Guessing a number knowing only if the number proposed is lower or higher?

yes binary search is the most effective way of doing this. Binary Search is what you described. For a number between 1 and N Binary Search runs in O(log(n)) time.

So here is the algorithm to find a number between 1-N

int a = 1, b = n, guess = average of previous answers;

while(guess is wrong) {

    if(guess lower than answer) {a = guess;}
    else if(guess higher than answer) {b = guess;}

    guess = (a+b)/2;

} //Go back to while

Well, you're taking the best possible approach without the extra information - it's a binary search, basically.

Exactly how you use the "average result of previous guesses" is up to you; it would suggest biasing the results towards that average, but you'd need to perform analysis of just how indicative previous results are in order to work out the best approach. Don't just use the average: use the complete distribution.

For example, if all the results have been in the range 600-700 (even though the hypothetical range is up to 1000) with an average of 670, you might start with 670 but if it says "guess higher" then you would probably want to choose a value between 670 and 700 as your next guess, rather than 835 (which is very likely to be higher than the real result).

I suggest you log all the results from previous enquiries, so you can then use that as test data for alternative approaches.