Sum to $100$ game

Question

In the sum to $100$ game between two players that can choose any number frm $1$ to $10$, it is given in the book 'Trimathalon' by Paul Sally.
The rule is simple that each player can choose in any round, any number from $1$ to $10$ (irrespective of chosing that number earlier, or not); and the one with reaching the cumulative sum of both players' values to be $100$ wins.
The book states that: if one player has chosen the numbers : $1,2,4,1,10,6,9,2,1,8$; then would win.
But, it is also known that for one player to win, he must have the following cumulative sum value for the other player, in reverse order as:
$89, 78, 67, 56,45,34,23,12,1$.

I want to know why the numbers given in the book are correct, as then need the other player to make choices as :
round 1: Player 1: $1$, desired sum total for player 2: $\,\,\,1$, Player 2: $9$
round 1: Player 1: $2$, desired sum total for player 2: $12$, Player 2: $7$
round 3: Player 1: $4$, desired sum total for player 2: $23$, Player 2: $10$

So, is the author assuming the above sequence of inputs by the Player 2, if so why?

Answer 1

What you say the book states is the solution to a sample game (at the top of page 6, opposite to the solution on page 7), in which your opponent has responded to your moves by picking $9,7,10,1,5,2,9,10$, and $3$, in that order. So after your (winning) first choice of $1$, what you're doing is responding with $11-k$ to each of your opponent's moves.

Answer 2

The point is that if you choose a number first I can choose a number to make the two add up to $11$. Since I want to make the sum $100$, if I make the sum $89$ I will win. You choose $n$, I choose $11-n$ and the sum becomes $100$. Since $100$ is a winning number, so is $89$. If I want to reach $89$ it suffices to reach $78$ because I can make the next two moves total $11$ and so on. If I get to start, I choose $1$ and then nine pairs summing to $11$ will make the sum be $100$.