $~3~$ of the latter $~5~$ dice to match the sum of the first $~3~$ dice is

$$\sum_{x = 3}^{18} \left[f(x) \times g(x)\right].$$

Therefore, the problem reduces to computing:

• $~f(3), f(4), \cdots, f(18)$

• $~g(3), g(4), \cdots, g(18).$

Stars and Bars theory will be used to assist in the computations. For the theory, see this article and this article.

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$\underline{\text{Computation of} ~f(x): ~\text{Introduction } ~: ~x \leq 8}$

The probability will be computed as $~\dfrac{N(x)}{D} ~: D = 6^3 = 216.$

To compute $~N(x) ~: ~x \leq 8,~$ you want the number of solutions to :

• $x_1 + x_2 + x_3 = x$
• $x_1, ~x_2, ~x_3 \in \{1,2,3,4,5,6\}.$

I will use the approach discussed in this answer.

Note that:

• The shortcuts discussed in the answer's Addendum are pertinent. For one thing, each of the three variables have the same upper bound of $~6.~$ For another, for $~x \leq 8,~$ the upper bounds may be ignored, since any satisfying solution of $~x_1 + x_2 + x_3 = x ~: 1 \leq x \leq 8,~$ will automatically prevent any of the three variables from being as large as $~7.$

• The method used in Addendum-2 is pertinent, because basic Stars and Bars theory assumes that each variable has a lower bound of $~0~$ rather than $~1.$

So, the first thing to do, when computing $~N(x),~$ is to use the change of variables $~y_i = x_i - 1.$

So, you are now computing the number of solutions to :

• $y_1 + y_2 + y_3 = (x - 3).$
• $y_1, ~y_2, ~y_3 \in \{0,1,2,3,4,5\}.$

Per Stars and Bars theory, For $~x \leq 8,~$ you have that $~\displaystyle N(x) = \binom{[x-3] + 2}{2} = \binom{x-1}{2}.$

Therefore, for $~x \leq 8,~$ you have that

$$f(x) = \frac{\binom{x-1}{2}}{216}.$$

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$\underline{\text{Subsequent} ~f(x) ~\text{Computations}}$

I will manually explore $~~f(9)~$ and $~f(10).~$ Then, I will use the fact that by symmetry, $~f(x) = f(21 - x).~$

For example, just as there is only 1 way of having the three dice sum to $~3,~$ there is only 1 way of having the three dice sum to $~18.$

Since the maximum value of $~x~$ that will be under direct scrutiny is $~x = 10,~$ it is impossible to have any solution to $~x_1 + x_2 + x_3 = x \leq 10,~$ that has more than one of the three dice violating the upper bound of $~x_i \leq 6.~$

Therefore, with respect to the use of Inclusion-Exclusion, as discussed in the linked answer, you will have a computation of

$$T_0 - T_1 = T_0 - \binom{3}{1} \times |S_1|. \tag1 $$

In (1) above, $~|S_1|~$ represents the number of solutions where $~x_1 \geq 7.~$ By symmetry, you will have that $~|S_1| = ~|S_2| = ~|S_3|,~$ since each of the three dice have the same upper bound of $~6.$

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$\underline{\text{Computation of} ~f(9)}$

As in the introductory section,
$~x_1 + x_2 + x_3 = 9 ~: ~x_i \in \Bbb{Z^+}$
bijects to
$~y_1 + y_2 + y_3 = 6 ~: ~y_i \in \Bbb{Z_{\geq 0}},$
which has $~\displaystyle \binom{6 + 2}{2} = 28~$ solutions.

From this you must deduct $~3~$ times the number of solutions to
$~x_1 + x_2 + x_3 = 9 - 6 = 3 ~: ~x_i \in \Bbb{Z^+},$
which bijects to
$~y_1 + y_2 + y_3 = 0 ~: ~y_i \in \Bbb{Z_{\geq 0}},$
which has $~\displaystyle \binom{0 + 2}{2} = 1~$ solution.

Therefore,

$$~f(9) = \frac{N(9)}{216} = \frac{28 - (3 \times 1)}{216} = \frac{25}{216}.$$

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$\underline{\text{Computation of} ~f(10)}$

As in the introductory section,
$~x_1 + x_2 + x_3 = 10 ~: ~x_i \in \Bbb{Z^+}$
bijects to
$~y_1 + y_2 + y_3 = 7 ~: ~y_i \in \Bbb{Z_{\geq 0}},$
which has $~\displaystyle \binom{7 + 2}{2} = 36~$ solutions.

From this you must deduct $~3~$ times the number of solutions to
$~x_1 + x_2 + x_3 = 10 - 6 = 4 ~: ~x_i \in \Bbb{Z^+},$
which bijects to
$~y_1 + y_2 + y_3 = 1 ~: ~y_i \in \Bbb{Z_{\geq 0}},$
which has $~\displaystyle \binom{1 + 2}{2} = 3~$ solutions.

Therefore,

$$~f(10) = \frac{N(10)}{216} = \frac{36 - (3 \times 3)}{216} = \frac{27}{216}.$$

Note that $~N(3) + \cdots + N(10) = 108 = \dfrac{216}{2},~$ as expected.

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$\underline{\text{Computation of} ~g(x) ~: ~\text{Introduction}}$

$~g(x)~$ will be much more difficult to compute, which is the main reason that I decided that it was not reasonable to expect the OP to show work here. As in the computation of $~f(x),~$ there is a symmetry in the dice throws.

For example, $~g(3) = g(21 - 3) = g(18).~$ That is, $~g(3)~$ is asking for the probability that at least three of the five remaining dice are 1's, while $~g(18)~$ is asking for the probability that at least three of the five remaining dice are 6's.

Similarly, $~g(4)~$ is asking for the probability that three of the five dice sum to $~4~$. This can only be achieved if at least two of the five dice show a 1, and simultaneously, at least one of the five dice shows a 2. $~g(21-4) = g(17)~$ will similarly require that at least two of the five dice show a 6, and simultaneously, at least one of the five dice shows a 5.

Therefore, it is sufficient to compute $~g(3), ~g(4), ~\cdots, ~g(10),~$ with the understanding that $~g(21 - x) = g(x).$

In general, $~g(x)~$ will be expressed as $~\dfrac{M(x)}{E} ~: ~E = 6^5 = 7776.~$ That is, you are throwing $~5~$ dice, and there are $~6~$ equally likely choices for each of the dice. So, $~M(x)~$ will represent the enumeration of the number of satisfying rolls of $~5~$ dice.

A roll will be deemed satisfactory if and only if it is possible to select three of the five dice such that the sum of these three dice is $~x.$

Note that per Multinomial Syntax, the expression

$~\displaystyle \binom{n}{(a_1,a_2,\cdots,a_r)}~: ~a_1 + \cdots + a_r = n ~: a_1, \cdots, a_r, n \in \Bbb{Z_{\geq 0}}$ denotes

$\displaystyle \frac{n!}{a_1! \times a_2! \times \cdots \times a_r!}.$

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$\underline{\text{Computation of} ~g(x) ~: ~\text{Advanced Considerations}}$

As $~x~$ increases from $~3~$ to $~4,~$ and up through $~10,~$ the computation of $~M(x)~$ becomes progressively more complicated.

The values of $~N(3), ~N(4), \cdots, N(10),~$ have already been computed. A brief overview of the computation of $~M(10)~$ will illustrate the technique that I will use. With $~N(10) = 27,~$ the corresponding dice rolls can be listed. I will use the syntax $~\langle a,b,c\rangle,~$ to represent that the dice rolls of $~a,b,c~$ occurred in some order. Then, the pertinent dice rolls are

$$D_1 ~: ~\langle 1,3,6\rangle, ~D_2 ~: ~\langle 1,4,5\rangle, ~D_3 ~: ~\langle 2,2,6\rangle, \\ D_4 ~: ~\langle 2,3,5\rangle, ~D_5 ~: ~\langle 2,4,4\rangle, ~D_6 ~: ~\langle 3,3,4\rangle.$$

In effect, I will need to compute

$$|D_1 \cup D_2 \cup D_3 \cup D_4 \cup D_5 \cup D_6|.$$

In general, for the higher values of $~x~$ in the set $~\{3,4,\cdots, 10\},~$ I will use Inclusion-Exclusion to enumerate $~M(x).~$

As partial shortcuts, in the next three sections, I will perform the pertinent calculations that represents each of the three possible patterns $~\langle a,a,a\rangle, ~\langle a,a,b\rangle,~$ and $~\langle a,b,c\rangle.$

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$\underline{\text{Computation of} ~g(x) ~: ~\text{Advanced Shortcuts - the} ~\langle a,a,a\rangle ~\text{Pattern}}$

For illustration, I will specifically focus on the Pattern $~\langle 1,1,1\rangle.$

Note that

$$6^5 = (1 + 5)^5 = \sum_{k=0}^5 \binom{5}{k} 1^k 5^{5-k} = \sum_{k=0}^5 \binom{5}{k}5^{5-k}.$$

Further, the expression $~\displaystyle \binom{5}{k} 5^{5-k}~$ represents the number of ways of throwing exactly $~k~$ 1's, with each of the other $~(5-k)~$ dice not equal to $~1.~$

The $~\langle a,a,a\rangle~$ pattern represents that there are at least 3 a's in the five dice rolls. Therefore, the enumeration of the $~\langle a,a,a\rangle~$ pattern is represented by

$$\sum_{k = 3}^5 \binom{5}{k} 5^{5-k} = 250 + 25 + 1 = 276.$$

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$\underline{\text{Computation of} ~g(x) ~: ~\text{Advanced Shortcuts - the} ~\langle a,a,b\rangle ~\text{Pattern}}$

For illustration, I will specifically focus on the Pattern $~\langle 1,1,2\rangle.$

This means that there must be at least two 1's thrown, among the five dice, and simultaneously, at least one 2 thrown among the five dice. The various possibilities are explored below.

In this section, the symbol $~A~$ will be used to represent any die throw that is not a $~1~$ or a $~2.~$ This implies that there will be exactly $~4~$ choices for each die that is specified as $~A.$

• $\langle 1-1-1-1-2 \rangle.$
  Enumeration is $~\displaystyle \binom{5}{4} = 5.$

• $\langle 1-1-1-2-2 \rangle.$
  Enumeration is $~\displaystyle \binom{5}{3} = 10.$

• $\langle 1-1-1-2-A \rangle.$
  Enumeration is $~\displaystyle \binom{5}{(3,1,1)} \times 4^1 = 80.$

• $\langle 1-1-2-2-2 \rangle.$
  Enumeration is $~\displaystyle \binom{5}{2} = 10.$

• $\langle 1-1-2-2-A \rangle.$
  Enumeration is $~\displaystyle \binom{5}{(2,2,1)} \times 4^1 = 120.$

• $\langle 1-1-2-A-A \rangle.$
  Enumeration is $~\displaystyle \binom{5}{(2,1,2)} \times 4^2 = 480.$

Therefore, the enumeration of the $~\langle a,a,b\rangle~$ pattern is represented by

$$5 + 10 + 80 + 10 + 120 + 480 = 705.$$

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$\underline{\text{Computation of} ~g(x) ~: ~\text{Advanced Shortcuts - the} ~\langle a,b,c\rangle ~\text{Pattern}}$

For illustration, I will specifically focus on the Pattern $~\langle 1,2,3\rangle.$

This means that among the five dice, there must simultaneously be at least one 1, at least one 2, and at least one 3. The various possibilities are explored below.

In this section, the symbol $~A~$ will be used to represent any die throw that is not a $~1, ~2, ~$ or a $~3.~$ This implies that there will be exactly $~3~$ choices for each die that is specified as $~A.$

• $\langle 1-1-1-2-3 \rangle, ~\langle 2-2-2-1-3 \rangle, ~\langle 3-3-3-1-2 \rangle.$
  Enumeration is $~3 \times \displaystyle \binom{5}{(3,1,1)} = 60.$

• $\langle 1-1-2-2-3 \rangle, ~\langle 1-1-3-3-2 \rangle, ~\langle 2-2-3-3-1 \rangle.$
  Enumeration is $~3 \times \displaystyle \binom{5}{(2,2,1)} = 90.$

• $\langle 1-1-2-3-A \rangle, ~\langle 2-2-1-3-A \rangle, ~\langle 3-3-1-2-A \rangle.$
  Enumeration is $~3 \times \displaystyle \binom{5}{(2,1,1,1)} \times 3^1 = 540.$

• $\langle 1-2-3-A-A \rangle.$
  Enumeration is $\displaystyle \binom{5}{(1,1,1,2)} \times 3^2= 540.$

Therefore, the enumeration of the $~\langle a,b,c\rangle~$ pattern is represented by

$$60 + 90 + 540 + 540 = 1230.$$

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$\underline{\text{Computation of} ~g(3)}$

This is represented by the enumeration of $~\langle 1,1,1\rangle. ~$ As previously illustrated, $~M(3) = 276.$ Therefore,

$$g(3) = \frac{M(3)}{7776} = \frac{276}{7776}.$$

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$\underline{\text{Computation of} ~g(4)}$

This is represented by the enumeration of $~\langle 1,1,2\rangle. ~$ As previously illustrated, $~M(4) = 705.$ Therefore,

$$g(4) = \frac{M(4)}{7776} = \frac{705}{7776}.$$

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$\underline{\text{Computation of} ~g(5)}$

$N(5)$ has already been computed as $~\displaystyle \binom{4}{2},~$
which is represented by $~D_1 ~: ~\langle 2,2,1 \rangle~$ and $~D_2 ~: ~\langle 1,1,3 ~\rangle.$

So, $~M(5) = |D_1 \cup D_2| = |D_1| + |D_2| - |D_1 \cap D_2|.$

Per previous discussion of the $\langle a,a,b\rangle ~$ pattern, you have that $~|D_1| = |D_2| = 705.~$

$~D_1 \cap D_2~$ requires $~\displaystyle \langle 1,1,2,2,3\rangle \implies |D_1 \cap D_2| = \binom{5}{(2,2,1)} = 30.$

Therefore,

$$g(5) = \frac{M(5)}{7776} = \frac{705 + 705 - 30}{7776} = \frac{1380}{7776}.$$

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$\underline{\text{Computation of} ~g(6)}$

$N(6)$ has already been computed as $~\displaystyle \binom{5}{2},~$ which is represented by
$~D_1 ~: ~\langle 2,2,2 \rangle, ~D_2 ~: ~\langle 1,1,4 \rangle, ~$ and $~D_3 ~: ~\langle 1,2,3 ~\rangle.$

So, $~M(6) = |D_1 \cup D_2 \cup D_3|$
$= |D_1| + |D_2| + |D_3| - |D_1 \cap D_2| - |D_1 \cap D_3| - |D_2 \cap D_3| + |D_1 \cap D_2 \cap D_3|.$

Per previous discussion of the $\langle a,a,a\rangle, ~\langle a,a,b\rangle, ~$ and $~\langle a,b,c\rangle~$ patterns, you have that
$~|D_1| + |D_2| + |D_3| = 276 + 705 + 1230 = 2211.~$

Then, $~|D_1 \cap D_2| = 0 = |D_1 \cap D_2 \cap D_3|.$

Also, $~D_2 \cap D_3~$ represents $~\langle 1,1,4,2,3 \rangle ~\implies ~|D_1 \cap D_3| = \displaystyle \binom{5}{(2,1,1,1)} = 60.$

Similarly, $~D_1 \cap D_3~$ represents $~\langle 2,2,2,1,3\rangle ~\implies ~|D_1 \cap D_3| = \displaystyle \binom{5}{3,1,1} = 20.$

Therefore,

$$g(6) = \frac{M(6)}{7776} = \frac{2211 - 60 - 20}{7776} = \frac{2131}{7776}.$$

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$\underline{\text{Computation of} ~g(7)}$

$N(7)$ has already been computed as $~\displaystyle \binom{6}{2},~$ which is represented by
$~D_1 ~: ~\langle 1,1,5 \rangle, ~D_2 ~: ~\langle 1,2,4 \rangle, ~D_3 ~: \langle 1,3,3\rangle, ~$
and $~D_4 ~: ~\langle 2,2,3 ~\rangle.$

So, $~M(7) = |D_1 \cup D_2 \cup D_3 \cup D_4|.$

$|D_1| + |D_2| + |D_3| + D_4| = 3345.$

$\displaystyle |D_1 \cap D_2| = \binom{5}{(2,1,1,1)} = 60.$

$\displaystyle|D_1 \cap D_3| = \binom{5}{(2,2,1)} = 30.$

$|D_1 \cap D_4| = 0.$

$\displaystyle|D_2 \cap D_3| = \binom{5}{(2,1,1,1)} = 60.$

$\displaystyle|D_2 \cap D_4| = \binom{5}{(2,1,1,1)} = 60.$

$\displaystyle|D_3 \cap D_4| = \binom{5}{(2,2,1)} = 30.$

All of the 3-way intersections and the 4-way intersection enumerate to $~0.$

Therefore,

$$g(7) = \frac{M(7)}{7776} = \frac{3345 - 60 - 30 - 60 - 60 - 30}{7776} = \frac{3105}{7776}.$$

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$\underline{\text{Computation of} ~g(8)}$

$N(8)$ has already been computed as $~\displaystyle \binom{7}{2},~$ which is represented by
$~D_1 ~: ~\langle 1,1,6 \rangle, ~D_2 ~: ~\langle 1,2,5 \rangle, ~D_3 ~: \langle 1,3,4\rangle, ~$
$~D_4 ~: ~\langle 2,2,4\rangle, ~$ and $~D_5 ~: ~\langle 2,3,3 ~\rangle.$

So, $~M(8) = |D_1 \cup D_2 \cup D_3 \cup D_4 \cup D_5|.$

$|D_1| + |D_2| + |D_3| + D_4| + |D_5| = 4575.$

$\displaystyle |D_1 \cap D_2| = |D_1 \cap D_3| = |D_2 \cap D_4| = \binom{5}{(2,1,1,1)} = 60.$

$\displaystyle |D_2 \cap D_5| = |D_3 \cap D_4| = |D_3 \cap D_5| = \binom{5}{(2,1,1,1)} = 60.$

$\displaystyle |D_2 \cap D_3| = \binom{5}{(1,1,1,1,1,)} = 120.$

$\displaystyle |D_4 \cap D_5| = \binom{5}{(2,2,1,)} = 30.$

All of the other set intersections enumerate to $~0.$

Therefore,

$$g(8) = \frac{M(8)}{7776} = \frac{4575 - 120 - (6 \times 60) - 30}{7776} = \frac{4065}{7776}.$$

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$\underline{\text{Computation of} ~g(9)}$

$N(9)$ has already been computed as $~\displaystyle 25,~$ which is represented by
$~D_1 ~: ~\langle 1,2,6 \rangle, ~D_2 ~: ~\langle 1,3,5 \rangle, ~D_3 ~: \langle 1,4,4\rangle, ~$
$~D_4 ~: ~\langle 2,2,5\rangle, ~D_5 ~: ~\langle 2,3,4\rangle$ and $~D_6 ~: ~\langle 3,3,3 ~\rangle.$

So, $~M(9) = |D_1 \cup D_2 \cup D_3 \cup D_4 \cup D_5 \cup D_6|.$

$|D_1| + |D_2| + |D_3| + D_4| + |D_5| + |D_6|= 5376.$

$\displaystyle |D_2 \cap D_6| = |D_5 \cap D_6| = \binom{5}{(3,1,1)} = 20.$

$\displaystyle |D_1 \cap D_3| = |D_1 \cap D_4| = |D_2 \cap D_3| = \binom{5}{(2,1,1,1)} = 60.$

$\displaystyle |D_2 \cap D_4| = |D_3 \cap D_5| = |D_4 \cap D_5| = \binom{5}{(2,1,1,1)} = 60.$

$\displaystyle |D_1 \cap D_2| = |D_1 \cap D_5| = |D_2 \cap D_5| = \binom{5}{(1,1,1,1,1,)} = 120.$

All of the other set intersections enumerate to $~0.$

Therefore,

$$g(9) = \frac{M(9)}{7776} = \frac{5376 - (2 \times 20) - (6 \times 60) - (3 \times 120)}{7776} = \frac{4616}{7776}.$$

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$\underline{\text{Computation of} ~g(10)}$

$N(10)$ has already been computed as $~\displaystyle 27,~$ which is represented by
$~D_1 ~: ~\langle 1,3,6 \rangle, ~D_2 ~: ~\langle 1,4,5 \rangle, ~D_3 ~: \langle 2,2,6\rangle, ~$
$~D_4 ~: ~\langle 2,3,5\rangle, ~D_5 ~: ~\langle 2,4,4\rangle$ and $~D_6 ~: ~\langle 3,3,4 ~\rangle.$

So, $~M(10) = |D_1 \cup D_2 \cup D_3 \cup D_4 \cup D_5 \cup D_6|.$

$|D_1| + |D_2| + |D_3| + D_4| + |D_5| + |D_6|= 5805.$

$\displaystyle |D_3 \cap D_5| = |D_5 \cap D_6| = \binom{5}{(2,2,1)} = 30.$

$\displaystyle |D_1 \cap D_3| = |D_1 \cap D_6| = |D_2 \cap D_5| = \binom{5}{(2,1,1,1)} = 60.$

$\displaystyle |D_2 \cap D_6| = |D_3 \cap D_4| = |D_4 \cap D_6| = \binom{5}{(2,1,1,1)} = 60.$

$\displaystyle |D_4 \cap D_5| = \binom{5}{(2,1,1,1)} = 60.$

$\displaystyle |D_1 \cap D_2| = |D_1 \cap D_4| = |D_2 \cap D_4| = \binom{5}{(1,1,1,1,1,)} = 120.$

All of the other set intersections enumerate to $~0.$

Therefore,

$$g(10) = \frac{M(10)}{7776} = \frac{5805 - (2 \times 30) - (7 \times 60) - (3 \times 120)}{7776} = \frac{4965}{7776}.$$

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$\underline{\text{Summary of Results}}$

$f(x)~$ denotes the probability that the sum of the first three dice will be $~x.~$

$g(x)~$ denotes the probability that you can select $~3~$ of the remaining $~5~$ dice so that the sum of the $~3~$ selected dice is $~x.$

Then, the overall probability that you can select $~3~$ of the latter $~5~$ dice to match the sum of the first $~3~$ dice is

$$\sum_{x = 3}^{18} \left[f(x) \times g(x)\right].$$

$f(x) ~$ equals

• $\dfrac{\binom{x-1}{2}}{216} ~: ~x \in \{3,4,\cdots,8\}.$
• $f(9) = \dfrac{25}{216}, ~f(10) = \dfrac{27}{216}.$
• $f(21-x) = f(x) ~: ~x \in \{3,4,\cdots,10\}.$

Then

• $g(3) = \dfrac{276}{7776}, ~~g(4) = \dfrac{705}{7776}, ~~g(5) = \dfrac{1380}{7776}, ~~g(6) = \dfrac{2131}{7776}.$

• $g(7) = \dfrac{3105}{7776}, ~~g(8) = \dfrac{4065}{7776}, ~~g(9) = \dfrac{4616}{7776}, ~~g(10) = \dfrac{4965}{7776}.$

• $g(21-x) = g(x) ~: ~x \in \{3,4,\cdots,10\}.$

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$\underline{\text{Addendum}}$

Just Java-sanity-checked $~M(3), ~M(4), ~\cdots, ~M(10).~$ Corrected mistakes in the computation of $~M(7)~$ and $~M(8).$