The median game

Five friends play a simple game with the following rules:

1. Players play consecutively one after the other.
2. Each player must call out a whole number between 1 and 10 (inclusive), such that it hasn't been called out already.
3. The winner is the player whose number is the median of all called out numbers.

What should be the optimal strategy for each player to maximise their chance of winning? Which player is the most likely to win if they all play optimally?

Note that if a player only has losing moves then they will just choose randomly. If they have multiple optimal moves then they will also choose one randomly.

• If a player has no optimal move (or indeed only losing moves), will they just choose randomly? Apr 1 at 11:19
• This is a good question. Yes I would say they just choose randomly. Apr 1 at 11:26

Consider that

1,2,9 and 10 are guaranteed to lose so no player will pick them unless forced into a random choice. This means that any player trying to win would pick from the set {3,4,5,6,7,8} and the median of five numbers from this set will be 5 or 6.

Given that, I see play proceeding as follows

Player 1 picks either 5 or 6 and Player 2 picks the other one. Given the symmetry of their choices, both players have an equal chance of winning.
Player 3 will pick either 4 or 7 (at random) because if they don't then Player 4 will fill the gap between Player 3's choice and 5/6 and guarantee a loss for Player 3.
From there, Player 3 can only win if Players 4 and 5 pick numbers which are both less than 4 (if 4 was picked) or greater than 7 (if 7 was picked) and this has probability $$\frac{3}{7} \times \frac{2}{6} = \frac{1}{7}$$.
Players 1 and 2 then both win with probabilites $$\frac{3}{7}$$ as it is guaranteed lose for Players 4 and 5 after Player 3's choice.

N.B. I'm assuming here that strategies which are equally optimal for a given player are equally likely to be executed. This seems to be implied from OP's comments but I'm adding here for clarification.

• @noedne I'm not sure what you mean. Even if Player 3 always chooses the same number, they don't know what Player 4 or 5 will do so still have a 1/7 chance of winning. Apr 1 at 15:35
• @noedne Ok, what you're essentially saying is that OP should clarify that equally optimal strategies are equally likely to be executed, right? I'm happy enough that we assume this, given my comment and the response under the question. Otherwise, the question has no definitive answer. Apr 1 at 16:11
• That's fine, but maybe it should be stated that this assumption is why 1 and 2 have equal probabilities of winning after choosing 5 and 6. A symmetry argument assumes that there is some well defined probability of winning in the first place, which has not yet been shown. Apr 1 at 16:23
• This is a great answer! Your assumptions are correct. Apr 1 at 23:18