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In the game of craps, people bet on the outcome of the throw of
two, six-sided dice. Specifically, wagers are placed on the sum of
the outcome on the upward facing number on each of the two dice. We
are going to discuss relative likelihoods by counting the number of
ways (the number of microstates) associated with specific sums (the
macrostate).
A. How many microstates are associated with a throw summing to
4?
B. How many microstates are associated with a throw summing to 5?
C. How many microstates are associated with a throw summing to 10?
D. Which is more likely, rolling 2 dice that sum up to 4 or rolling 2 dice that sum up to 5?
E. Which is more likely, rolling 2 dice that sum up to 4 or rolling 2 dice that sum up to 10?
The total number of different possible outcomes will be:
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)
(2,1),(2,2,),(2,3),(2,4),(2,5),(2,6)
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
A. Macrostate = 4
Possible microstates = (1,3), (2,2), (3,1)
Hence, there are 3 microstates associated with the macrostate 4.
B. Macrostate = 5
Possible microstates = (1,4), (2,3), (3,2), (4,1)
Hence, there are 4 microstates associated with the macrostate 5.
C. Macrostate = 10
Possible microstates = (4,6) ,(5,5), (6,4)
Hence, there are 3 microstates associated with the macrostate 10.
D. Total outcomes = 36
The number of different ways to get a sum of 4 = 3 (from part A)
Probability (sum = 4) = 3/36 = 1/12 = .083333
The number of different ways to get a sum of 5 = 4 (from part B)
Probablity (sum = 5) = 4/36 = 1/9 = .111111
Since the probabilty of getting a sum of 5 is higher, rolling 2 dices that sum up to 5 is more likely.
E. Total outcomes = 36
The number of different ways to get a sum of 10 = 3
Probability (sum = 10) = 3/36 = 1/12 = .083333
Also, Probabiltiy (sum = 4) = .083333
Since the probabilty of getting a sum of 4 and the probabilty of getting a sum of 10 are equal, both are equally likely.