I shall present my question and then add some of the work I have tried, unfortunately I haven't had much luck so far.
The game
The game is played between $3$ players. You and $2$ opponents who we shall call $I$ and $R$ (for intelligent and random) You each select an integer from $1-100$ (inclusive) and then money is exchanged as follows:
Whoever selected the Largest number is the looser, they must pay the two winners what they each selected.
Example: You pick $42$, $I$ picks $12$ and $R$ picks $51$ then $R$ must pay you $\$42$ and he must pay $I$$\$12$
In the event of a two or more players selecting the largest number then a loser is selected uniformly between them.
$R$ plays uniformly randomly, that is he selects his pick uniformly between the first $100$ integers.
$I$ Plays perfectly to maximise their own E.V (all standard game theory assumptions)
What should your strategy be to maximise your expected value?
My workings so far:
For what it is worth in the 2-player scenario where it is just you v.s $R$ then you should pick $33$ as
$\mathbb{E}[\delta_x ] = \frac{100-x}{100}(x) - \frac{x-1}{100}(-\frac{x}{2})$
(where $\delta_x$ is the strategy of picking $x$)
This function is maximised at $33$ from simple calculus.
No the answer is not $1$ as picking $2$ performs much better when you know someone is picking $1$
I asked my teacher and they said a mixed strategy but I am struggling with formulating this.
My idea was to determine $f(x,y) = \mathbb{E}[\delta_x | \text{other player is picking y } ]$ and then find the maxima of this function but I am struggling.
Any help would be great :)