I'm trying to understand some maths homework, but the meaning of one particular part is not clear to me.
The relevant parts of the task description go as follows (translated from its original language):
Matte and Ida are playing a game of money. Matte has $a$ coins and Ida has $b$ coins in the start. Each turn, Matte wins one coin from Ida at a probability of $p$, or else he loses a coin to Ida with a probability of $q = 1 - p$, where $0 < p < 1$. The game is over when one of the two have won all of the money. We are looking for the probability that Matte wins all of the coins and thus the game.
Let $A_k$ be the probability that Matte wins the game when he has $k$ coins. Then $A_0 = 0$ and $A_{a+b} = 1$ and we want to find $A_a$. You can also see that $A_k$ fulfils the recursive relation $A_k = p * A_{k+1} + q * A_{k-1}$, if $1 ≤ k < a + b$ since if Matte eventually wins the game he must either have $k+1$ coins (with a probability of $p$) or $k - 1$ coins (with a probability of $q$) in the round after he had $k$ coins.
Now my question is, later in the text (omitted here for brevity) it is claimed that $A_1 = p * A_2$ as $A_0 = 0$.
This is not clear to me at all. Why does this relation hold in the given context?
There are 2 ways to look at this :
[[1]] Elaborating what @J.W.Tanner has rightly indicated , nearly 14 hours ago : We have the recursive relation $A_k=p×A_{k+1}+q×A_{k-1}$ , where we can plug-in $k=1$ & use $A_0=0$ , to get $A_1$ in terms of $A_2$.
We will get $A_1=p×A_{1+1}+q×A_{1-1}=p×A_2+q×A_0=p×A_2$
[[2]] Alternately : Probability of winning with 1 Coin ( $A_1$ ) involves winning 1 more Coin ( $p$ ) & then winning the game with those 2 Coins ( $A_2$ ) ; there is no other way to win the game because losing a Coin here is losing the last Coin , which will lose the game too.
In other words , $A_1=p×A_2$
Both ways give Same Conclusion.