Someone gets to choose a number between $1$ and $100$. I than have to try and guess by guessing a number and being told if the number is higher lower or equal. If I get the number on my first try I get $50$ on my second try $40$ and so on. Now if the other person chooses the number randomly the game is pretty clear to me. But will the other person choose the number randomly? The obvious guessing pattern is $50, 75$(or $25$) and so on. So it seems maybe the person wouldn't want to choose the numbers $50,75,25$ and maybe also not $12,13,87,88,62,63,37,38$. This logic could then be continued to deliver even more numbers that maybe shouldn't be chosen as often.
So my question is where does the equillibrium lie? Is a number chosen at random and if not what is the probability distribution and what is my guessing strategy?
We can compute the expected value if the number is chosen randomly. You have $0.01$ chance to win $50$, $0.02$ chance to win $40$, $0.04$ chance to win $30$, $0.08$ chance to win $20$, $0.16$ chance to win $10$, $0.32$ chance to win $0$, and $0.37$ to lose $10$, giving an expected value of $2$, so you should play.
I don't see an easy way to compute the payoff if your opponent chooses the number to hurt you. You can start with any number in the range $[36,64]$ and guarantee success in $7$ rounds. If you follow bisection, both $1$ and $100$ are guaranteed to take the maximum number of choices, so your opponent should choose one of those. Of course, if you recognize it, you should guess those first, despite the chance you have to go through $9$ guesses.