Im scretching my head about the following roll-under-game:
Choose a bet-amount $B$ and a number $N$ between $2$ and $96$. Then the bank picks a Number $R$ between $1$ and $100$ discretly uniformly distributed. If $N< R$ you won the game and get $w(N)\cdot B$ back, where ... $$w(N)=\frac{\beta}{P(R<N)}=\frac{\beta}{\frac{N-1}{100}}=\frac{100\cdot\beta}{N-1}$$ ... with $P(R<N)=\frac{N-1}{100}$ and $0\leq\beta\leq1$. On the other hand, if the result is $N\geq R$ you loose a portion $\gamma\cdot B$ of your bet-amount.
In the case considered here the following applies: $\beta=0.985$ and $\gamma\in[0,1]$ (loss-ratio) with $\gamma:=1-\delta$, where $\delta\in[0,1]$ (cashback-rate). So for $\delta=0$ or else $\gamma=1$ we get the classical roll-under-game with no cashback-bonus.
I'll give two quick example to illustrate the game:
i) [$\beta=0.985$, $\gamma=1$ (no chashback)]: If you pick $N=51$ you will win with a probability of $50\%$ and get $1.97\cdot B$ back. In the event of a loss the entire bet-amount is lost.
ii) [$\beta=0.985$, $\gamma=0.7$ (30% chashback)]: If you pick $N=51$ you will win with a probability of $50\%$ and get $1.97\cdot B$ back. In the event of a loss you will receive $0.3\cdot B$ back.
We choose a fixed number $N\in\{2,3,\ldots,96\}$, a bet-amount $B>0$ and play the game with a seed-capital of $M_0\in\mathbb{R}_+$ multiple times in a row.
Set $\omega:=W(N)-1$. Lets define be a sequence of r.v.'s with $X_n\in\{1,0\}$ such that $X_n=1$ describes a win and $X_n=0$ represents a loss of the $n$-th game played. Obviously $P(X_n=1)=:p=P(R<N)=\frac{N-1}{100}$ and $P(X_n=0)=:q=1-p$ holds for all $n\in\mathbb{N}$, therefore $X\sim Ber(p)$. Because $X_n$ is bernoulli distributed we know that $E[X_n]=p$.
Now we take a look at the earnings in each round. Obviosly they can be described by the sequence of r.v. $Y_n=X_n\cdot\omega B+(X_n-1)\cdot\gamma B$. The expected value is:
$$\begin{align}E[Y_n]&=E[X_n\cdot\omega B+(X_n-1)\cdot\gamma B]\\&=E[X_n]\cdot\omega B+(E[X_n]-1)\cdot\gamma B\\&=p\cdot\omega B+(p-1)\cdot\gamma B=B(p\cdot\omega-(1-p)\cdot\gamma)\\&=B(p\cdot\omega-q\cdot\gamma)\end{align}$$
So it is a fair game with $E[Y_n]=0$ if $\omega=1$, $\gamma=1$ and $p=0.5$. We continue to simplify the expression from above with $\delta:=1-\gamma$ (Cashbackrate), $\Delta:=\beta-1$ and $\omega=W(N)-1=\frac{\beta}{p}-1$. Leading to:
$$\begin{align}E[Y_n]&=B\cdot(\omega p-\gamma q)=B\cdot((\frac{\beta}{p}-1)p-\gamma q)\\&=B\cdot(\beta-p-\gamma(1-p))=B\cdot(\beta-p-\gamma+\gamma p)\\&=B\cdot(\beta-p(1-\gamma)-\gamma)=B\cdot(\beta-p\delta-(1-\delta))\\&=B\cdot((\beta-1)+\delta(1-p)=B\cdot((\beta-1)+\delta q)\\&=B\cdot(\Delta+\delta q) \end{align}$$
In the case described at the beginning with $\beta=0.985$ and no cashback (meaning $\gamma=1$) and $N=51$ (which gives $p=0.5$) we get $E[Y_n]=B\cdot(\Delta+\delta q)=B(-0.015+0\cdot 0.5)=-0.015\cdot B$. So in a roll-under-game without any cashback on losses we expect a loss of 1.5% of the bet-amount each game. Now let's check how the cashback rate has to be changed in order to expect a profit:
$$E[Y_n]\geq 0\Leftrightarrow B\cdot(\Delta+\delta q)\geq0 \Rightarrow (\Delta+\delta q)\geq0 \Leftrightarrow q\geq\frac{-\Delta}{\delta} $$
So we expect a profitable game when $\frac{-\Delta}{\delta}<q$. For $N=51$, $\beta=0.985$ and with a cashback-rate of $\delta=5\%,7\%,9\%,11\%,13\%,15\%$ we achieve a expected profit of $1\%,2\%,3\%,4\%,5\%,6\%$ of the bet-amount $B$ each game. On the other hand if we choose a cashback-rate of $\delta=3\%$ we achieve a fair game.
Is this calculation correct so far?
For the sake of completeness we look at the capital stocks r.v.'s $M_1,M_2,\ldots$ after each round. $M_0$ describes the overall capital at the beginning of the game. It is clear that $M_n=M_0+\sum_{k=1}^n X_n$ applies. For the expected value we get:
$$\begin{align}E[M_n]&=E[M_0+\sum_{k=1}^n Y_n]=M_0+\sum_{k=1}^n E[X_n]=M_0+B\sum_{k=1}^n(\Delta+\delta q)\\&=M_0+n\cdot B(\Delta+\delta q)\end{align}$$
Now i try to find the probability that i am runied after $n$ games. For this we define $W_n=\sum_{k=1}X_k$ the number of wins in $n$ trails, which we know is binomial distributed $W_n\sim Bin(n,p)$. Further we define $L_n=n-W_n$ the number of lost games in $n$ trails. So we are bankrupt if:
$$\begin{align}M_0+W_n\omega B-L_n\cdot\gamma B\leq 0 &\Leftrightarrow M_0+W_n\omega B \leq L_n\cdot\gamma B \\ &\Leftrightarrow M_0+W_n\omega B \leq (n-W_n)\cdot\gamma B \\ &\Leftrightarrow M_0+W_n\omega B \leq n\gamma B-W_n\gamma B \\ &\Leftrightarrow M_0+W_n B(\omega-\gamma) \leq n\gamma B \\&\Leftrightarrow W_n\leq\frac{n\gamma B-M_0}{B(\omega-\gamma)} \end{align}$$
So the probability to get ruined after $n$ games is given by $P(W_n\leq K)$ with $K:=\frac{n\gamma B-M_0}{B(\omega-\gamma)}$. Because $W_n\sim Bin(n,p)$ we know the cumulativ distribution function $F_{W_n}(x)$. This gives us:
$$P(W_n\leq K)=F_{W_n}(K)=\sum_{k=0}^{\lfloor K \rfloor}\binom{n}{k}p^kq^{n-k}$$
Is this correct? Are there any flaws? Especially the last result seems somehow strange to me.
How can I determine the probability that I have a higher capital $M_n$ after $n$ games than at the beginning $M_0$? Which means nothing else than, how do I calculate the probability $P(M_n\geq M_0)$?
Is a approach through random walks better? Basically it is a random walk with different step-sizes. It walks $\gamma B$ to the left side with probability $q=1-p$ and walks $\omega(N) B$ to the right with the probability $p=\frac{N-1}{100}$. Is there any literature for this special case of a random-walk?
Any assistance, thoughts or comments would be much appreciated.