Probability of sum of four dice being greater than or equal to $10$

Question

Four fair dice are rolled, the probability of getting a sum of at least 10 on the dice is

Here, I am not able to figure out a suitable approach to this question. Please enlighten me.

Answer 1

Via stars-and-bars and inclusion exclusion:

The probability of rolling a $10$ or greater as the sum of four dice will be $1$ minus the probability of having rolled a $9$ or less. Let us attempt this problem instead for simplicity.

First, let us assume that the dice are all rolled in sequence or are otherwise distinguishable. This allows us to work in an equiprobable sample space allowing us to use counting techniques to calculate probabilities. Let $x_1,x_2,x_3,x_4$ be the results of the first, second, third and fourth die rolls respectively.

We are interested in counting the number of positive integral solutions to the system:

$$\begin{cases}x_1+x_2+x_3+x_4\leq 9\\ 1\leq x_1\leq 6\\ 1\leq x_2\leq 6\\ 1\leq x_3\leq 6\\1\leq x_4\leq 6\end{cases}$$

To accomplish this, let us do two things. First... let us temporarily ignore the upper bounds on the variables. That is, rather than looking at $1\leq x_1\leq 6$ we look instead at $1\leq x_1$. We will take care of the upper bounds later. Second, let us introduce a placeholder variable which accounts for the leftover amount needed to make the sum a total of $9$ exactly rather than the sum simply being at most $9$. That is, introduce $x' = 9 - x_1-x_2-x_3-x_4$

We are now then trying to find the number of solutions to the system:

$$\begin{cases}x_1+x_2+x_3+x_4+x' = 9\\1\leq x_1\\1\leq x_2\\1\leq x_3\\1\leq x_4\\0\leq x'\end{cases}$$

One more change of variables, letting $y_i = x_i-1$ for each $1\leq i\leq 4$ and $y'=x'$ we are trying to find the number of solutions to the system:

$$\begin{cases}y_1+y_2+y_3+y_4+y' = 5\\ 0\leq y_i~~~\forall i\in\{1,2,3,4\}\\0\leq y'\end{cases}$$

You mentioned "balls in bins" in the comments so this should be in a problem form that you know how to solve. Do so.

The number of non-negative integer solutions to the system $\begin{cases}x_1+x_2+\dots+x_k = n\\0\leq x_i~~\forall i\end{cases}$ is via stars-and-bars equal to $\binom{n+k-1}{k-1}$

Now... among these arrangements that we just counted, there is the possibility that there are some of those which violated the upper bound conditions that we had tried to exclude... that is, when there were one or more of the $x_i$ who were greater than or equal to $7$ (equivalently, one or more of the $y_i$ who were greater than or equal to $6$). Apply inclusion-exclusion based on whether or not one or more of the upper bound conditions were violated. Count how many arrangements we counted were "bad" because $x_1$ specifically was too large. Recognize that this is the same amount who were bad because $x_2$ was too large, similarly for $x_3$ and $x_4$. Finally, note that it is impossible for multiple to have been too large while simultaneously having a sum of $9$ or less (as four dice with two dice rolling $7$ or greater would have the sum be strictly greater than $15$) so the inclusion-exclusion can be finished quite early rather than having to deal with intersections of two, three, or four events (like would have been necessary had we approached the problem directly rather than indirectly like we are doing here).

In fact, if you observe closely, you will find there were not any bad arrangements at all. If $x_1\geq 7$ and $x_i\geq 1$ for the other $i$, then $x_1+x_2+x_3+x_4\geq 7+1+1+1=10$. Still, it is worth pointing out the general technique with inclusion-exclusion that should have been followed if there had been some "bad" arrangements we counted.

Now, after having arrived at a corrected count of how many different possible outcomes there are corresponding to a sum of $9$ or less, divide by the total number of outcomes of rolling four dice there are overall. Finally, subtract this value away from $1$ in order to get the result to the original problem.

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Final answer:

$1 - \dfrac{\binom{5+5-1}{5-1}}{6^4} = \dfrac{65}{72}$