Possible variation on Bertrand's ballot theorem, chance you lose all your money on a particular turn of game Here is yet another problem from my probability textbook:

A man possessed of $a + 1$ ponds plays even wagers for a stake of $1$ pound. Find the chance that he is ruined at the $(a + 2x + 1)$th wager and not before.

Let me go through the answer at the back of my book:

He must lose $a + x$ and win $x$ out of the first $a + 2x$ games, his losses never exceeding his gains by more than $a$, and then he must lose the $(a + 2x + 1)$th game.

I understand all this. But I don't understand this next step:

The $a + 2x$ games may occur in $\binom{a + 2x}{x}$ orders and among these, $\binom{a + 2x}{x - 1}$ are unfavorable.

How did the book get that $\binom{a + 2x}{x - 1}$ are unfavorable? If we assume this, then I understand the rest of the answer:

$$\text{Chance} = \left(\binom{a + 2x}{x} - \binom{a + 2x}{x - 1}\right)\left({1\over2}\right)^{a + 2x + 1}.$$

So my question is, why does it follow that the number of orders in which $x$ gains and $a + x$ losses can occur so that the number of losses is never $a + 1$ more than the number of wins, is $\binom{a + 2x}{x} - \binom{a + 2x}{x - 1}$?
Which is to say, among the $\binom{a + 2x}{x}$ orders in which $x$ gains and $a + x$ losses can occur, how does one see that there will be $\binom{a + 2x}{x - 1}$ orders in which the losses will at some time be at least $a + 1$ in excess of the gains?
I have tried messing around with Bertrand's ballot theorem https://en.wikipedia.org/wiki/Bertrand%27s_ballot_theorem, but not to much success. Ideally, there should be a easy way via mimicking the reflection proof of Bertrand's ballot theorem, but I don't see it.
 A: From the set of sequences counted by $\binom{a+2x}{x}$, consisting of $x$ wins and $a+x$ losses starting which take the total from $a+1$ down to $1$, we need to subtract the sequences that reach $0$ at some point in the middle. To do this, we use the reflection principle. Given a bad sequence which reaches zero at some point, reflect it at the point where it first reaches zero. This means that all wins after the reflection point are replaced with losses, and vice versa. Since the original path started at $a+1$, went down to $0$, and then went back up to $1$, the reflected path will instead start at $a+1$, go down to $0$, and then go down to $-1$. It follows that the reflected path will have $x-1$ wins and $a+x+1$ losses.
Since the reflection operation is a bijection, in order to count the number of bad paths starting from $a+1$ and ending at $1$, we can instead count the total number of paths starting at $a+1$ and ending at $-1$. The number of ways to choose a path with $x-1$ wins and $a+x+1$ losses is $\binom{a+2x}{x-1}$.
