Probability and Expectation ```Name: Tuan C. Status: student Age: N/A Location: N/A Country: N/A Date: N/A ``` Question: Given two boxes, and told that one contains double the amount of money of the other. When one box is chosen at random and opened, there were 200 dollars. Most would reason that the other box will contain either 100 or 400, and therefore the expectation is 250, and hence it is superior to the first box. Since the first box was chosen at random, is there a fallacy in applying probability considerations here? Thanks. Replies: Tuan, To use the expectation of 250 as a definition of what is in the box is the fallacy. Consider the two options, as compared to taking the first box. If you choose the second rather than the first, you might lose 100 dollars. On the other hand, you might gain 200 dollars. By choosing the other box, you have a 50% chance of losing \$100 and a 50% chance of gaining \$200. If the same option with the first box containing \$200 comes up 10,000 times, always choosing the first box results in a gain of \$2,000,000. Always choosing the second box results in a gain of \$2,500,000. A person who likes risks will choose the second box because he stands to win more than he can possibly lose. Dr. Ken Mellendorf Physics Instructor Illinois Central College This apparent paradox is resolved by: 1. recognizing that the information "one contains double the amount of money of the other." removes the condition that amounts of money are randomly distributed. and 2. the proper statement of the "expectation value" is that "There is a 50% chance that the second box will contain either \$100 or \$400." That there is a misapplication of the term "expectation value" consider the analogous "experiment". A coin has a 50% chance of landing "heads" or "tails". If the coin is tossed again there is still a 50% chance of the coin landing "heads" or "tails". The events must also be independent, which is not the case if you know the amounts in the second box. Finally, restate the problem, "The second box contains either \$1 or \$999." You have a 50/50 chance of getting either \$1 or \$999. It is the "boxes" you are averaging, not the "amounts" in the boxes. That has no relevance regarding the probability of choosing the second box. Vince Calder Click here to return to the Mathematics Archives

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