Expected value of a sure win game Suppose that there are 2 players playing the following game:

Given 10 coins, each player takes turn to collect either 1, 2 or 3 coins. The winner is the one who collects the last coin.

Assume that both players are smart. Then in this game, the player who start first by picking 2 will always wins as he can always picks 2, 6, 10 to collect the last coin.

If the winner gets 100 dollars, then what is the fair price of this game?

My attempt is that the winner has expected value 100 since he will always wins.

Is my reasoning above correct?
You have indeed found the correct strategy to win this game, which is to always present your opponent with a multiple of four coins. Taking two coins in the first round will leave your opponent with eight coins in their first turn, after which you will be able to leave four coins in their second turn and take the last coin yourself.

When it comes to determining the fair price of this game, it rather depends on who gets to start. If a player is certain of getting the first turn, then yes, the game has an expected return of \$100. If the starting player is assigned at random, however, it has an expected return of \$50 only.
Let's say that you have a probability $p$ with $p \in[0,1]$ to be the the starting player, then the fair price of the game should be $p\cdot 100$ .