Question

If you choose (retain right) (retain left) in your turns, the best move sequence for your friend to reduce your gain to a minimum will be:

• A. (retain upper) (retain lower)
• B. (retain lower) (retain upper)
• C. (retain upper) (retain upper)
• D. (retain lower) (retain lower)
• A. Rs. 4
• B. Rs. 3
• C. Rs. 2
• D. None of these
• A. Rs. 3
• B. Rs. 6
• C. Rs. 4
• D. None of these

Answer 1

The Correct Option is B

We start with the 4×4 board given in the question:
\[ \begin{matrix} 2 & 1 & 2 & 4
5 & 1 & 6 & 7
9 & 1 & 3 & 2
6 & 1 & 8 & 4 \end{matrix} \] Your first move: (retain right) → Split vertically, keep right 2 columns:
\[ \begin{matrix} 2 & 4
6 & 7
3 & 2
8 & 4 \end{matrix} \] Friend’s move: His goal is to minimize your final cell’s value, so he splits horizontally and chooses (retain lower) → keep bottom half:
\[ \begin{matrix} 3 & 2
8 & 4 \end{matrix} \] Your second move: (retain left) → Split vertically, keep left column:
\[ \begin{matrix} 3
8 \end{matrix} \] Friend’s final move: Splits horizontally and chooses (retain upper) → keeps top cell:
\[ [3] \] Thus your final gain = Rs. 3. Testing other friend moves shows this sequence minimizes your gain under your stated moves, confirming answer (B).

Answer 2

When both players are optimal, the game becomes a perfect-information zero-sum game.
From the initial board, the first player (you) will choose the split (horizontal or vertical) that maximizes the minimum gain you can force, assuming the opponent will minimize. By testing all initial possible moves:
- If you start with (retain right), opponent can force you down to 3 (as in Q82).
- If you start with (retain left), similar analysis yields maximum possible gain 4 if opponent plays poorly, but against optimal opponent, your gain is reduced to 3.
- If you start with horizontal splits, the best you can guarantee is still 3.
Therefore, regardless of the starting optimal path, optimal play from both sides ends with your gain being Rs. 3.

Answer 3

From the initial board, first move (retain right) yields:
\[ \begin{matrix} 2 & 4
6 & 7
3 & 2
8 & 4 \end{matrix} \] Now, no matter which half your friend chooses horizontally, you will have at least one path leading to a final cell of value 3 or greater.
Case 1: Friend chooses upper → You can select the column with 4, ensuring at least Rs. 4.
Case 2: Friend chooses lower → The best friend can do is force you into a cell with value 3 via minimization (as in Q82), but never below 3.
Therefore, with optimal play from your side starting with (retain right), you can always guarantee at least Rs. 3.