Is this casino promotion exploitable? The promotion is like this:
Starting credit: 500 dollars
Maximum bet: 500 dollars
Win up to 10000 dollars and get 10000 dollars free.
House edge 52.5%.
Is this exploitable?
 A: I will use the Gambler's ruin method to solve this problem.
The gambler stops playing after winning or getting ruined, whichever happens first.
Let $P(k)$ denote the probability that you lose all ýour money when your initial capital is $k \cdot \$500$.
$P(k) = pP(k+1) + qP(k-1), \:\: k = 1, 2, ... ,N-1 \:\:\:\: (*)$
where the total capital of you and the house is $\$10000$, i.e you need to win $\$9500$ from the house, and $10000/500 = N$, i.e. $N = 20$. 
Rewriting $(*)$ we have 
$P(k+1) - \frac{1}{p} P(k) + \frac{q}{p} P(k-1) = 0, \:\: k = 1, 2, ... ,N-1 \:\:\:\: (**)$
which is a second-order homogenous linear-coefficient difference equation.
Note that we also have 
$P(0) = 1$ and $P(N) = 0$
So to find $P(k)$ reduces to solving $(**)$ subject to these boundary conditions.
Let $P(k) = r^k$, $(**)$ becomes
$r^{k+1}-\frac{1}{p}r^k + \frac{q}{p} r^{k-1} = 0, \:\:\: p + q = 1$
Setting $k=1$ we get a second order equation with solutions
$r_1 = 1$ and $r_2 = q/p$.
Then
$P(k) = c_1 + c_2(q/p)^k$
Using the boundary conditions
$P(0) = 1 \Rightarrow c_1 + c_2 = 1$
$P(N) = 0 \Rightarrow c_1 + c_2\left(\frac{q}{p}\right)^N = 0$ 
Solving for $c_1$ and $c_2$, we obtain
$$c_1 = \frac{-(q/p)^N}{1-(q/p)^N}, \:\:\:\: c_2 = \frac{1}{1-(q/p)^N}$$
Hence
$$P(k) = \frac{(q/p)^k - (q/p)^N}{1-(q/p)^N}$$

In our case we are looking for $P(k)$, with $k = 1$, $N = 20$, $p=0.475$, $q =0.525$.
$$P(\text{lose the game}) = P(1) = \frac{(0.525/0.475)^1 - (0.525/0.475)^{20}}{1-(0.525/0.475)^{20}} = 0.9836$$
$$P(\text{win the game}) = 1 - P(\text{lose the game}) = 0.0164$$

Let $G$ be the random variable taking on values $\$19500$ if the gambler wins the game and $-\$500$ if the gambler loses the game.
The casino promotion is exploitable if the expected value of the game is greater than zero.
$$E[G] = 0.0164 \cdot \$19500 + 0.9836 \cdot (-\$500)= -\$172$$
Hence, not exploitable. 
A: I found this online "Huygens' Result"

According to the formula it would mean that the player's odds of success in reaching the 10000 dollars is around 5%.
I don't know the exact math but if the bonus given is 2000% and the player's odds of success is 5%, wouldn't that mean the bonus is exploitable even with the 2.5% house advantage?
