An Unfair Game

In every round, the probability of earning one dollar is $\frac{2}{6} = \frac{1}{3}$ , and the probability of losing one dollar is $\frac{4}{6} = \frac{2}{3}$ . In the first round you have $2$ dollars. If you win, you will have $3$ dollars and the game ends inmediately. If you lose, you will have $1$ dollar, and you will be forced to win the next round, or you will end with no dollars. If you win the next round, you will have $2$ dollars again, and as each dice roll is independent, your probability of ending with $3$ dollars at this point will be the same it was at the start of the game. Let $P$ be the probability of ending the game having three dollars. $P$ can be expressed as: $P = \frac{1}{3} + \frac{2}{3} \cdot \frac{1}{3}P \\ P - \frac{2}{9}P = \frac{1}{3} \\ P = \frac{{1/3}}{{1 - 2/9}} \\ P = \frac{3}{7}$ Then $a=3$ and $b=7$ . Therefore, the answer is: $a^b = 3^7 = \boxed{2187}$ Another way of solving this problem is calculating the sum of the probabilities of all the possible ways of ending up with $3$ dollars. The most obvious way is winning the first round. Another way is losing in the first round and then winning twice (after losing it is necessary to win the next round). In general, all the ways are of the form lose-win-lose-win-lose- ... -win-lose-win-win. The sum of probabilities is then: $P = \frac{1}{3} + \left( {\frac{2}{3} \cdot \frac{1}{3}} \right)\frac{1}{3} + \left( {\frac{2}{3} \cdot \frac{1}{3}} \right) \cdot \left( {\frac{2}{3} \cdot \frac{1}{3}} \right)\frac{1}{3} + \ldots \\ P = \frac{1}{3} + \left( {\frac{2}{9}} \right)\frac{1}{3} + \left( {\frac{2}{9}} \right)^2 \frac{1}{3} + \ldots \\ P = \frac{1}{3}\sum\limits_{k = 0}^\infty {\left( {\frac{2}{9}} \right)^k } \\ P = \frac{1}{3} \cdot \frac{9}{7} = \frac{3}{7}$ And the rest of the problem follows.