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#1 2012-10-10 07:42:56

Registered: 2012-01-10
Posts: 232

4 couples

4 couples are going on vacation to a 5-star hotel in Martinique. The receptionist gave them four consecutive rooms with numbers from 1 to 4. The women went upstairs to choose rooms (and each of them stayed to her own room), while their husbands were having a drink at the bar. After some drinks, they decided to go upstairs to their rooms. 
"If we all go upstairs and start opening the 4 doors at random until we find our wife, what are the chances we get to the correct room with maximum 2 attempts each?" the first man asked.

"This is easy", says the second. "Since we will open 2 out of 4 doors, each of us has 50% probability to locate his wife. So, in order for all four of us to make it, the probability is (1/2)^4, that is, 1/16."

"Wrong!" says the third. There is a way to get to the correct room with probability greater than 40% and without havng to communicate with our wives or between us, after the first of us goes upstairs."

What is the method he is suggesting?


#2 2012-10-14 06:52:48

Registered: 2009-11-13
Posts: 224

Re: 4 couples

Let's call the men A, B, C, and D. Their plan is as follows: A will go to Room 1, B to Room 2, C to Room 3, and D to Room 4. If a man does not see his own wife in the first room, he will then go to the room corresponding to the man whose wife he sees in the first room – e.g. if A sees B's wife in Room 1, he opens Room 2 next; if it's C's wife he sees, he goes to Room 3 next, etc.

This strategy will improve their chances of finding the right wives in two attempts or fewer from 1/16 = 6.25% to 10/24 ≈ 41.67%.

Last edited by scientia (2012-10-14 06:54:03)


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