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Imagine three doors where behind one door $\text{A}$ there is a new car, behind door $\text{B}$ there is a goat, and behind door $\text{C}$ there is a new car and a goat.

The problem is that each door is labeled incorrectly...

If you can open only one door, is it possible to label all the doors correctly?

I'm trying to answer this in logical form, but I am unsure if my answer is good enough:

Denote doors as $D_1, D_2, D_3$

$D_n $ is equal to either $A,B$ or $C$.

If we assume that the correct solution is $$D_1 = A $$ $$D_2 = B $$ $$D_3 = C $$


$$\begin{align} D_3 = A &\implies D_1=B \implies D_2 = C\\ D_3 = B &\implies D_1=C \implies D_2 = A\\ \end{align}$$


$$\begin{align} C = D_1 \implies D_2 = A \implies D_3 = B\\ C = D_2 \implies D_1 = B \implies D_3 = A\\ \end{align}$$

Would you consider this a valid solution?

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I thought this was going to be a Monty Hall problem at first. – Sujaan Kunalan Jun 22 '13 at 20:39
Why turn a cute little puzzle into mystery math? – André Nicolas Jun 22 '13 at 20:48
I don't understand your solution. The problem is: if you see three doors labeled $A,B,C$ and open the door labeled $A$ and inside is $B$ then what is inside the doors labeled $A$ and $C$? What if inside the door labeled $A$ is $C$? What if first you open the doors labeled $B$ or $C$? – P.. Jun 22 '13 at 20:48
It's not valid since nobody understands what you're saying. Speak English next time. – Ink Jun 22 '13 at 20:57
up vote 32 down vote accepted

The key is the assertion that each door is labelled incorrectly. If capitals stand for the correct labels and lower case letters for the incorrect ones currently present, you will find there are just two possibilities:

$$Ab: Bc :Ca$$

$$Ac :Ba :Cb$$

Each of the doors is labelled differently in the two cases. Whichever door you open, you will therefore know which case applies, and will be able to apply correct labels without opening a second door.

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Nicely (and concisely) put! (+) – amWhy Jun 22 '13 at 20:58

I must confess I cannot follow your notation with the implications and equals. But the key point should be “each door is labeled incorrectly”. Suppose, without loss of generality, that you opened a door which was labeled $A$ but actually contained $B$. Then you know that the door labeled $B$ can't actually contain $A$: if it did, then door $C$ would be labeled correctly, which is against the premise. So you know that the door labeled $B$ must contain $C$ and the door labeled $C$ must contain $A$. This argumentation serves for any door you opened and any possible content, simply by exchanging the names $A$ through $C$.

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I don’t follow your solution; words would make it much clearer. Here is one solution.

Open the door labelled Car and Goat.

  • If you find only a goat, you know that that the door labelled Car must hide a car and a goat (since it doesn’t hide just a goat and can’t hide just a car); the remaining door, labelled Goat, must hide the car.

  • If you find only a car, you conclude similarly that behind the door labelled Goat you’ll find the car/goat pair and hence that the goat is behind the Car door.

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