How many Nash equilibrium in the furthest from the average game Game: 3 persons pick a non-negative number up to one hundred (inclusive). The one who has the furthest from the mean of these three numbers wins 1000$.
My idea:
I think that we need to select cases when players cannot improve their situation to avoid losing.
(1) If one player chooses a number less than 50, and the other two choose the numbers greater than 50, no one is willing to change the situation (as the player is either won or cannot win). (2) And vice versa, if the first number is $>50$ and the other two $<50$ results in Nash eq. as well. So, I reckon the result is
$$ ( (1) + (2))*3 =  (50*50*50 + 50*50*50)*3 = 750000$$
I multiplied by 3 as each of 3 players could win. If one plays 50, another one would definitely change his choice to split the 1000$ among two persons, so the inequalities were strict.
edit
Initially, I thought that (48, 52, 80) (the first player picks 48, the second picks 52, the third picks 80) is Nash eq., however, the first player still wishes to pick a smaller number. He can choose 0, 1, 2,...,$y$. To find  $y$, I solve the following inequality: distance of player 1 from the average is greater than the distance from 80 to the average
$$ \frac 13 ( y + 52 + 80) - y  \geq 80 - \frac 13 ( y + 52 + 80)$$
Solving it gives $24 \geq y$
However, I knew that 80 was the largest number in the right part. How can I solve it in the general case? In other words, how can I write Nash equilibrium as a triplet (f, s, t) to eventually find the number of equilibria?
 A: 
If one player chooses a number less than 50, and the other two choose the numbers greater than 50, no one is willing to change the situation (as the player is either won or cannot win).

False.
If player 1 picks $48$, player 2 picks $52$ and player 3 picks $80$, then the mean is $60$ and the winner is player 3. However, in this situation, if player $1$ picked $0$, the mean would be $44$, and player 1 would have won the game. So player 1 is willing to change the situation, as they can go from losing to winning by changing their choice from $48$ to $0$.

Your analysis of the position $(48, 52, 80)$ is correct, yes, but you are focusing too much on what player 1 could pick.
The thing is that if player $1$ had any chance of winning, then the choices he had to switch to were $0$ and $100$. If switching to $0$ or $100$ did not bring him victory, nothing would.
So, when doing the general analysis, I would focus on the case when:

*

*Player $1$ picked the smallest number

*Player $3$ won with the largest number

*Player $2$ picked something in between.

and then I would consider when exactly switching to $0$ will help player $1$ to win.
