Random Colourings of a Chessboard

The expected value of $|B-W|$ refers to a weighted mean of all the possible values which $|B-W|$ can take on.

Since $B$ and $W$ can only take positive integers, ranging from $0$ to $9$ , $|B-W| = 1,3,5,7,9$ only.

For example, for $|B-W| = 9$ , either $B=9, W=0$ or $B=0, W=9$ . Thus, the probability of $|B-W|=9$ can be expressed as $\frac {2{9 \choose 0}}{2^9}=\frac {2}{512}$ .

In general, if $|B-W|=m$ , then its probability is $\frac {2{9 \choose {\frac {9-m}{2}}}}{2^9}=\frac {9 \choose {\frac {9-m}{2}}}{256}$ .

Note: The probabilities are calculated by multiplying the number of ways to have $n$ squares of one colour by 2, where $n$ is an integer and $1 \leq n \leq 9$ , since there are 2 possible colours. Also, as there are $n$ squares of one colour, there will be $9-n$ squares of the other colour.

Thus, the expected value of $|B-W|$ is given as:

$9 \cdot \frac {9 \choose {\frac {9-9}{2}}}{256}+7 \cdot \frac {9 \choose {\frac {9-7}{2}}}{256}+5 \cdot \frac {9 \choose {\frac {9-5}{2}}}{256}+3 \cdot \frac {9 \choose {\frac {9-3}{2}}}{256}+1 \cdot \frac {9 \choose {\frac {9-1}{2}}}{256}$

$={\frac {1}{256}}({9 \cdot {9 \choose 0}}+{7 \cdot {9 \choose 1}}+{5 \cdot {9 \choose 2}}+{3 \cdot {9 \choose 3}}+{1 \cdot {9 \choose 4}})$

$=\frac {1}{256}(9+63+180+252+126)$

$=\frac {630}{256}$

$=\frac {315}{128}$

Thus, $a+b=315+128=443$ .

If the board squares were randomly colored black and white, then $|B-W|$ can take only five discrete values, namely $9 , 7 , 5 , 3 , 1$ . This can be explained as follows:-

In case of all squares are black or white, then $|B-W|=9$

In case only one square is black or white, then $|B-W|=7$ , and so on ...

The next step we calculate the frequency of each value:-

The frequency of each of the five discrete values can be expressed as a combination ${a \choose b}$

where:- $a=$ the total number of squares, $b=$ the number of squares of one color

For example:- the frequency of the value $5$ can be expressed calculated as ${9 \choose 2}\times2$ , where the two factor comes from the fact that $|B-W|=5$ occurs when there are two white squares (rest black) of two black squares (rest white)

The calculated frequencies are as follows:- $2$ for $9$ , $18$ for $7$ , $72$ for $5$ , $168$ for $3$ , $252$ for $1$ , totaling $512=2^9$

Finally, the weighted mean (expected value) is $\frac {9\times2+18\times7+72\times5+168\times3+252\times1}{512}=\frac {315}{128}$ , making the solution $315+128=443$

We know that the total possible ways of coloring will be 2^9 (either black of white)

We have 5 cases

9 white, 0 black - can be done in 9C0=1 way, therefore value of |B−W|= 9

8 white, 1 black - can be done in 9C1 = 9 ways, therefore value of |B−W|= 7

7 white 2 blacks - can be done in 9C2= 36 ways, therefore value of |B−W|= 5

6 white 3 blacks - can be done in 9C3 = 84 ways, therefore value of |B−W|= 3

5 white 4 blacks - can be done in 9C4 = 126 ways, therefore value of |B−W|= 1

The same will happen for 4 white 5 blacks, 3 white 6 blacks, 2 white 7 blacks, 1 white 8 blacks, and 9 white 0 blacks.

Therefore the expected value is of |B−W| is calculated from;

(1/2^9) x (2x9x1 + 2x9x7 + 2x36x5 + 2x84x3 + 2x126x1)

=1260/512 =315/128

Therefore, the value of a+b

Notice the value of $|B - W|$ equals 9 when there are 9 black or 9 white squares, 7 when there are 8 of one color and one of the other, 5 when there are 7 of one color and 2 of the other, 3 when we have 6 of one color and 3 of the other, and 1 when we color 5 of one color and 4 of the other. Notice that these colorings can be done by choosing a certain number of squares to paint black, and leave the remaining squares white, we can do this in ${9 \choose x}$ ways where x is the number of black squares we need. We then calculate the expected value by multiplying each value of $|B - W|$ by its probability, considering there are $2^9$ possible outcomes in all since each of the nine squares can be colored black or white, we then obtain that the expected value equals $\displaystyle \sum_{i=0}^9 \frac{{9 \choose x}}{2^9} \cdot |x - (9-x)|$ , which works out to $\frac{315}{128}$ , from which $a + b = 315 + 128 = 443$

There are 9 squares. So total number of ways in which the board can be colored will be 2^9.

Different possible values of |B - W| are:- i) 9 No of ways = C(9, 0) + C(9, 9) = 2 ii) 7 No of ways = C(9, 1) + C(9, 8) = 18 iii) 5 No of ways = C(9, 2) + C(9, 7) = 72 iv) 3 No of ways = C(9, 3) + C(9, 6) = 168 v) 1 No of ways = C(9, 4) + C(9, 5) = 252

Expected value = (9 2 + 7 18 + 5 72 + 3 168 + 1*252)/512 = 315/128

a + b = 443

the various cases possible are: (denoting black by B and white by W) difference Product(p) 1. 9B - 9C9 ways 9 9C9* 9 2. 8B,1W - 9C8 ways 7 9C8* 7 3. 7B,2W - 9C7 ways 5 9C7* 5
4. 6B,3W - 9C6 ways 3 9C6* 3 5. 5B,4W - 9C5 ways 1 9C5* 1 and others similarly
Then expected value = (Sum of all p's)/2^9=315/128 Thus a+b=315+128=443

To find the value of $|B-W|$ , we simply have to find the average of each value multiplied by its probability.

Without loss of generality, assume $B \geq W$ since the values are interchangeable.

Expected value of $|B-W|$ is therefore $\frac{\sum [(value of B-W)\times(number of ways to color B number of squares out of 9 black)]}{total number of ways to color the board}$ $=\frac{(5-4)\times{9 \choose 5}+(6-3)\times{9 \choose 6}+(7-2)\times{9 \choose 7}+(8-1)\times{9 \choose 9}+(9-0)\times{9 \choose 9}}{{9 \choose 5}+{9 \choose 6}+{9 \choose 7}+{9 \choose 8}+{9 \choose 9}}$ $=\frac{126+252+180+63+9}{126+84+36+9+1}=\frac{630}{256}=\frac{315}{128}$

Thus $a+b=315+128=443$

For any colouring, $B + W = 9$ . Suppose there are $k$ white squares. This happens with probability $\frac{\binom{9}{k}}{2^9}$ , and the value of $\vert B - W \vert$ is $\vert (9-k)-k \vert = \vert 9 - 2k \vert$ . We can calculate the expected value of $\vert B - W\vert$ by summing over all $k$ . This gives $\begin{aligned} \frac{1}{2^9} \sum\limits_{k=0}^{9} \binom{9}{k} \vert 9-2k \vert &= \frac{1}{2^9} (1 \times 9 + 9 \times 7 + 36 \times 5 + 84 \times 3 + 126 \times 1 + 126 \times 1 + 84 \times 3 + 36 \times 5 + 9 \times 7 + 1 \times 9) \\ &= \frac{1260}{2^9} \\ &= \frac{315}{128}. \end{aligned}$

Hence $a + b = 315 + 128 = 443$ .