# One drunk receptionist and four brilliant mathematicians

Suppose that there is one hotel with nine floors (first floor = ground floor + 1) where the math seminar takes place, four brilliant mathematicians who are guests of the hotel, one drunk receptionist and four invitation letters for night party each addressed to single one of the mathematicians. All mathematicians are in the conference room which is on the ground floor . At the one moment one of the mathematicians comes out of the conference room and goes to the floor where his room is by elevator. Since he is drunk, receptionist forgets to deliver a invitation letter to mathematician. Ten minutes later, since he is drunk, he takes randomly one of the letters instead of all four and goes after mathematician. Receptionist enters the elevator but he doesn't know on which floor mathematician's room is. Only thing that he knows for sure is that room isn't on second, fourth, sixth and ninth floor .

What is probability that receptionist on the first try goes to the right floor where the mathematician is and that letter which he carries is addressed exactly to that mathematician?

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Seems like it should be 1/20 but this question is too twisty to make sense of. Isn't it possible the drunk receptionist will enter and then immediately exit the elevator and then attempt to find the mathematician on the first floor where the reception also presumably is? What floor is the conference room on and does it matter? Not asking for spoilers of course. :) – Dan Brumleve Sep 15 '11 at 5:31
This had such an alluring title, I'm a little disappointed in the delivery. :) – anon Sep 15 '11 at 5:46
@pedja, so why complicate the question by adding them? – Peter Taylor Sep 15 '11 at 8:09
This question is entirely missing motivation: where does it come from? Why do you want to know the answer? What makes this a question of general interest rather than something localized to a very particular (rather contrived) situation? – Pete L. Clark Sep 15 '11 at 9:14
@pedja: If you are actually interested in a number theoretic question, why are you asking only about a single value of $n$ and phrasing it in terms of a hotel? – Pete L. Clark Sep 15 '11 at 13:29

Presumably the receptionist could have gone to one of the floors 3, 5, 7, or 8 with equal probability (though if he's so plastered he didn't bother checking which letter he picked, one wonders how he pushed a button corresponding to one of the intended floors). And presumably the mathematician is on one of these floors with equal probability. Then the problem of being on the same floor is equivalent to asking: if two people each pick a number from $\{3,5,7,8\}$, what's the probability they are the same? There's $4$ ways they could each pick the same number, and $4\times4=16$ ways they could each pick some number independently, so there is a $4/16=1/4$ chance of collision. Now the chance the letter is the correct one is $1/4$, and completely independent of whether or not the receptionist got the right floor, so the chance that the receptionist got both the floor correct and the letter correct is $1/4\times1/4=1/16$.
Alternatively, if both the mathematician and the receptionist may end up on the first floor, the numbers are instead $\{1,3,5,7,8\}$, giving a final probability of $\frac{5}{5\times5}\times1/4=1/20$ by our previous reasoning. If the mathematician's floor may be the first but the receptionist is destined to go up, then for every admissible floor there is a $1/5$ chance the mathematician winds up there, and for every nonfirst admissible floor there is a $1/4$ chance of the receptionist winding up there, so the chance they collide is $(1/5)\times(1/4)$ times four (one for each nonfirst admissible room), or $1/5$. The letter is again a $1/4$ chance of being correct, and independent of floor, so the chance the floor and letter are correct is $1/5\times1/4=1/20$.