OK, here's a nice variant:
"You are playing the Monty Hall problem. However, you secretly know one of the goats is the former pet of an eccentric billionaire who lost it and is willing to pay an enormous amount for its return, way more than the car is worth. You really want that goat.
"The host is unaware of this. After you pick your door, as is traditional, the host opens one door, which he knows doesn't have the car. He reveals a goat, which you can tell is the ordinary goat and not the secretly valuable one. The host offers to let you switch doors.
"Should you?"
Vikram Punathambekar
in reply to Colin the Mathmo • • •Colin the Mathmo
in reply to Vikram Punathambekar • • •@vpunt It's not identical, because Monty's choice is based around the existence of the car. In the original it's the car you want, in this version it's one of the (indistinguishable to Monty) goats you want, so there is a difference.
Whether the calculation produces the same result or not, it's a different calculation that you have to perform.
Vikram Punathambekar
in reply to Colin the Mathmo • • •but if Monty had revealed the billionaire's goat there's no puzzle right? The revealed goat being the dud is baked into the puzzle.
In both puzzles, a goat that you *don't want has been revealed already*.
In the original, you want the car, in this version, you want the other goat - mutually exclusive outcomes.
josh g.
in reply to Colin the Mathmo • • •mpark
in reply to Colin the Mathmo • • •Does this actually change anything? Just like in the original, there is a 1 in 3 chance you made the correct choice. Once Monty reveals the unwanted choice, there's a 2 in 3 chance the desired prize is behind the unchosen door. So you switch.
All this seems to do is reduce the penalty for being unlucky.
Sarah Brown
in reply to Colin the Mathmo • •like this
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