One unfortunate similarity between this puzzle and the original "Monty Hall Puzzle" is that when someone who doesn't know the solution paraphrases it, it can easily turn into a different (and possibly unsolvable) puzzle! After reading the original wording, I could find a strategy that yields a 2/3 success rate. Here is the original statement from the Mathematical Intelligencer. One strategy is given at the very end of this email. The Return of Monty Hall In a new ("Let's-Make-A-Deal"-type) game show for couples, there are 3 curtains behind which are hidden a car, a car key, and a goat. One member of the couple is designated "the carmaster" -- the car-master's goal is to find the car. The other member is designated "the key-master" -- the keymaster's goal is to find the key. If both partners succeed in their respective tasks, the couple drives away in their new car. If either one fails, the couple receives the booby prize, the goat. The game begins with the car-master (at this point, the key-master is led out of the room and cannot observe the proceedings). The car-master has two tries to find the car (i.e., open any curtain; if the car isn't there, then open another curtain). If the car-master succeeds in finding the car, all open curtains are reclosed, and the keymaster is brought back into the room. No communication whatsoever is permitted between the car-master and key-master at this point. The keymaster now has two tries to find the key (i.e., open any curtain; if the key isn't there, open another curtain). Assuming the couple plays optimally, what are their odds of winning the car? Regards, Nick Bernie Cosell wrote:
In a recent Math Intelligencer, AS Landsberg posed a variation on Monte Hall [paraphrased]:
There are three doors, behind which are a car, a car key and a goat. There are two players, the car-player and the key-player. Each player gets to pick two doors to see if they can find their corresponding prize. Each player acts independently and with no information from the other. If they can each find their corresponding prize, the two drive off in their new car.
He mentions a proof that with a sufficiently clever strategy the two will be able to drive off in their car 2/3rds of the time. I just can't see how to make that work! If they each pick randomly, I get the odds of their winning the car as 4/9. It is fairly easy to come up with strategies that up the odds to 1/2. But I just don't see how to get it all the way to 2/3rds.
/Bernie\
======================= Solution For the car-master, first pick the first curtain. If it's the car, then done. If it's the key, then pick the third curtain next. If it's the goat, then pick the second curtain next. For the key-master, first pick the third curtain. if it's the key, then done. If it's the car, then pick the first curtain next. If it's the goat, then pick the second curtain next. This strategy ensures that the two players fail in exactly the same two cases, thus achieving an overall success rate of 2/3. This assumes that the two players can establish a reliable labeling of the curtains. =======================