Problem Statement
A particle starts at \((4, 4)\). Each turn, it moves either:
- 1 unit in the \(-x\) direction,
- 1 unit in the \(-y\) direction, or
- 1 unit diagonally in the \(-x\) and \(-y\) directions (i.e. \(-1, -1\)).
Each move has equal probability \(\frac{1}{3}\).
The particle continues moving until it hits either the x-axis or the y-axis for the first time.
What’s the probability that it hits the origin?
Original Problem Link: Click here
Solution
Let’s try to visualize what’s happening.
At each step, the particle chooses one of 3 directions:
←, ↓, or ↙ (i.e., move left, down, or diagonally left-down).
Now, we stop the moment we hit either axis. That means:
- The particle can only reach the origin \((0, 0)\) if it never touches the x-axis or y-axis before that.
- This is a strict constraint: you either go all the way to the origin without hitting any edge, or the game ends.
Key Insight
To reach \((0, 0)\), the last point before that must be \((1, 1)\). Why?
- Because if we ever touch a point like \((1, 0)\) or \((0, 1)\), the process ends.
- So the only way to reach the origin is by moving from \((1, 1) \to (0, 0)\), which is a diagonal move (only allowed move from \((1, 1)\)).
Thus, we break the problem into two parts:
- What is the probability of reaching \((1, 1)\) from \((4, 4)\)?
- Then, with a \(\frac{1}{3}\) chance, we go diagonally to \((0, 0)\).
Step 1: Getting to \((1, 1)\)
Let’s count the number of paths from \((4, 4) \to (1, 1)\) using only the allowed moves:
We define three types of moves:
- L (left): \((-1, 0)\)
- D (down): \((0, -1)\)
- Diag (diagonal): \((-1, -1)\)
We need to move a total of:
- 3 units left
- 3 units down
→ But we can also substitute some L+D pairs with diagonal moves.
Let’s enumerate all valid move combinations that result in reaching \((1, 1)\) from \((4, 4)\):
Case 1: 3 Left, 3 Down → 6 moves total
- No diagonals
- Ways = \(\binom{6}{3} = 20\)
- Each path has probability \(\left(\frac{1}{3}\right)^6\)
Case 2: 2 Left, 2 Down, 1 Diagonal → 5 moves
- Total ways = \(\binom{5}{1} \cdot \binom{4}{2} = 5 \cdot 6 = 30\)
- Each path has probability \(\left(\frac{1}{3}\right)^5\)
Case 3: 1 Left, 1 Down, 2 Diagonal → 4 moves
- Total ways = \(\binom{4}{2} = 6\) (choose 2 out of 4 to be Diagonal, rest are L & D)
- Then multiply by 2 for L and D order → \(6 \cdot 2 = 12\)
- Each path has probability \(\left(\frac{1}{3}\right)^4\)
Case 4: 3 Diagonal moves → 3 moves total
- Only 1 way
- Probability = \(\left(\frac{1}{3}\right)^3 = \frac{1}{27}\)
Step 2: Multiply with probability to go from (1,1) → (0,0)
That can only happen if the particle chooses Diagonal again → Probability = \(\frac{1}{3}\)
Final Probability
\[P = \left[ 20 \cdot \left(\frac{1}{3}\right)^6 + 30 \cdot \left(\frac{1}{3}\right)^5 + 12 \cdot \left(\frac{1}{3}\right)^4 + 1 \cdot \left(\frac{1}{3}\right)^3 \right] \cdot \frac{1}{3}\] \[= \left[ \frac{20}{729} + \frac{30}{243} + \frac{12}{81} + \frac{1}{27} \right] \cdot \frac{1}{3}\]Converting all to a common denominator (say 729), we get:
\[= \left[ \frac{20}{729} + \frac{90}{729} + \frac{108}{729} + \frac{27}{729} \right] \cdot \frac{1}{3} = \frac{245}{729} \cdot \frac{1}{3} = \frac{245}{2187}\]The required probability is:
\[\boxed{\frac{245}{2187}}\]💡 Did you enjoy this problem? Check out more puzzles in the Problems section!