Probabilistic Treasure hunt game I am designing a treasure hunt game where users start of at a fixed number of steps (x) away from the treasure. The user responses are either a or b. A correct answer places the user one step closer to the treasure, a wrong answer takes the user one step away from the treasure. (maximum steps away from the treasure is x.). The two options have a 50% chance of being right on each question.
The Game ends when player is 0 steps from the treasure.
Is it correct that the user will always get to the treasure in 2x moves? Or is it 2^x (2 power x)?
Is there anything a player can do to increase their chances of winning? Is there a way I  can make the game harder. (Apart from increasing x)? 
Please keep the answer simple.
 A: The stochastic process you are describing is well known to be a random walk. I know OP said that he doesn't know about stochastic processes, but for the audience, consider the transition matrix in his problem, with state space $\{0,1,\cdots,x\}$ : 
$$
P = \underset{x+1 \text{ columns }}{\underbrace{\begin{pmatrix}
1/2 & 1/2 & 0 & \cdots & 0 \\
1/2 & 0 & 1/2 & 0 & \vdots \\
0 & 1/2 & 0 & \ddots & \vdots \\
\vdots & \vdots & \ddots & \ddots & \vdots \\
0 & \cdots & \cdots & 0 & 1 \\
\end{pmatrix}
}}
$$
From this matrix we get the matrix $Q$ of probabilities in transient states :
$$
Q = \underset{x \text{ columns }}{\underbrace{
\begin{pmatrix}
1/2 & 1/2 & 0 & \cdots & 0 \\
1/2 & 0 & 1/2 & 0 & \vdots \\
0 & 1/2 & 0 & \ddots & \vdots \\
\vdots & \vdots & \ddots & \ddots & 1/2 \\
0 & \cdots & \cdots & 1/2 & 0 \\
\end{pmatrix}
}}
$$
and thus the fundamental matrix becomes the inverse of
$$
I-Q = \begin{pmatrix}
1/2 & -1/2 & 0 & \cdots & 0 \\
-1/2 & 1 & -1/2 & 0 & \vdots \\
0 & -1/2 & 1 & \ddots & \vdots \\
\vdots & \vdots & \ddots & \ddots & -1/2 \\
0 & \cdots & \cdots & -1/2 & 1 \\
\end{pmatrix}
$$
One can guess by looking at the inverses of the matrix for small $x$ (using a computer to compute inverses for greater $x$ if his eyes are blind) that
$$
(I-Q)^{-1} = \begin{pmatrix}
2x & 2(x-1) & 2(x-2) & \cdots & 2 \\
2(x-1) & 2(x-1) & 2(x-2) & \cdots & \vdots \\
2(x-2) & 2(x-2) & 2(x-2) & \ddots & \vdots \\
\vdots & \vdots & \ddots & \ddots & 2 \\
2 & \cdots & \cdots & 2 & 2 \\
\end{pmatrix}
$$
which means that the expected number of steps to go from $0$ to $x$ is
$$
\sum_{k=1}^x 2k = 2\frac{x(x+1)}2 = x^2 + x.
$$
Furthermore, if it is of interest in your case, if your player has reached state $n$ and you want to know the expected number of steps before he reaches state $x$, that would be
$$
\sum_{k=1}^n 2k + (x-n)2n = 2n(x-n) + n^2 + n.
$$
Hope that helps,
A: It may happen that the (2k) first answers are abababa... in which case you will not have reached 0 after (2k) steps, hence no uniform deterministic bound on the time needed to reach 0 can hold. But the expected time needed to reach 0 starting from x is x^2+x (this can be explained rather elementarily if you wish). 
If the probabilities of a step in each direction are the same at each site but unequal, the expected time becomes linear in x if the chances to make a step towards 0 are more than 50% but it becomes exponential in x if the chances to make a step towards 0 are less than 50% (and I guess some physicists would call this a phase transition).
A: I hope I understand your problem correct. On each step the right answer is either $a$ or $b$ with probability $0.5$ - so the player can be at positions $0,1,2,...,x$. 
First of all, player cannot do anything here since his choice does no influence at all - the random number generator in average will do the same. Second, there is always non-zero probability not to find the treasure in $n$ steps, no matter how great is $n$. Third, to make the game more harder you may want to make it more interesting - namely incorporate the nice dependence on the player's decisions.
