If someone goes to the casino with 10 euro and each time he gambles 1, where every time he has a chance of either losing or gaining find the chance If someone goes to the casino with 10 euro and each time he gambles 1 euro, where every time he has a 40% chance of gaining 1 euro and a 60% chance of losing it, find the probability of him eventually reaching 100 euro.
I found that the recursive function for this is $P_n=\frac{3P_{n-1}+2P_{n+1}}{5}$ however this is as far as I got. Based on this recursive function, could you please show me how to solve the question?
 A: The best way to solve the problem avoids solving the recurrence relation directly.
Let's say that the value of having $x$ euros is $(\frac32)^x$. I have chosen this function because now, the game is fair: from a state with value $(\frac32)^x$, the expected value of the next state is $\frac25 \cdot (\frac32)^{x+1} + \frac35 \cdot (\frac32)^{x-1}$, which simplifies to $(\frac32)^x$ because $\frac25 \cdot \frac94 + \frac35 \cdot 1= \frac32$.
Now, intuitively, if the game is fair, and you start with a value of $(\frac32)^{10}$, your expected value at the end should also be $(\frac32)^{10}$. But if you have a probability of $p$ of getting to $100$ euros, then:

*

*with probability $p$, you have a value of $(\frac32)^{100}$ at the end;

*otherwise, with probability $1-p$, you have a value of $(\frac32)^0 = 1$.

The expected value at the end is $(\frac32)^{100} \cdot p + (1-p)$. Solving $(\frac32)^{100} \cdot p + 1 - p = (\frac32)^{10}$ gives us $p = \frac{(\frac32)^{10}-1}{(\frac32)^{100}-1}$.
A way to make the intuition that the final expected value should equal the starting value is the optional stopping theorem. This also gives us some conditions to check for such an argument to work (which are satisfied in this case).
