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Using probability to make fair decisions

Jake, Daniel, Timmy, Amy, and Pam are going on a fishing trip. They are taking a pick-up truck which only holds 33 people inside the truck. The other 22 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 55 numbers. Each number is a random integer from 11 to 10001000.
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 1\text{Method }1: If the first number is 11 - 200200, Jake and Daniel ride in the back. If the first number is 201201 - 400400, Daniel and Timmy ride in the back. If the first number is 401401 - 600600, Timmy and Amy ride in the back. If the first number is 601601 - 800800, Amy and Pam ride in the back. If the first number is 801801 - 10001000, Pam and Jake ride in the back.
Method 2\text{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 11 - 333333, Timmy rides in the back. If the second number is 334334 - 666666, Amy rides in the back. If the second number is 667667 - 10001000, Pam rides in the back.
Method 3\text{Method }3: Jake is assigned the number 11. Daniel is assigned the number 250250. Timmy is assigned the number 500500. Amy is assigned the number 750750 . Pam is assigned the number 10001000. The 22 people whose numbers are closest to the fourth number read will ride in the back.
Decision-making method
Fairness assessment
  • Method 1\text{Method }1
  • Method 2\text{Method }2
  • Method 3\text{Method }3
  • Not fair - Jake and Daniel have the greatest chance of riding in the back.
  • Fair
  • Not fair - Jake and Pam have the least chance of riding in the back.