Player A and B start with a capital of $n$ Dollars each, $n\in \mathbb{N}$. A coin (has not to be fair) is flipped. The probability of "Head" is $p\in[0,1]$. If "Head" occurs, B gives 1 Dollar to A and if "Tail" occurs, A gives 1 Dollar to B. Show that the game will end after finite steps.
We did not have random walks yet!
Let $p\in(0,1)$ because if $p\in\{0,1\}$ the result is trivial because the game will end after exactly $n$ steps. We first look at
$A_n$="Head occurs in $n$th toss".
Since $\mathbb{P}[A_i]=p>0$ we have $$ \sum_{i=1}^\infty \mathbb{P}[A_i]=\sum_{i=1}^\infty p = \infty $$ Since the events $\{A_i\}_{i\in \mathbb{N}}$ are independent we can use the scond Borel-Catelli Lemma $$ \mathbb{P}[\lim \sup A_n]=1 $$ That means that the if we repeat the experiment infinite times, the probability that the event "Head" occurs infinite times is 1. The same can be shown for "Tail" since $1-p>0$ for $p\in(0,1)$. With a little bit of calculus we can shot that$$ \mathbb{P}[\lim \inf A_n]=0 $$ The probability that the coin will show only "Head" after several times when we flipp the coin infinite times is 0.
From here on I am not really sure how to proceed and show that the game will end in finite steps. Do I have to use Kolmogorovs 0-1 law? Your help would be very much appreciated!
It suffices to show that, with probability 1, B gives 1 dollar to A $n$ times in a row (or less if the game will end). We can split up $A_1,A_2,\dots$ into blocks of $n$, i.e., let $E_1$ be the event that $A_1,\dots,A_n$ are all heads, $E_2$ be the event that $A_{n+1},\dots,A_{2n}$ are all heads, etc. The point is that $E_1,E_2,\dots$ are independent and $P(E_i) = p^n$, so $\sum_i P(E_i) = \sum_i p^n = +\infty$ implies that, with probability $1$, there are infinitely many $n$ heads in a row, so in particular one set of $n$ heads in a row.