Hint: $2^7=128>100$. That is, you can simply guess in the middle and bisect until you get to the solution.
In general, just find the smallest power of $2$ greater than the max number in your set. For $1000$, you should be able to get it in $10$ guesses, because $2^{10} = 1024$.
The answer is that $2^7=128>100$.
To be more precise:
Say that you start by guessing $50$. Either you're right, in which case you're done, or you're wrong; if you're wrong, then you know that the number is either in $[1,49]$ or $[51,100]$, so that there are only $49$ or $50$ possibilities left.
Now, assuming that you aren't already done, you have a collection of $49$ or $50$ possibilities left; guess the middle one. That is, if you are on $[1,49]$, guess (say) $25$; if you're on $[51,100]$, then guess (say) $75$. If you got it right: great. If not, then you find out whether it should be higher or lower; in particular, if you previously knew it was in $[1,49]$, then you either now know it is in $[1,24]$ or you now know it is in $[26, 49]$. If you previously knew it was in $[51,100]$, then either you know that it is in $[51,74]$ or that it is in $[76,100]$.
In any case, you have either already guess right, or you have narrowed down the possibilities to one of $[1,24]$, $[26,49]$, $[51,74]$, or $[76,100]$. In any one of these cases, there are at most $25$ possibilities left, and it must be one of them.
Continue in this way: cut your current interval of possibilities in half by a guess, so that you are either right or you can discard roughly half of the possibilities based on the announcement of "higher" or "lower".
By continuing this process, in the third step you narrow it down to at most $12$ possibilities; in the fourth, to at most $6$; in the fifth, to at most $3$; in the sixth, to at most $1$; and voila! In your seventh guess, assuming that you're unlucky enough to have not guessed it yet, there's only one number that could possibly be it.
Best Answer
Your first guess is just one number. For the second guess, you have two numbers that you will get correct depending on the high or low of the first. For the $n^{th}$ guess you have $2^{n-1}$, so after $6$ guesses you have had the possibility of guessing $63$ numbers. Assuming your opponent chooses his number uniformly, your chance of guessing it is $\frac {63}{100}$
Added: I assumed the player was smart enough to keep the guesses far enough apart so there is room for the strategy to play out. If the player randomly picks among the numbers that are still available, I think you would have to simulate it. I don't see an easy way to calculate it.