Combinatorics: endless series I have the following problem: 
In an urn, you have 1 blue and 9 white balls. You pull out one ball a time; if it is the blue one, you win. If it it is white, you throw it back in and pull again. Imagine 2 people are playing this game. Who has a better chance to win, the person who goes first or the second?
 A: There is a $\frac{1}{10}$ chance that the first player wins on the first pull. There is a $\frac{9}{10}\cdot\frac{1}{10}=\frac{9}{100}$ chance that the second player wins on their first pull. Finally, there is an $\frac{81}{100}$ chance that the game returns to the "start", i.e. it is the first player's turn again. So the probability $P$ that the first player wins satisfies
$$P=\frac{1}{10}+\frac{81}{100}P$$
and therefore
$$\frac{19}{100}P=\frac{1}{10}$$
$$P=\frac{10}{19}$$
Thus the first player has a slight advantage (the probability that the second player wins is only $\frac{9}{19}$).
A: Let $q$ be the probability of the next player winning this turn, and let $P$ be the probability of the next player eventually winning.  The next player eventually wins if he wins on this turn, or if he fails to win on this turn (with probability $1-q$) and then the other player eventually loses (with probability $1-P$).
So $$P=q + (1-q)(1-P)=1-P+qP,$$
or
$$P = \frac{1}{2-q}.$$
Note that for small $q$ (long games), $P$ approaches $1/2$, while for large $q$ (short games), $P$ approaches $1$.  For this particular game, $q=1/10$, so $$P=\frac{1}{2-1/10}=\frac{1}{19/10}=\frac{10}{19}.$$
A: Complete solutions have already been given, but you can see without calculation that in any case the first to draw has a better chance to win. Let $S$ be the chance that the second player eventually wins, and reasoning for the aburd assume that $S\geq\frac12$. For the first player two scenarios are possible: with some (small) positive probability $p$ she wins immediately, and otherwise she finds herself in a position equivalent to the initial positon of the second player. In the first scanrio she has a 100% chance to win, and in the second scenario she has probility $S$ to eventually win. Her actual odds are a weighted average of these two probabilities $1$ and $S\geq\frac12$ (the weights depend on $p$, but both are positive) with definitely lies in the interval $(\frac12,1)$. So adding $S$ to it gives more than $1$, contradicting the fact that both players cannot win at the same time (in principle one could imagine that neither wins and the game goes on for ever; this turns out to have probability $0$, but the reasoning does not depend on this).
