TITLE: Best strategy for this game. [Nintendo Wii game]
QUESTION [5 upvotes]: Which is the best strategy for this game? Actually is a Nintendo Wii game.
It's a 4 people game.
There is a ladder of 10 steps.
Each player says a number (they can chose between those numbers "1","3","5").
If their number is unique (if he's the only one who said that number in that round), the player may advance of a number of steps equivalent of the number.
If there's more than one player who said the same number they cannot advance.
E.G.
Player A says "5"
Player B says "1"
Player C says "3"
Player D says "3"
Player A advances 5 steps, player B advances 1 step. The remaining 2 cannot advance.
The one who reach the top of the ladder wins.
Usually I chose to say always "5", but I'd like to know what's your point of view on this.

REPLY [2 votes]: The situation is symmetric in the players (at least in the beginning or when they all are at the same step of the ladder), so if there is any best strategy, we may assume that all players implement it and pick $1,3,5$ with probabilities $p_1,p_3,p_5$, respectively, where $p_1+p_3+p_5=1$.
The following can happen:

All pick the same number and no one advances (happens with $p_1^3+p_3^3+p_5^3$)
Three pick the same and one advances by five (happens with $3(p_1^2+p_3^2)p_5$)
Three pick the same and one advances by three (happens with $3(p_1^2+p_5^2)p_3$)
Three pick the same and one advances by one (happens with $3(p_3^2+p_5^2)p_1$)
Two pick "1" and the others advance by "3" and "5" (happens with $12p_1^2p_3p_5$)
Two pick "3" and the others advance by "1" and "5" (happens with $12p_3^2p_1p_5$)
Two pick "5" and the others advance by "1" and "3" (happens with $12p_5^2p_1p_3$)

The expected total progress is then $$ \tag115(p_1^2+p_3^2)p_5+9(p_1^2+p_5^2)p_3+3(p_3^2+p_5^2)p_1+96p_1^2p_3p_5+72p_3^2p_1p_5+48p_5^2p_1p_3$$
and by symmetry the indivdual expected progress is $\frac13$ of this.
You may want to determine the maximum of exporssion (1) subject to the constraints $p_1+p_3+p_5=1$, $p_1,p_3,p_5\ge0$.