# Nice Designs

By Stephen Meskin

We are looking for nice designs that are 4 units square. We say that a 4 × 4 square has a nice design if each of its rows and columns (which of course are of length 4) have nice designs. We say that a 1 × 4 row is nicely designed if it is one of the following four:

And we say that a 4 × 1 column is nicely designed if it is the transpose of a nicely designed row.

Problem 1

Find all nicely designed 4 × 4 squares.

For example, the following is nicely designed:

OXXO
OXXO
XOOX
XOOX

Problem 2

Show that if a nicely designed 4 × 4 square is

2. the 0-degree line or
3. the 45-degree line or
4. the 90-degree line or
5. the 135-degree line or
6. rotated 90-degrees or
7. dualized (Xs changed to Os and vice versa),

then the result is again nicely designed.

We say that two designs are in the same orbit if there is a sequence of Problem 2 actions taking one to the other. The orbits form a partition of nicely designed 4 × 4 squares.

Problem 3

Exhibit the orbits of nicely designed 4×4 squares.

Solutions may be emailed to puzzles@actuary.org.
In order to make the solver list, your solutions must be received by October 1, 2024.

Solutions to Previous Puzzles:
It Never Rains in Southern California

What is the probability the server wins a game? The probability a server wins four points in a row is (0.6)4=0.1296. The probability the server wins in five points is (0.6)3*0.4*4*0.6=0.2074, as the server must win three of the first four points and then win the fifth point. The probability the server wins in six points is (0.6)3*(0.4)2*10*0.6=0.2074, as the server must win three of the first five points and then the sixth point. For a game to get to deuce, the probability is (0.6)3*(0.4)3*20=0.2765. Given a game hits deuce, the server wins the game with probability 0.36/(0.36+0.16) = 0.6923. Hence the server wins a deuce game with probability 0.2765*0.6923=0.1914. Summing these four numbers gives the server a 73.57% probability of winning the game.

What is the probability of a set going to a tiebreak? For a set to go to a tiebreak, the first 10 games must be split five games each. There are six different ways for a set to reach five games aside: There are no breaks of serve, or each player loses serve once, twice, three, four, or five times each. We need to compute the probabilities of these events separately using similar techniques of the above, knowing the probability of a server winning a game = 73.57%. Doing this leads to a 28.40% chance of us reaching five games a piece. Given that we are at 5-all, the only way to reach a tiebreaker is for the next two games to both hold serve or both break serve. The probability of this is (07357 * 0.7357) + (0.2643*0.2643) = 61.11%. Multiplying 0.6111 * 0.2840 = 17.36% chance of getting to a tiebreak.

What is the mean number of points in a tiebreak? Here we have to use our binomial theorem again to figure out the probability of the tiebreaker ending in 7 points (7-0), 8 points (7-1), and so on. However, the probabilities are slightly different of the first person serving to win the first 7 point compared to the second player, as the first player serves 3 of the first serves, while the other player serves 4 times. Doing this math leads to a mean number of points to right around 11.84 points.

What is the probability of a player winning the set but winning a minority number of points? There are two ways to compute this. One, we can come up with probabilities of winning each non-tiebreak game by 2, 3, or 4 points, and do something similar for the tiebreaker. Next, we can see that the only way this phenomenon can occur is if the score of the set is 6-4, 7-5, or 7-6 (at 6-3 you can only tie in the number of points won). We can use this information to cut down on a lot of the grunt-work and use the binomial theorem many times over to get our answer. The second method is to perform a simulation and see how often the winner of the set wins the most points.  Both methods leads us to see that right around 2.1% of the time the person who wins the set won the fewest number of points.

Solvers

Bob Conger, Bernie Erickson, Bill Feldman, Clive Keating, Rui Gio, David Promislow, Al Spooner, and Daniel Wade.

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