The game "Rock-Paper-Scissors" is played between two players with each simultaneously choosing one of three options: "Rock", "Paper", or "Scissors". If both choose the same option, the outcome of the game is a draw. If they choose differently, the winner is determined by the rules:

• Rock beats Scissors
• Scissors beats Paper
• Paper beats Rock

Let us suppose that the winning player receives a payment from the losing player. No money changes hands if the game is drawn.

Each player devises a mixed strategy: they choose "Rock", "Paper", or "Scissors" at random, independently of any previous games. For example, one strategy may be to play "Rock" with probability 30%, "Paper" 50%, and "Scissors" 20%.

For a symmetric zero-sum game such as this, a Nash equilibrium is a mixed strategy for which, if employed by one player, there exists no counter-strategy that would give the opponent expected positive net winnings.

For example, in the case of Rock-Paper-Scissors with the loser paying the winner one dollar, there is one Nash equilibrium strategy: choose each of the three options with equal probabilities $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$.

Suppose we adjust the payoff structure of Rock-Paper-Scissors so that a player who wins with "Rock" receives 5 dollars, a win with "Paper" awards them 3 dollars, and a win with "Scissors" awards them 1 dollar, in each case paid by the loser. Now the equilibrium strategy is to choose "Rock", "Paper", "Scissors" with probabilities $(\frac{1}{9},\frac{5}{9},\frac{1}{3})$ respectively. Counterintuitively, the equilibrium strategy assigns the smallest probability to "Rock" despite it yielding the greatest reward.

We generalise Rock-Paper-Scissors to a game with $n$ options as follows, with each player simultaneously choosing a number from $1,\dots,n$:

• If both choose the same number, the outcome of the game is a draw.
• If the difference between the two numbers is odd, the winner is the player who chose the smaller number.
• If the difference is even (and non-zero), the winner is the player who chose the larger number.
• If the winning player's number is $m$, they receive $2m-1$ dollars from the losing player.

We refer to this generalised Rock-Paper-Scissors game as $RPS(n)$. Observe that $RPS(3)$ corresponds to the payoff structure described earlier.

It can be shown that for all positive integers $n$, the game $RPS(n)$ has precisely one Nash equilibrium strategy. Define $P(n)$ to be the probability that a player, using the Nash equilibrium strategy for $RPS(n)$, chooses the number $n$. For example, $P(3)=\frac{1}{9}$, $P(4)=\frac{1}{5}$, and $P(10)\approx 0.0479638009$.

Define $S(N)=\displaystyle\sum_{n=3}^N P(n)$. You are given $S(10)\approx 1.1546112276$ and $S(100)\approx 4.8779925686$.

Find $S(10^5)$, giving your answer rounded to ten places after the decimal point.