To find the probability that the third roll is double the sum of the first two rolls, we need to first find the possible outcomes of each roll and then calculate the probability of each outcome. Let's the first roll as X1, the second roll as X2, and the third roll as X3.
The only way for each roll are {1,2,3,4,5,6}, and since the die is fair, each outcome is equally likely.
Now, we need to find the cases where X3 is double the sum of X1 and X2. This can happen in the following ways:
If X1+X2 = 1, then X3 must be 2.
If X1+X2 = 2, then there is no possible value of X3 that satisfies the condition.
If X1+X2 = 3, then there is no possible value of X3 that satisfies the condition.
If X1+X2 = 4, then X3 must be 8.
If X1+X2 = 5, then there is no possible value of X3 that satisfies the condition.
If X1+X2 = 6, then X3 must be 12.
If X1+X2 = 7, then there is no possible value of X3 that satisfies the condition.
If X1+X2 = 8, then X3 must be 16.
If X1+X2 = 9, then there is no possible value of X3 that satisfies the condition.
If X1+X2 = 10, then X3 must be 20.
If X1+X2 = 11, then there is no possible value of X3 that satisfies the condition.
If X1+X2 = 12, then X3 must be 24.
Therefore, there are 5 possible outcomes that satisfy the condition, out of a total of 6^3 = 216 possible outcomes (since each roll has 6 possible outcomes). So the probability of X3 being double the sum of X1 and X2 is:
P(X3 is double the sum of X1 and X2) = 5/216 ≈ 0.0231
So, the probability of the third roll being double the sum of the first two rolls is approximately 0.0231.