You're all set, diagonally!

Probability Level 2

If three $1\times1$ squares are randomly chosen from a standard $8\times8$ chessboard, the probability that they all lie on the same diagonal is $\frac ab$ for coprime positive integers, find the value of $a+b$ .

The answer is 751.

1 solution

Brian Charlesworth
Sep 17, 2015

In total there are $30$ diagonals on a chessboard, ( $15$ ascending from left to right and $15$ descending from left to right).

Since $8$ of these have fewer than $3$ squares we need only consider the $30 - 8 = 22$ diagonals that have $3$ or more squares. Of these, $4$ each have lengths $3,4,5,6,7$ and $2$ have length $8.$

For a diagonal of length $n,$ there are $\dbinom{n}{3}$ combinations of $3$ squares that can be chosen from that diagonal. Thus the total number of combinations of $3$ squares that lie on the same diagonal of the chessboard is

$4*\left(\dbinom{3}{3} + \dbinom{4}{3} + \dbinom{5}{3} + \dbinom{6}{3} + \dbinom{7}{3}\right) + 2*\dbinom{8}{3} = 392.$

Without restrictions there are $\dbinom{64}{3} = 41664$ combinations of $3$ squares that can be chosen,

so the desired probability is $\dfrac{392}{41664} = \dfrac{7}{744},$ and thus $a + b = 7 + 744 = \boxed{751}.$

Is it possible to generalize this?

Pi Han Goh - 5 years, 9 months ago

What kind of? Can u post the generalized question?

Shyambhu Mukherjee - 5 years, 8 months ago

For $n\times n$ chessboard.

Pi Han Goh - 5 years, 8 months ago

@Pi Han Goh – 1,2 length diagonals are cancelled. 3-(n-1) length diagonals are chosen,each of no4.n length are of only 2. Now we can follow Brian's solution.

Shyambhu Mukherjee - 5 years, 8 months ago

@Shyambhu Mukherjee – Can you post the question? Thanks.

Pi Han Goh - 5 years, 8 months ago

@Pi Han Goh – I have just substituted n n in place of 8 8 in the question. However for m squares to select we can still follow Brian's solution but there we have to put mcm=Mac(m+1) then series will shorten.

Shyambhu Mukherjee - 5 years, 8 months ago

For the $n \times n$ board,

The total ways to select favorable squares is, $2 \times [ {3 \choose 3} + {4 \choose 3} + {5 \choose 3}+.....+{n-1 \choose 3} + {n \choose 3} + {n-1 \choose 3} ....... {5 \choose 3} + {4 \choose 3}+ {3 \choose 3}]$

$2 \times [ ({3 \choose 3} + {4 \choose 3} + {5 \choose 3}+.....+{n-1 \choose 3} + {n \choose 3}) +({3 \choose 3} ....... {n-2 \choose 3}+ {n-1\choose 3})]$ ... $(1)$

Using the binomial identity,

${r \choose r} + {r+1 \choose r} + {r+2 \choose r} +........ + {m \choose r} = {m+1 \choose r+1}$

$(1)$ , Simplifies to

$2 \times [{n+1 \choose 4} + {n \choose 4} ]$

Total ways to select squares would be, ${n^2 \choose 3}$

Hence, probability = $\dfrac{2 \times [{n+1 \choose 4} + {n \choose 4}]}{{n^2 \choose 3}}$

You could simplify it further if you wish.

A Former Brilliant Member - 5 years, 7 months ago

THANKS! Simplified form: $\dfrac{(n-2)(n-1)}{n(n+1)(n^2-2) }$

Pi Han Goh - 5 years, 7 months ago

@Pi Han Goh – Awesome! We could also do it for choosing x squares instead of 3, but that would take a bit longer to simplify.

A Former Brilliant Member - 5 years, 7 months ago

@A Former Brilliant Member – True True! ^_^

Pi Han Goh - 5 years, 7 months ago

@Pi Han Goh – For an $n\times n$ chessboard and $2\leq r\leq n$ squares, we get $P=2\cdot\frac{2n-r+1}{r+1}\cdot\frac{\binom{n}{r}}{\binom{n^2}{r}}\quad\sim\quad\frac{4n^{1-r}}{r+1}$

as the probability for all $r$ squares to lie on one diagonal.

Carsten Meyer - 1 year, 1 month ago

@Carsten Meyer – Beautiful expression!

Pi Han Goh - 1 year, 1 month ago

You're all set, diagonally!

The answer is 751.

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