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Tue, 21 Sep 2010

Monty Hall V2

Today I have a small puzzle for you. It was originally told to me by logix and is a nice variation of the Monty Hall problem.
If you've never heard about the Monty Hall problem before, you could check out the linked Wikipedia article.

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 car-master:
- the car-master's goal is to find the car.
The other one is designated the key-master:
- the key-master'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 gets nothing.

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 re-closed, and the key-master is brought back into the room. No communication whatsoever is permitted between the car-master and key-master at this point.

The key-master 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?

Randomly picking works out to 4⁄9, i.e. about 44% – You can surely improve this, but how? And what is the best solution? Post your suggestions in the comments…

– Sec


posted at: 09:57 | Category: /pastimes | permanent link to this entry | 6 comments (trackback)
 

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