University of Toronto stats prof. Jeffrey S. Rosenthal wrote a nice paper called "Monty Hall, Monty Fall, Monty Crawl" (https://web.archive.org/web/20230706235

University of Toronto stats prof. Jeffrey S. Rosenthal wrote a nice paper called "Monty Hall, Monty Fall, Monty Crawl" (https://web.archive.org/web/20230706235720/https://probability.ca/jeff/writing/montyfall.pdf https://web.archive.org/web/20230706235720/https://probabili...) in Math Horizons (Sep 2008). It discusses two variations of the Monty Hall problem that are likely to be answered incorrectly by those who used flawed logic to arrive at the correct answer to the original: > Monty Fall Problem: In this variant, once you have selected one of the three doors, the host slips on a banana peel and accidentally pushes open another door, which just happens not to contain the car. Now what are the probabilities that you will win the car if you stick with your original selection, versus if you switch to the remaining door? > Monty Crawl Problem: As in the original problem, once you have selected one of the three doors, the host then reveals one non-selected door which does not contain the car. However, the host is very tired, and crawls from his position (near Door #1) to the door he is to open. In particular, if he has a choice of doors to open (i.e., if your original selection happened to be correct), then he opens the smallest number available door. (For example, if you selected Door #1 and the car was indeed behind Door #1, then the host would always open Door #2, never Door #3.) What are the probabilities that you will win the car if you stick versus if you switch?