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For more details see A387260 and A387261.

The total number of different strategies is given in A387260. See there for more information.

The game "Risk or Safety" consists of two competing players. A player whose turn it is tosses a coin. If it is heads they earn a point, and they can choose to go for safety, put the point aside in a save box, and give the turn to their opponent. Or they take the risk to continue and toss the coin again. If it is heads again, the number of "open" points increases by one, and they can choose again to continue or stop, converting the "open" points to "saved" points. Whenever it is tails, all the open points are lost, and it is the turn of the other player. Who collects n points first wins.

If a game situation is described by a triple (a, b, c) with a = open points, b = saved points of the player whose turn it is, and c = saved points of the opponent, then the number of possible game situations is n^2(n + 1)/2 = A002411(n). One possibility to calculate the optimal strategy is to examine each triple with a > 0, derive the possible follow-on triples for each possible strategy at this situation, and determine for each stop/continue combination the one with the largest winning probability at this point in the decision tree of all strategies. It is assumed that one optimal strategy exists that both players are playing. For all possible strategies, a system of A002411(n) linear equations needs to be solved. Another approach is to simulate the game iteratively, selecting the solution with the highest winning probability after k turns, and continuing this process until no further improvements are made. Both methods have given the same results, although the matrix approach breaks down for n >= 6, since the inversion of hundreds of billions of 126 X 126 matrices is just not feasible.

The optimal solution for n = 3 is to stop at (1, 0, 0), else always continue. The average game duration is 6 (n = 2), 120/11 (n = 3), 258/17 (n = 4), and 5532/275 (n = 5).

A cat is randomly hiding in one of 8 boxes, and a single player tries to catch the cat. In each step, the player opens a box. If the cat is found, the game ends; if not, the cat will move randomly either to the box on the left or to the box on the right. If the cat is in box 1 or 8, it has a 50% probability of escaping. By using the optimal strategy, the cat's escape rate is minimized to 0.223305988. The average duration of the game is 5.35 steps. Note that, since the problem is symmetric, a mirrored strategy also exists.

A cat is randomly hiding in one of 8 boxes and a single player tries to catch the cat. In each step, first the player opens a box, if the cat is found, the game is finished, if not the cat will move randomly one box to the left or one box to the right. If the cat is in box 1 it can only move to box 2 in the next step. A safe strategy to find the cat in no more than 12 steps is to open the boxes in this order: 2,3,4,5,6,7,7,6,5,4,3,2. On average this strategy takes 25963/4096 (6.339) steps to catch the cat. However, applying the strategy to open box a(n) at step n reduces the average duration of the game to 4.749593 which is a factor of 1.335 faster. Note, since the problem is symmetric also a mirrored strategy exists.

Conjecture: a(11)..a(16) are 5, 6, 8, 9, 11, 12 with at least 99.7% confidence (based on numerical simulation discussed below).

Let L be the ratio of vector length to side length of a hypercube. If L <= 1 the probability P that a vector with random start and random orientation falls completely inside one hypercube can be calculated with one integral, e.g. in case of four dimensions: P_4(L) = (8/Pi^2)*Integral_{0..Pi/2} Integral_{0..Pi/2} Integral_{0..Pi/2} (1 - L * cos(x) * cos(y) * cos(z)) * (1 - L * sin(x) * cos(y) * cos(z)) * (1 - L * sin(y) * cos(z)) * (1 - L * sin(z)) * cos(y) * cos^2(z) dx dy dz = 1 - 16*L/(3*Pi) + 3*L^2/Pi - 32*L^3/(15*Pi^2) + L^4/(6*Pi^2). If L > 1, the probability needs to be integrated over multiple parts which leads to elliptical integrals in the case of three dimensions. Therefore for L > 1, numerical simulations were applied (see Python code below).

Teams play each other twice for a total of M = n*(n-1) matches.

A victory is awarded 3 points, a draw 1 point and a defeat 0 points.

The total number of possible match outcomes is 3^M = A053764(n) and a(n) is how many of them result in all teams finishing with the same points score.

If all matches were randomly assigned a result, the probability that all teams would end up with the same number of points is a(n)/A053764(n), which in a typical league of 18 or 20 teams is very small.

A007080(n) is the number of ways if there are no draws.

Fig. 1.1 in the paper "An aperiodic monotile" by Smith et al. (2023) shows an example of tiling the plane with "hat" monotiles. A "hat" monotile covers two thirds of a hexagon and one third of two neighboring hexagons, respectively. A hexagon is assigned a tile if the tile covers two thirds of the hexagon. Since the surface of a "hat" tile is 4/3 the surface of a hexagon, every forth hexagon on average is covered by parts of three different tiles. In this case we assign a zero to this hexagon.

We assign the number "1" to a tile that has the orientation of the dark grey tile in Fig. 1.1 of the paper by Smith. Tiles can only be rotated by multiples of 60 deg and for each counterclockwise rotation of the tile by 60 deg, we increase the count by 1 such that unreflected tiles are assigned the numbers from 1 to 6. Reflected tiles are assigned accordingly the numbers from 7 to 12.

Without loss of generality we place a tile "1" in the central hexagon like in the quoted Fig 1.1 (any other tiling can be transformed into this tiling by rotation or reflection). As next hexagon to be filled we choose the right upper neighbor hexagon. The tiling continues through the surrounding hexagons in counterclockwise direction.

A detailed description of the counting of the tiling options is given in the pdf in the links. a(n) is the number of possible assignments of the first n hexagons such that the plane can be filled at least up to heaxagon 919 (end of ring 17). It would be desirable to demand that the tiling can be continued to infinity, but this is much more difficult to prove.

An open question is: What is the asymptotic behavior of a(n) when n tends to infinity?

a(n)/8^n is the probability that the knight falls off the chess board at step n. The average number of steps it takes the knight to fall off the board is Sum_{n>0} n*a(n)/8^n = A376736(8)/A376737(8) = 101338/51901 or approximately 1.953 steps.

Because of the mirror symmetry of the problem to the board diagonal, all terms are even.

a(n)/4^n is the probability that the 1-step rook falls off the chess board at step n. The average number of steps it takes this piece to fall off the board is Sum_{n>0} n*a(n)/4^n = A376606(8)/A376607(8) = 4374/901 or approximately 4.855 steps.

Because of the mirror symmetry of the problem to the board diagonal, all terms are even.