2 are people 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): https://i.stack.imgur.com/ytYKp.jpg
A contest comprises of 3 jars labelled one, two, three. How do I prove that if Person A picks jar two on his 1st go, his probability of victory is 1/3, while if he picks jar one his probability of victory is 2/3. Help would be much appreciated :)
If A chooses jar 2, he has of course a chance of $1/3$ to be right. If he is not right, he looses the game, because it remains only one jar per side. When the anchor says "smaller" or "bigger", person B win the game.
If A chooses jar 1, he has a chance of $1/3$ of winning; in the others $2/3$ of cases, the anchor has to say "bigger", and person B has to choose between jar 2 and 3, with $1/2$ probability for each jar. The total win probability for A is $\frac{1}{3}\textrm{ (A wins at 1st try)} + \frac{2}{3} \times \frac{1}{2} \textrm{ (A looses at 1st try and B looses at his try)} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$