Probability of getting a total less than 8 from three dice given that the total when two dice are rolled is less than 4.

Question

I've been struggling with this question involving probability and dice rolls:

A die is rolled three times and the scores are added. The total obtained after it has been rolled twice is 4 or less. Given that this is the case, find the probability that after the die has been rolled for the third time the total will be less than 8. Give your answer in its simplest exact form.

I've calculated the probability of getting a total of 4 or less from two dice rolls, which is 1/6. However, I'm not sure how to extend this to the three dice problem. If the total from two is 4 or less then the only ways of getting a total of 8 or more from three dice are to get a 4, 5 or 6. But the lower the total from the first two dice is, the higher the probability of getting less than 8.

I've tried using the formula $P(A|B)=P(A ∩ B)/P(B)$, making A the probability of getting a total of less than 8 from 3 dice and B the probability of getting a total of 4 or less from 2 dice. However, I can't seem to get the right answer. I got P(A) = 25/216 (by considering all the possible combinations of three dice rolls that would result in a total less than 8, and the different possible permutations of each combination), and P(B) = 1/6.

Answer 1

HINT

• in the first two rolls, one/two way(s) to roll sum $2/3$,

• If the first two rolls yielded $2/3,\;$ five/four ways the third die can be rolled to meet conditions.

Can you now put the things together ?

Answer 2

Let event $E_A$ represent getting a sum of $4$ or less in the first two rolls. In how many ways can this happen? Well, the sum could be $2$, $3$, or $4$.

$$\begin{array}{c|c|c} \text{Sum after $E_A$} & \text{Options} & \text{# Ways}\\ \hline 2 & (1, 1) & 1 \\ \hline 3 & (1, 2) \, (2, 1) & 2 \\ \hline 4 & (1, 3)\, (3, 1) \, (2, 2) & 3\\ \hline & & \mathbf{6} \end{array}$$

A total of $6$ ways.

Now, let $E_B$ represent $E_A$ and then a third roll of the die happening. The third roll of the die can happen in $6$ ways, so altogether we have $6 \times 6 = 36$ ways.

Finally, let $E_C$ represent $E_A$ happening and then after the third roll, the sum being less than $8$.

Here's a table showing how $E_C$ can play out in $22$ ways.

$$\begin{array}{c|c|c} \text{Sum after $E_A$} & \text{Acceptable range ($3$rd roll)} & \text{# Ways (Adjusted}\\ \hline 2 & 1 - 5 & 5 \times 1 \\ \hline 3 & 1 - 4 & 4 \times 2\\ \hline 4 & 1 - 3 & 3 \times 3\\ \hline & & \mathbf{22} \end{array}$$

[Understanding the table - Take the second row for example, there are $2$ ways in which the sum could be $3$ and on the third roll, $4$ outcomes are acceptable. so, a total of $2 \times 4 = 8$ ways]

So, the probability of getting a sum less than $8$ after the third roll when the first two rolls had yielded a sum of $4$ or below is $\, 22/36 \,$ or $\, 11/18 \,$.