If I think of a number between 1 and 1,000, guessing it in 10 yesno questions is easy... so that's not the puzzle!
Guessing it in 10 yesno questions, that must be all asked in advance, is also relatively easy... so that's not the puzzle either!
How many questions would you need to guess my number, if you had to ask all questions beforehand, and I also could lie once?
(In reply to
Beginning of a solution by Tristan)
When I read the other comments, I realized that I had approached this
problem from a purely theoretical perspective, while others had tried
to actually come up with the questions. I also realized that I had
made an error in my reasoning.
I didn't think to ask questions about lies, such as "Did you lie on
question 1?" and that type of thing. Because of questions like these,
lying may change more than one answer. The rest of my conclusions were
thrown off. I still have confidence in my 2^N/(N+1) sequence, since
each number has N+1 possible combinations of answers that correspond to
it. Armando has shown a great way reach this limit (at least for numbers 11000). Congratulations!

However, I've though of a tricksy way to bypass this limit. We
can design a question in a way that lying will not change any
answers. For example:
Consider each number in their binary form.
Is the following statement true? You will answer this question
with a lie or the last digit of the number is a 1, but not both.
All the questions can be in the above form, creating a 10 question solution.

There is yet another better solution, but it is even tricksier. I
can design a question that "forces" the number to be, say, a 1.
For example, we could easily use this paradox. This solution only requires 1 question!

Posted by Tristan
on 20050415 22:48:02 