Different probabilities with ten dice

Question

Given 10 dice:

1. What probability do I have to throw ten dice, so that the sum of each gives 24?
   • I have no idea how to calculate this one as simple as possible.
2. What are the chances of $3$ dice in a row $(1, 2, 3)$, $(2, 3, 4)$, $(3, 4, 5)$, $(4, 5, 6)$? (You throw $10$ dice and at least $3$ of them have to be the triplets.)
   • Can this be calculated by saying $4 \times \left({1\over6}\right)^7$, since we have four combinations, each requiring $3$ given digits?
3. What are the chances of $5$ dice in a row $(1, 2, 3, 4, 5)$ or $(2, 3, 4, 5, 6)$
   • Analogous to the above problem, I suggest $2 \times \left({1\over6}\right)^5$.
4. What are the chances of a pair $(1, 1)$, $(2, 2)$, $(3, 3)$, $(4, 4)$, $(5, 5)$, $(6, 6)$?
   • Since there are $10$ dice, this should be $100\%$. Or not?
5. What are the chances of two, three, or four pairs (separately)?
   • Since there are $10$ dice, you always have $4$ pairs, right?
6. What are the chances of one, two or three "drillings" (three of the same value)?
   • The first die can be any of $6$ options, then two are a fixed option, and the rest are, again, one of $6$ options. I think the answer should then be $\left({1\over6}\right)^7$.
7. What are the chances of rolling seven dice with the same value?
   • I think this is $\left({1\over6}\right)^7$.

Answer 1

Hints: For (1): Let $x_i$ denote the value rolled on the $i$-th die.

You have $\displaystyle \sum_{i=1}^{10} x_i = 24$ and $\forall i, 1\le x_i \le 6$.

We can translate this problem to: $$\sum_{i=1}^{10}y_i = 14, \forall i, 0\le y_i \le 5$$

Each solution to this Diophantine Equation maps to an arrangement of dice that add up to 24, and over all possible solutions, you map to all possible arrangements of dice. A single arrangement of dice has probability $\dfrac{1}{6^{10}}$.

I need to think more about problems (2) and (3).

For (4), you are correct. This is the Pigeonhole Principle.

For (5), you need more clarification. Can $(1,1,1)$ be considered three pairs?

I do not understand what you are looking for with problem (6).

For (7): Assuming you want exactly seven dice that are the same, choose seven of the dice. Choose the value for those dice. The remaining three dice can be any of the other five values.