This problem involves logic-based math, I tried making truth tables for this problem but I don't think you can because there are 9 doors!

Below is what I came up with but I want to know if there is a better way of figuring this out.


Base on the tip from the hostess, there is something behind door8; then it is either 1 dollar or the price; since the sign on the price door is always true; therefore Door8 can only be false with \$1 behind it.

  1. Since door8 is false, then there is something behind door9. Similarly, door9 can only be false with \$1 behind it.
  2. Since door9 is false, door6 is true.
  3. Since door6 is true, door3 is wrong.
  4. Since door3 is wrong, then door5 is wrong, and door7 is true.
  5. Since door5 is wrong, door2 is wrong, and door4 is wrong.
  6. Since door7 is true, the Prize is behind door1.
  7. Since door2 is wrong, there is something behind door2; it is \$1 since the sign is false.
  8. Since door4 is wrong, the sign on door1 is true.

Base on the information above, we can create a table for all the doors.


Since sign on door one is true, and sign on door 6 is true, then there is nothing behind door 6 and 7.

Door 3/4/5 can be either \$1 or nothing.

  • $\begingroup$ Since door 7 is right, the prize is NOT behind door 1; you win Nothing. $\endgroup$ – peterwhy Jul 11 '15 at 19:46
  • $\begingroup$ Have you tried drawing a directed graph for the doors, where each door is a node and there is an arrow from Door A to Door B if A makes a statement about B (and there can be self-loops for self-referential doors)? Then it is visually clear that Door 2 might be a good place to start as it has no outbound arrows (except to itself). And then Door 5, because it has an outbound edge only to Door 2. $\endgroup$ – Marconius Jul 11 '15 at 20:45
  • $\begingroup$ try puzzlestack $\endgroup$ – Bhaskar Vashishth Jul 12 '15 at 23:15

The first two steps in your reasoning could not have been done more simply.

You could use boolean algebra (e.g. see http://www.allaboutcircuits.com/textbook/digital/chpt-7/introduction-boolean-algebra/ - I will use this notation - or http://mathworld.wolfram.com/BooleanAlgebra.html) to simplify some of the rest of the analysis. First indicate what each door says:

$\begin{align} D1 &: \text{"D1,D3,D5,D7 or D9 has GP"} \\ D2 &: \text{"I have \$0"} \\ D3 &: D5 + \overline{D7} \\ D4 &: \overline{D1} \\ D5 &: D2 + D4 \\ D6 &: -D3 \\ D7 &: \text{"D1 has \$0 or \$1"} \\ \end{align}$

So given that you know D6 is true, you can use the rules of boolean algebra to re-express this knowledge

$D6 = \overline{D3} = \overline{D5 + \overline{D7}} = \overline{D5} \times D7 = (\overline{D2 + D4}) \times D7 = \overline{D2} \times \overline{D4} \times D7 = D1 \times \overline{D2} \times D7$

So you know that D1 and D7 are true, and hence D1 has the grand prizeD7 has the grand prize. [edited 13 July UTC]

  • $\begingroup$ As I noted in my comments above, I don't understand how you can claim $D1$ has grand prize if $D7$ is also true. All I can see is that $D1$ is true and $D1$ has either \$0 or \$1, which means $D1$ has \$0. $\endgroup$ – peterwhy Jul 12 '15 at 11:11
  • $\begingroup$ Oh, that's a silly mistake. The prize is behind D7, and D1 is empty and is telling the truth. $\endgroup$ – Marconius Jul 12 '15 at 16:59

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