Science:Math Exam Resources/Courses/MATH103/April 2012/Question 02 (e)
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Question 02 (e)  

Short answer question. Show in detail how you arrive at your answer (work will be considered for this problem). In (a simplified version of) the dice game Pass the Pigs two model pigs serve as dice. Each model pig can either land on the side (with probability p(S) = 4/8), on the feet (with probability p(F) = 3/8), or on the snout (with probability p(N) = 1/8). In each turn both model pigs are rolled. The scoring is as follows:
Find the average score per turn. You can leave your answer unsimplified as a sum of fractions. 
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint. 
Hint 1 

Remember that the average number of points should be where is a random variable representing the number of points. Find the probabilities that X is equal to each value, and compute the sum. 
Hint 2 

We can find the probability that is equal to a given value by looking at all of the different combinations of rolls that can give a value for . For example: Now, perform the same calculation for the other possible values of . 
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution  

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Please rate my easiness! It's quick and helps everyone guide their studies. This game has 9 outcomes, but due to symmetry only 6 of them are distinct. The probabilities are distributed as follows (multiples of 1/64)
Let X be the random variable that returns the score. Then and thus the average is found to be 