Start the week with a little 'reverse' puzzle-thinking... There are lots of variations of the Monty Hall “paradox,” but as probably all readers here know in its standard form it is statistically better (indeed twice as good) for the contestant to ‘switch’ doors in the usual game as to stay with their initially chosen door. But one can make a small change in the set-up to make it better for the contestant to stick with their chosen door.
Instead of the usual premise of say a car behind one door, and billy goats behind the other two, consider the case of a goat behind one door and cars behind the other two (continue to assume the contestant prefers a car... even though billy goats can be downright appealing… and host Monty still knows what’s behind all 3 doors).
Now, contestant Sheldon picks say Door #1, and Monty in turn opens Door #3 revealing a car… should Sheldon switch to Door #2 or remain with #1 (or, does it make any difference)?
The quick math shows that at the outset Sheldon had a 2/3 chance of selecting a car by random chance, better than the chance that either one of the remaining doors hide a car, and thus Sheldon is better off staying with his 2/3 first pick in this instance (essentially, exactly the reverse argument of the normal set-up).
(Though, IF, as a prior, we know that Sheldon is a telepathic extraterrestrial, or one with X-ray vision, than that could alter the odds considerably…;)
