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#1 2012-10-11 06:42:56

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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 2012-10-15 05:52:48

Full Member


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 (2012-10-15 05:54:03)

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