Frank is a cold-hearted economist who has been given a chance to enter a 2-die l
ID: 1210531 • Letter: F
Question
Frank is a cold-hearted economist who has been given a chance to enter a 2-die lottery with an amazing prize. The lottery works as follows. Before two fair dice are rolled, Frank can choose an integer number n between 2 and 12, that is 2 le n le 12. If the sum of the rolled dice matches Frank's selection (e.g., the roll is (1, 2) and he chose n = 3), then Frank wins the prize. Otherwise he loses. Before making a decision, Frank wants to contemplate this problem in the light of the expected utility model. For that, he needs to construct the outcome space and assign probabilities to the different numbers he can select. Write down the outcome space, i.e., the space of all possible valid outcomes in this lottery. For example, a valid outcome is the draw (1,4), which signifies that the roll of the first die ended in 1 and the roll of the second ended in 4. What is the probability that Frank wins if he chooses n = 3? What is the probability that Frank wins if he chooses n = 5?Explanation / Answer
A) SINCE BOTH THE DICE HAVE 6 NUMBERS ON THEM FROM 1 TO 6, SO FOR EVERY NUMBER ON FIRST DICE WE HAVE SIX POSSIBLE OUTCOMES ON THE OTHER DICE. IT MEANS THAT FOR SIX OUTCOMES ON FIRST WE HAVE SET OF SIX DIFFERENT OUTCOMES ON SECOND DICE.
BY USING MULTIPLICATION PRINCIPLE WE SE THERE ARE 6 PROBABILITIES FOR DICE 1 AND 6 FOR DICE 2 OR 6* 6 = 36 OUTCOMES.
B) NOW SINCE WE KNOW THAT THERE CAN BE TOTAL OF 36 OUTCOMES THAN FOR ANY ONE SET WE WILL HAVE A CHANCE OF ONLY ONE OUT OF 36 OR 1/36.
n = 3 CAN BE OBTAINED IN ANY OF THE TWO WAYS (1,2) OR (2,1) OR WE CAN SAY THERE ARE ONLY TWO SETS WITH A TOTAL OF 3 OR 1/36 + 1/36 = 2/36 OR 1/18 ARE THE CHANCES THAT FRANK WINS IF HE CHOOSES 3. P(E) = 1/18
C) NOW SINCE WE KNOW THAT THERE CAN BE TOTAL OF 36 OUTCOMES THAN FOR ANY ONE SET WE WILL HAVE A CHANCE OF ONLY ONE OUT OF 36 OR 1/36.
n = 5 CAN BE OBTAINED IN ANY OF THE FOUR WAYS (1,4) OR (2,3) OR (3,2) OR (4,1) WE CAN SAY THERE ARE ONLY FOUR SETS WITH A TOTAL OF 5 OR 1/36 + 1/36 + 1/36 + 1/36 = 4/36 OR 1/9 ARE THE CHANCES THAT FRANK WINS IF HE CHOOSES 5. P(E) = 1/9
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