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In the game up or down, in each round your capital either increases by 1 euro or decreases by $1$ euro, each with probability $\frac{1}{2}$. The game stops when your capital is $10$ euro (you leave happy) or $0$ euro (you leave frustrated). You start with $1$ euro. Is the expected gain in your capital at the end of the game positive, zero or negative? Prove your answer!

Intuitively I'd say the probability of hitting zero before hitting $10$ euro is greater than the reverse since it takes less throws to reach zero than to reach $10$ from our starting capital. Hence I'd say our expected gain is negative. Am I correct in this and is there a way to give a clean proof of this?

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    The payout is 10x the buy in. While you are definitely right that the probability of hitting zero is higher than hitting 10, but if 9/10 times you hit zero and one time you hit 10, you break even. Can you use this information to figure out exactly what you want to know about the expected value here? – 2017-01-07

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The gain from the game is always either $+9$ or $-1$ (except if you manage to keep playing forever, but the probability of that happening is $0$).

However, the probability of leaving the game with a $+9$ gain is $\frac{1}{10}$, whereas the probability of leaving the game with a $-1$ loss is $\frac{9}{10}$. So the expected gain is $$ \frac1{10}\times 9 + \frac{9}{10}\times(-1) = 0 $$

Even though one of the probabilities is larger than the other, the size of the potential loss is smaller than the size of the potential gain, and these differences cancel each other out.

The easiest way to compute these probabilities is probably to work backwards from the knowledge that the expected gain over the entire game must be $0$. We can compute that by using additivity of expectation: The expected total gain is the expected gain in round 1, plus the expected gain in round 2, plus the expected gain in round 3, etc. In each round the expected gain is $$ p_n\times\frac{(+1)+(-1)}{2} + (1-p_n)\times 0 $$ where $p_n$ is the probability that the game lasts long enough that round $n$ is played at all -- but we don't need to compute the $p_n$s to see that each of the per-round expected gains is $0$ no matter what.