From Andrew Lin comes a game for getting yourself home:
You are stranded in a casino (lucky you!) and need to purchase a flight home. Flights cost $250, but you have only $100 at the moment. However, as I just said, you’re in a casino! Surely, you can gamble your way to $250.
The casino has a game called “Riddler’s Delight,” in which you can bet any amount of money in your possession for an even greater amount of money. You can even bet fractional (i.e., you can bet fractions of a penny), irrational or infinitesimal amounts if you so desire.
The catch is that the odds are not in your favor. In Riddler’s Delight, whenever you bet A dollars in an attempt to win B dollars (with B > A), your probability of winning is not A/B, which you would expect from a fair game. Instead, your probability of winning is always 10 percent less, or 0.9(A/B).
What should your betting strategy be to maximize your probability of getting home, and what is that probability?
If one bets $A$ in order to win $B$, the probability is as given, $\dfrac{9A}{10B}$.
Given $B > A$ and some positive value $m$:
$$B\cdot m > A\cdot m$$
$$-A\cdot m > -B\cdot m$$
$$AB-A\cdot m > AB-B\cdot m$$
$$A \cdot(B-m) > B \cdot(A-m)$$
$$\dfrac{A}{B} > \dfrac{A-m}{B-m}$$
It follows that if you only have one chance to play "Riddler's Delight", you should bet all $100 and try to win all $250. This yields a probability of $0.36$.
Here are a couple comparisons to consider.
Assume one has $A$ and splits the winnings to $B-m$ the first time and $B$ the second time.
The probability of ending up with $B$ is:
$$\dfrac{9A}{10(B-m)} \cdot \dfrac{9(B-m)}{10B}$$
$$ = \dfrac{81A}{100B}$$
$$< \dfrac{9A}{10B}$$
It follows that if one bets all they have, it is better to go for the full winnings than split it into two or more plays of the game.
Assume one has $A$ and only bets $A-n$. You still have a $\dfrac{1}{10}$ chance of betting with the remaining $n$ if you lose.
The probability of ending up with $B$ is:
$$\dfrac{9(A-n)}{10B} + \dfrac{1}{10} \cdot \dfrac{9n}{10B}$$
$$ = \dfrac{9}{10B} \left(A-n + \dfrac{n}{10}\right)$$
$$< \dfrac{9A}{10B}$$
It follows that when trying to win an amount, it is better to bet all you have instead of saving some for successive bets in case you lose.
Using the above, any variation of splitting will lead to a suboptimal probability. The maxium probability of 0.36 is obtained by betting all $100 and trying to win all $250.