My last post dealt with the probability curiosity referred to as The Monty Hall Problem. I though it would be useful to call out two ways of conceptualizing its highly counterintuitive solution. Neither of these concepts are mine. I simply offer them to enhance appreciation of the problem’s solution.
Concept 1
Step throught the scenarios:
Behind one door is the car, behind the other two are goats (Nanny and Fanny).
Scenario 1: The contestant initially selects the door with the car. The host then reveals Nanny. If the contestant switches doors, hu will select Fanny (lose).
Scenario 2: The contestant initially selects the door with Nanny. The host then reveals Fanny. If the contestant switches doors, hu will select the car (win).
Scenario 3: The contestant initially selects the door with Fanny. The host then reveals Nanny. If the contestant switches doors, hu will select the car (win).
Hence, switching results in a winning outcome in two out the three possible scenarios.
Concept 2
Increase the number of doors to 100. The player picks a door, then the host opens 98 of the other doors, revealing goats (they have names, but not as good as Nanny and Fanny). The host offers the contestant the opportunity to switch doors. The original odds that the contestant had of picking the door with the car behind it remain 1 in 100. Only one other door remains unopened. The odds of the car being behind that door are therefore 99 in 100.