Is this casino promotion exploitable?

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

$$P(k) = \frac{(q/p)^k - (q/p)^N}{1-(q/p)^N}$$

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In our case we are looking for $P(k)$, with $k = 1$, $N = 20$, $p=0.475$, $q =0.525$.

$$P(\text{lose the game}) = P(1) = \frac{(0.525/0.475)^1 - (0.525/0.475)^{20}}{1-(0.525/0.475)^{20}} = 0.9836$$

$$P(\text{win the game}) = 1 - P(\text{lose the game}) = 0.0164$$

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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.