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 they 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 up to four. Why is it that regardless of which jar Person A picks on his first go, Person B has a method for guaranteeing her probability of victory as 1/2. Help would be appreciated :)
HINT:
The question is asking for a comprehensive strategy for B, i.e. a recipe for what to do in all cases when B has a choice. In this simple game there are only 2 situations where B even has to choose (the rest are equivalent by symmetry):
If A picks 1 and does not win right away, what should B pick? There are 3 choices: 2, 3, or 4.
If A picks 2 and does not win right away, and if the anchor says "bigger", then what should B pick? There are 2 choices: 3 or 4.
So a strategy for B is simply a pair $(s,t)$ where $s\in \{2, 3, 4\}$ and $t \in \{3, 4\}$. There are 6 strategies, and each can be analyzed to find B's overall probability of winning (if using that strategy). If you do this carefully you will find that some of them indeed guarantees 1/2 chance of winning.