It's easy to compute the optimal strategy for $N$ reasonably small (say, up to $N \le 10000$). You just iterate backwards from the end of the game. There are three possible doubling states:
- Both players have the right to double (i.e. nobody has doubled yet);
- You have the right to double;
- Your opponent has the right to double.
Note that the number currently showing on the doubling die -- to use the backgammon term -- is irrelevant to your strategy (although not irrelevant to your expected gain/loss).
So for each doubling state $s \in \{1,2,3\}$ you want to compute the expected gain $E(s,x,y)$, expressed as a multiple of the number on the doubling die, where $s$ is the doubling state, $x$ is the number of points you need to win, and $y$ is the number of points your opponent needs to win.
$x$ and $y$ can't both be zero, so we can start the ball rolling with $E(s,0,y) = 1$, $E(s,x,0) = 0$. Then you have to set up a number of simple but tedious equations expressing $E(s,x,y)$ in terms of $E(s', x', y)$ for $x' < x$ and all values of $s$ and $s'$. Then you can work backwards to calculate $E(s,x,y)$ for all $s,x,y$.
The details are, as I said, tedious, and I'm not going to spell them out unless you pay me. Just be aware that you have to store the intermediate values in a table of size $3N^2$. Don't be seduced by the obvious recursive solution -- its asymptotic complexity is exponentially worse.
Fairness in a game is usually defined as: both (all $n$) players have the same theoretical chance of winning. That is, if both players play with perfect strategy - both players are perfectly skilled.
In reality, fairness doesn't matter, because players may play imperfectly (i.e. make a mistake)
Draws do not really count towards fairness, because neither player wins.
I've tried to analyse each game that you suggest.
Backgammon: Not known, but likely unfair.
It's very likely that one of the players has a slight advantage. It could be that the first player has first-move advantage, or it could be that the second player has an advantage from having more information about his opponents move. These advantages are unlikely to balance perfectly.
This is difficult to analyse, however, because of the large number of random dice rolls that are possible during a game.
The randomness of Backgammon makes this different than the other games - which are 'deterministic'.
In Backgammon, even if the first player has an advantage, they might still lose through bad luck. In the deterministic games below, if the first player has an advantage, they won't give it up unless they make a mistake.
Russian Roulette: Fair (with 2, 3 or 6 players, but only theoretically).
Both/all players have an equal chance of death.
However, the consequences of Russian Roulette are so dire that players are highly motivated to cheat.
There may be out-of-game consequences that could be asymmetrical for the surviving players, making it difficult to say that the game is fair in reality.
Tic-tac-toe: Fair
Tic-tac-toe, played perfectly, will always end in a draw. So, both players have a 0% chance of winning (the same chance!)
Abacus game/Nim: Unfair
There are no draws in Nim, and no randomness, so one player will always win.
If both players play perfectly, either the first player will always win, or the second player will always win, depending on the exact setup.
Checkers: Fair
Checkers has been analysed enough that we know perfect play will always end in a draw - same situation as Tic-Tac-Toe.
Go: not known, but almost certainly unfair
Go has a system of compensating for a perceived level of unfairness to try and make the game "fair". One player receives a chosen number of komi stones at the start of the game.
It would be very unlikely that a whole number of komi stones will ever make the game fair. But, Go is hard to analyse so we don't know for sure.
They also might add stones to compensate for difference in your level - it might seem odd to say this, but this is designed to deliberately make the game "unfair", because it gives the weaker player a theoretically better chance of winning!
It's "unfair" because the weaker player should win if they play perfectly. But, of course, the weaker player won't play perfectly - the unfairness compensates for the player's weakness!
Reversi: not known, but believed to be fair
Analysis so far suggests that perfect-play games will end in a draw. This would put it in the same category as Tic-Tac-Toe.
But, this is not fully analysed yet.
Chess: not known
As you say, in reality, it seems that White wins more often, so it's commonly said that first player (White) has an advantage.
However, these games are not played perfectly, so that actually doesn't help us to decide if this game is theoretically "fair".
If the first player really did have a theoretical advantage, then White should be able to win every time, not only some of the time! Because, there's no randomness - only a mistake will change the result.
How to make an unfair game fair
In each case, I have considered whether or not the first player is in a fair position against the second player.
So, we can make each game theoretically "fair" by flipping a (fair) coin to decide who goes first! Then, even if the first player has an advantage, you have a fair chance of playing first :)
Best Answer
The analysis in quarague’s answer isn’t correct because it only takes the possibility of doubling once into account, whereas future doubling opportunities in fact increase the expected payoff.
Denote by $x_k$ the expected value of the game for $A$ when $A$ has score $k$ and has the right to challenge and $B$ doesn’t. We can guess that $A$ challenges when the score is $1$, and $B$ accepts, and then check whether this is self-consistent. Under these assumptions, we have
$$ 2x_0=x_{-1}+x_1=x_{-1}-2x_{-1}=-x_{-1} $$
and
$$ 2x_{-1}=x_0+x_{-2}=x_0-1\;, $$
and thus $x_{-1}=-\frac25$, $x_0=\frac15$ and $x_1=\frac45$. The assumptions turn out to be self-consistent, since the expected return for $A$ at score $1$ would only be $\frac12\left(\frac15+1\right)=\frac35\lt\frac45$ without challenging, and the expected return for $B$ upon refusing would be $-1\lt-\frac45$.
It remains to find the optimal strategy and expected return in the initial state of the game, where both players have the right to challenge. Denote these expected returns by $y_k$. Then $y_0=0$ by symmetry, and $y_1$ is $\frac12$ if $A$ doesn’t challenge, $1$ if $A$ challenges and $B$ refuses and $-2x_{-1}=\frac45$ if $A$ challenges and $B$ accepts, so $A$ challenges at score $1$ and $B$ accepts.
To summarize, the initial value of the game is $0$ by symmetry; a player challenges exactly if they have a score of $1$, the other player always accepts, and the value of the game for the player with the right to challenge is $-\frac25$, $\frac15$ and $\frac45$ at a score of $-1$, $0$ and $1$, respectively.