Probability brain teaser In a repeatable bet, the odds of winning a dollar are 51% (odds of losing are 49%). If you start with $100, what is the chance you will eventually go broke?
-This is the only information given. 

In fact the 51% and 49% are probabilities.  The odds of winning a dollar are $\frac{51}{49}$ and the odd of losing are $\frac{49}{51}$.
There is an expression (I think originally found by Huygens) that if the player starts with $a$ dollars and the bank with $b$ dollars, and the odds of the bank winning one game is $d$, then the probability player reaches zero before the player reduces the bank to zero  is $$\frac{\,\, d^{a+b} - d^{a}}{d^{a+b}-1} \text{ or equivalently } \frac{ d^{a}-d^{a+b}\,\,}{1 - d^{a+b}} .$$  If $d \gt 1$ as in most gambling games, then as $b$ increases without limit this tends to $1$ , but if $d \lt 1$ as here, then as $b$ increases without limit this tends to $d^a$, which in this case is  $$\left(\tfrac{49}{51}\right)^{100} \approx 1.83\%.$$
The Wikipedia article on the gambler's ruin and unfair coin flipping has more detail.
