Two squares are chosen at random on a chessboard. The probability that they have a side in common is

Q: Wait, why do we use $^{64}C_{2}$? Can't we just pick one square and then another?

Great question! If you pick one square and then another, you are essentially counting the order of selection (e.g., picking square A then B is different from B then A). However, the question asks for the probability of 'choosing two squares,' which implies a set of two. Using $^{64}C_{2}$ correctly counts the number of unique sets of two squares, which is the standard approach for this type of probability problem.

Q: What about diagonal adjacency? Do those squares share a side?

No, they do not. The problem specifically asks for squares that have a 'side in common.' Diagonally adjacent squares only share a corner (vertex), not a side. Therefore, we only count horizontal and vertical adjacencies.

Q: Is there a faster way to calculate the total combinations without doing $^{64}C_{2}$?

You could think of it as picking the first square (64 options) and then the second (63 options), but that would count each pair twice (A,B and B,A). So, you would have to divide by 2. That leads you to $\frac{64 \times 63}{2}$, which is exactly what $^{64}C_{2}$ represents. It is the most robust way to avoid double-counting!

Question

Two squares are chosen at random on a chessboard (see figure). The probability that they have a side in common is :

Options:

(A) $\frac{2}{7}$

(B) $\frac{1}{18}$

(C) $\frac{1}{7}$

(D) $\frac{1}{9}$

Sigma Percentile · Accepted Answer

### The Geometry of the Chessboard

Imagine you are standing before a standard $8 	imes 8$ chessboard. It is a classic, elegant grid of $64$ squares.

When we talk about probability in this context, we are essentially exploring the 'universe' of all possible ways to select two squares from this grid. To find the probability that two randomly chosen squares share a side, we must first define our sample space and then isolate the specific, favorable configurations.

### Phase 1: Defining the Sample Space

Our first task is to determine the total number of ways to choose any two squares from the $64$ available. This is a classic combinatorics problem.

We are not concerned with the order in which we pick the squares, just the final pair. Thus, we use the combination formula $^{n}C_{r}$, where $n=64$ and $r=2$.

$$^{64}C_{2} = \frac{64 	imes 63}{2 	imes 1} = 32 	imes 63 = 2016$$

This number, $2016$, represents every possible pair of squares you could pick on the board. This is our denominator—the total number of outcomes.

### Phase 2: The Geometry of Adjacency

Now, let us find the favorable outcomes. Two squares share a side if they are adjacent. This can happen in two ways: horizontally or vertically.

Consider a single horizontal row of $8$ squares. If you label them $1, 2, 3, 4, 5, 6, 7, 8$, the pairs are $(1,2), (2,3), (3,4), (4,5), (5,6), (6,7),$ and $(7,8)$. That is exactly $7$ pairs per row.

Since there are $8$ rows on the board, the total number of horizontal adjacent pairs is:

$$8 	imes 7 = 56$$

Now, apply the exact same logic to the vertical columns. Each column has $8$ squares, yielding $7$ vertically adjacent pairs. With $8$ columns in total, the number of vertical adjacent pairs is also:

$$8 	imes 7 = 56$$

### Phase 3: The Final Calculation

To find the total number of favorable outcomes, we simply add these two counts together:

$$	ext{Total Favorable} = 56 + 56 = 112$$

We have $112$ pairs of squares that share a side. Now, we calculate the probability $P$ by dividing the favorable outcomes by the total outcomes:

$$P = \frac{112}{2016}$$

To simplify this fraction, let us use the factors we already found. We know $2016 = 32 	imes 63$ and $112 = 16 	imes 7$. Substituting these back into our probability equation:

$$P = \frac{16 	imes 7}{32 	imes 63}$$

Notice the elegance of the cancellation: $16$ goes into $32$ twice, and $7$ goes into $63$ nine times. This leaves us with:

$$P = \frac{1}{2 	imes 9} = \frac{1}{18}$$

The probability that two randomly chosen squares on a chessboard share a side is exactly **$\frac{1}{18}$**. It is a beautiful result, born from simple geometric counting and careful combinatorial logic.

Animated Solution for Mathematics - Probability: Two squares are chosen at random on a chessboard (see figure). The probability that they have a side in common is :

Select Answer:

Visualized Solution

The Chessboard Grid

Total Ways to Choose 2 Squares

Expanding the Combination

Calculating Total Outcomes

Defining "Side in Common"

Horizontal Adjacent Pairs

Total Horizontal Pairs

Vertical Adjacent Pairs

Total Vertical Pairs

Total Favorable Outcomes

Probability Formula

Substituting the Values

Simplifying the Fraction

Final Answer

The Sigma Insight: Classical Definition of Probability

The Geometry of the Chessboard

Phase 1

Phase 2

Phase 3

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