2 people are in a competition. Jars are placed in a row and are labelled from left to right 1, 2, 3... A ball is concealed in one of these jars and all jars have the same chance of having the ball inside of them.
Each contestant takes turns to guess which jar the ball is inside of. If one of them guesses the right jar, the anchor will say 'correct', which indicates that the contestant has won. However, if the guess is wrong, the anchor will state 'smaller' or 'bigger' (referring to the numbers) to show the direction the ball is hidden. The contestants always obey the anchor's directions. For example, if ten jars are presented (labelled one to ten) and the ball is concealed in jar seven, the contest may go like the following (follow the link to see the example):
A contest has jars labelled one to nine. Why is it that regardless of which jar Person B picks, Person A has a method where he picks jar five on the first go and guarantees his probability of victory is higher than $1/2$.
I thought that if Person A picks 5 on the first go, he would have $2/3$ chance of winning. This is the equation I used to arrive at that: $1/9 + 8/9 \cdot 3/4 \cdot 1/3 + 2/3 \cdot 1/2 \cdot 1/1$
However, I do not know if this is correct.
The correct probability will be 5/9, but since you do not provide an explanation how you arrived at the various fractions in your answer, I can't tell you what the mistake is that you made. But here is a possible way to arrive at the answer.
It is important for these type of problems to mention how both participants play the game. Since there is an optimal strategy, we are going to assume that both players use it.
Let $n \geq 1$ be the number of jars that is used in the game and $P(n)$ be the probability that the first player that is allowed to choose a jar will win using the optimal strategy. It is clear that in the case $n=1$ we have $P(1)=1$ and also the case $n=2$ is easy and we have $P(2)=\frac{1}{2}$ regardless which jar is chosen.
Now consider a general game with $n$ jars and the first player chooses the $k$th jar ($1 \leq k \leq n$). What is the probability $P(k|n)$ that this will result in winning the game? There is a probability $\frac{1}{n}$ for this to be the correct one. More likely it will be wrong and it is either too high or too low. It could be too high with a probability $\frac{k-1}{n}$, which will give the other player the chance to win with a probability $P(k-1)$, because he/she is the first player in a game with only $k-1$ jars. Alternatively, the number $k$ was too low with a probability $\frac{n-k}{n}$ and the second player has probability $P(n-k)$ to win using the optimal strategy. So we find:
$P(k|n) = \frac{k-1}{n} \left[1 - P(k-1) \right] + \frac{1}{n} + \frac{n-k}{n} \left[1 - P(n-k) \right]$
where we used the probabilities $1 - P(k-1)$ and $1 - P(n-k)$ for the second player to loose. From this we obtain the value of $P(n)$ by finding the maximum value of $P(k|n)$ among all $1 \leq k \leq n$.
What I will leave as an exercise is to show that:
1) The winning probability with optimal strategy: $P(n) = \left\{ \begin{array}{ll} \frac{1}{2} & \text{for $n$ even} \\ \frac{n+1}{2n} & \text{for $n$ odd} \end{array} \right. $
2) For even values of $n$, we have $P(k|n) = \frac{1}{2}$ for all $k$
3) For odd values of $n$, we have $P(k|n) = \frac{n+1}{2n}$ for odd values of $k$ and $P(k|n) = \frac{n-1}{2n}$ for even values of $k$
You can prove this by induction and considering the four cases $n,k$ being even/odd.
In the particular case $n=9$ and $k=5$, you only need the value of $P(4)$ and it is a bit easier.