Question

Jake, Daniel, Timmy, Amy, and Pam are going on a fishing trip. They are taking a pick-up truck which only holds 3 people inside the truck. The other 2 people will have to ride in the back of the truck. No one wants to ride in the back.

They decide to let fate determine who has to ride in the back, by using this week's winning lottery ticket, which is about to be announced. The winning lottery ticket is a list of 5 numbers. Each number is a random integer from 1 to 1000.
Match each method for deciding who rides in the back with the correct assessment of its fairness.
(Consider a system to be fair when the probabilities of each event are equal.)Method Method 1: If the first number is 1- 200, Jake and Daniel ride in the back. If the first number is 201- 400, Daniel and Timmy ride in the back. If the first number is 401 - 600, Timmy and Amy ride in the back. If the first number is 601- 800, Amy and Pam ride in the back. If the first number is 801- 1000, Pam and Jake ride in the back.
Method 2: If the first number is even, Jake rides in the back. If the first number is odd, Daniel rides in the back. If the second number is 1 - 333, Timmy rides in the back. If the second number is 334 - 666, Amy rides in the back. If the second number is 667 - 1000, Pam rides in the back.
Method 3: Jake is assigned the number 1. Daniel is assigned the number 250 ,Timmy is assigned the number 500. Amy is assigned the number 750 . Pam is assigned the number 1000. The 2 people whose numbers are closest to the fourth number read will ride in the back.

Which method is fair? which method will cause Daniel and Jake to have the greatest chance of riding in the back? which method causes jake and pam to have the least chance of riding in the back

Answer 1

Explanation:

Answer 2

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Answer:

• fair: Method 1
• p(D&J) highest: Method 3
• p(J&P) lowest: Method 3

Explanation:

Method 1 gives a probability of riding in back of 0.2 +0.2 = 0.4 to each person. This method is fair. P(J&D) = P(J&P) = 0.2.

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Method 2 gives a probability of 1/2 each to Jake and Daniel, and about 1/3 each to Timmy, Amy, and Pam. Selection of both Jake and Daniel is impossible as one requires an even number, and the other requires an odd number. This method is not fair. The probability of both Jake and Daniel riding in back is zero. P(J&D) = 0; P(J&P) ≈ 0.167.

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Method 3 assigns probability 1/4 to Jake (who will always ride with Daniel) and Pam (who will always ride with Amy), and 1/2 to Daniel, Timmy, and Amy. It is not fair. The probability that Jake and Daniel will ride in back is 1/4 (maximum). The probability of both Jake and Pam riding in back is zero (minimum). P(J&D) = 0.25; P(J&P) = 0.

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The attached graph shows the distances of a possible number from each of the assigned numbers. (Method 3) The two closest are the two that are below 250 on the graph. For example, for lotto ticket number 250, Jake and Daniel are the closest; for ticket number 251, Daniel and Timmy are the closest. There is some ambiguity when the ticket number is 500, as both Daniel and Amy are 250 away. Hence, there is a tie for 2nd closest.

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Additional comment

Lottery ticket numbers are often listed in increasing order. One would need to be careful to use the numbers in the order drawn, not the order reported for the drawing. The minimum of the 5 numbers will not be uniformly distributed between 1 and 1000.