Question

After a boating mishap, you find yourself stranded on a deserted island with a pirate and...

After a boating mishap, you find yourself stranded on a deserted island with a pirate and a crate of shoes. There are ten pairs of shoes in the crate. You manage to grab 10 left shoes and 5 right shoes before being chased away by “Captain Hook”.

(a) Suppose your pirate friend has only one good leg – his right leg (his left leg is a wooden stump). Luckily, your feet are intact. Let x1 denote right shoes and x2 denote left shoes. Draw the corresponding Edgeworth box with indifference curves for you and Captain Hook.

Hint: Since you have two good feet suppose that your utility from shoes is

u1 (x1; x2) = min {x1; x2} (3)

while your pirate friend’s utility from shoes is given by

u2 (x1; x2) = x1: (4)

(b) If and when you trade with Hook (assuming he does not just simply kill you for your shoes) what will be the outcome? Illustrate the set of Pareto efficient outcomes of trade.

(c) After a few days on the island Hook loses his good leg. Thus he is left with only two stumps on his legs. In any case, Hook finds shoes useful, but given his condition, he is indifferent between right and left shoes. (As long as they fill the stump they do the same job!) Draw the corresponding Edgeworth box along with indifference curves.

(d) Suppose again that you own 10 left shoes and 5 right ones. What will be the outcome when you are allowed to trade with Hook when he has two stumps for legs? Illustrate the set of Pareto-efficient outcomes of trade.

Homework Answers

Answer #1

First let us calculate the number of ways 2 pairs can be picked. Since there are 10 pairs, this can come in 10C2 = 45 ways.

Now let us calculate the number of ways only one pair out of the four can be picked. The one pair can come in 10 ways. The remaining two shoes can come in 18C2 ways. However since we want only one pair which has already been selected, we subtract 9 (for 9 pairs). Thus the total number of ways = 10*(18C2-9) = 10*(153-9) = 1440.

Total number of combinations with atleast one pair = 45 + 1440 = 1485.

Total number of combinations = 20C4 = 4845.

Probability = 1485/4845 = 0.3065

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