Stay or Search?
Jesse drives a lot from Chicago to New York for a business trip. On his way, he usually spends a night at one of 4 motels--A, B, C, D in order of appearance--in Youngstown, Ohio. Once he passes a motel, he never goes back. From his long experience, he knows that their rates per night are all distinct and evenly apart ($10) when arranged in ascending order.
One day, Jesse passes motel A whose rate is $80 and then arrives at motel B whose rate is $70. To spend as little money as possible for the night, should Jesse stay at B, or move on for C or D, or does it cost the same either way on average?
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1 solution
If he chooses to stay at B, he spends $70 for the night.
If he chooses to move on, he faces one of the following 3 possibilities, with equal probabilities:
Suppose that he moves on and finds the rate at motel C to be $100 with a probability of 3 1 × 2 1 = 6 1 . Then that means motel D's rate must be $90, which he will end up paying that night by moving on further.
Similarly, suppose that he finds the rate at motel C to be $50 with a probability of 3 1 × 2 1 = 6 1 . Then that means motel D's rate must be $60, so he will definitely stay at C, paying $50 for the night.
Now, suppose that he sees a rate of $90 at motel C, which could have come from ($70, $80, $90, $100) or ($60, $70, $80, $90) with equal chances. Then, motel D must have a rate of $100 or $60, the average of which is $80, which is less than the rate $90 at motel C. Therefore, whenever he sees a rate of $90 at motel C with a probability of 3 1 × 2 1 + 3 1 × 2 1 = 3 1 , he chooses to take chances and move on, paying an average amount of $80 for the night.
Finally, suppose that he sees a rate of $60 at motel C, which could have come from ($60, $70, $80, $90) or ($50, $60, $70, $80) with equal chances. Then, motel D must have a rate of $90 or $50, the average of which is $70, which is more than the rate $60 at motel C. Therefore, whenever he sees a rate of $60 at motel C with a probability of 3 1 × 2 1 + 3 1 × 2 1 = 3 1 , he chooses to stay, paying $60 for the night.
Overall, by moving on, he pays on average (in dollars)
6 1 × 9 0 + 6 1 × 5 0 + 3 1 × 8 0 + 3 1 × 6 0 = 7 0 .
So, staying at B or moving on for C or D, it costs the same amount $70 on average. □