2 Players playing a game to guess a number between 1 and 100 such that the higher of the two guesses wins $100 - \max(guess_{1}, guess_{2})$. What should be the ideal guess in this game?
My thought is first to assume uniform distribution for the opponent as we don't know anything how they might guess and in turn have a expected payout:
$E[x_{1}] = a(100 - a)$ for the guess a between 1 and 100 yielding the optimal guess to be 50 but not sure how to continue form here as I am not convinced this is the optimal answer.
Okay I'm considering that you two are choosing at random. Let you be $A$ and $B$ be the other friend. I'm using my basic high school combinatorics knowledge. Please correct me if there is a mistake.
Let $x_A,x_B$ be the numbers you choose. $x_A,x_B ∈${${1,2,3,4,...100}$}
So there are total $$100×100=10000$$ possible pairs of $x_A,x_B$. Out of which $$99×100=9900$$pairs have unequal values of $x_A,x_B$. Out of these, exactly half pairs have $x_A>x_B$. Thus total pairs in this case are $$\frac{9900}{2}=4950$$
Hence if $E$ is the event of having $x_A>x_B$ or getting $100, then
$$P(E)=\frac{4950}{10000}=0.495$$
Hence there is $49.5\%$ chance of you getting a hundred bucks, $49.5\% $ of your friend getting it, and $1\%$ for equality.
Regarding the ideal choice, you get this probability for any random choice. But by studying data collections, you can check that people tend to choose a value between 30 to 70 usually. It gives them a sense of "Not very much but not very less" thinking that the other person might give a lower number. And $37,73,77,etc$ are the most chosen numbers worldwide. Hence you can try giving higher values than these.