You start with a bankroll $x.
You play until your capital reaches N or 0 (those would be absorbing states in Markov chains lingo). We'll solve for the infinity case later.
Each time you play, you have a probability p of winning $1 and a probabilty q of losing $1, with q=1-p.
Let
be the probability that you go busto before reaching N given you start with $k. We're looking to find the general form of
We have
and
.
If your roll is at $k and you play once, with probability p you're going to be at $k+1 and with probability q at $k-1. In the first case, the probability of eventually busting is
and in the second case
. We can write :
, for
.
We can rewrite this as :
 p_k = p p_{k+1} + q p_{k-1})
(because p+q=1)
 = q(p_k-p_{k-1}))
 = \frac{q}{p}(p_k-p_{k-1}))
By replacing the right term using the same equation until you reach
it is easy to see that gives
 = (\frac{q}{p})^k (p_1-p_0))
Now we're going to sum the left term for all k from N to 0 :
 = (p_N-p_{N-1}) + (p_{N-1}-p_{N-2}) + \ldots + (p_2-p_1) + (p_1-p_0) = p_N - p_0 = -1)
 = \sum_{k=0}^{N-1} (\frac{q}{p})^k (p_1-p_0) = (p_1-p_0) \sum_{k=0}^{N-1} (\frac{q}{p})^k = -1)
For the special case p=q=1/2 this gives
 N = -1)
For p not equal to q, we get
 \frac{(\frac{q}{p})^N-1}{(\frac{q}{p})-1} = -1)
(the barely readable thing in ( ) is q/p)
 = -\frac{(\frac{q}{p})-1}{(\frac{q}{p})^N-1})
 = \sum_{k=0}^{x-1} (p_{k+1}-p_k) = (p_1-p_0) \sum_{k=0}^{x-1} (\frac{q}{p})^k = (p_1-p_0) \frac{(\frac{q}{p})-1}{(\frac{q}{p})^x-1})
Replacing with what we obtained for p1-p0 above :
 = -\frac{(\frac{q}{p})-1}{(\frac{q}{p})^N-1} \frac{(\frac{q}{p})-1}{(\frac{q}{p})^x-1} = -\frac{(\frac{q}{p})^x-1}{(\frac{q}{p})^N-1})
And finally
^x-1}{(\frac{q}{p})^N-1})
For p=q, it's very easy to find the same way
Now, if we look at the probability
of reaching N before going busto, we can do it the same, but going towards
instead of
and we find
^x-1}{(\frac{q}{p})^N-1})
and for p = q = 1/2
Note that it was not (completely) obvious that
. As it is the case, we have proven the game ends with probability 1 (we reach either N or 0 at some point or another).
Now, back to our original problem. We play against the house who has infinite money. We find the probability of busting by letting N go to infinity in our expression above. 3 cases appear :
1. q > p : that means we have more chance to lose a flip than win a flip
When the house has an edge and infinity money, we busto with probability 1 (no kidding).
2. q = p : we're flipping 50/50
Even when the house has no edge, as long as it has infininte money and we don't have an edge either, we busto with probability 1 (no shock either).
3. q < p : we have an edge
^x-1}{-1} = 1+(\frac{q}{p})^x-1 = (\frac{q}{p})^x)
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