Maths Olympiad Prep

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Problem 988

AMC 12 late, AIME early
Combinatorics Difficulty 4.9 Find the answer HMMT November · United States · 2021

Two distinct squares on a 4×44 \times 4 chessboard are chosen, with each pair of squares equally likely to be chosen. A knight is placed on one of the squares. The expected value of the minimum number of moves it takes for the knight to reach the other square can be written as mn\frac{m}{n}, where m,nm, n are positive integers and gcd(m,n)=1\operatorname{gcd}(m, n)=1. Find 100m+n100 m+n.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

Solution:

We can do casework based on the position of the knight: corner, edge, or center. In each case, we can quickly compute all 15 distances by writing a 1 down in all squares reachable from the original square, then writing a 2 down in all blank squares reachable from a square with a 1, writing a 3 down in all blank squares reachable from a square with a 2, and so on. The resulting tables are below:

0325
3412
2143
5232
3032
2321
1214
2323
4321
3032
2321
1214

The expectation can be computed by weighing the sum of the distances in each of these tables by the number of squares of that type:

11615(4(21+52+43+24+25)+8(31+62+53+14)+4(41+52+43+24))=1240(168+272+136)=125 \begin{aligned} \frac{1}{16 \cdot 15} & (4(2 \cdot 1+5 \cdot 2+4 \cdot 3+2 \cdot 4+2 \cdot 5)+8(3 \cdot 1+6 \cdot 2+5 \cdot 3+1 \cdot 4)+4(4 \cdot 1+5 \cdot 2+4 \cdot 3+2 \cdot 4)) \\ & =\frac{1}{240}(168+272+136) \\ & =\frac{12}{5} \end{aligned}

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.