Asymmetric Bingo

In a game of “asymmetric bingo,” you and your opponent have two differently sized boards: You play on a 5×5 board, while your opponent has an 8×8 board. The 8×8 board has 64 squares, collectively marked with the numbers 1 through 64 in some random arrangement. Meanwhile, the 5×5 board is populated with 25 numbers chosen and arranged randomly (without replacement) from 1 through 64. There are no “free” squares like there are in traditional bingo.

Here’s how the game works: One at a time, a number from 1 to 64 is drawn randomly, without replacement. If that number appears on your 5×5 board, you place a marker on the corresponding square. Otherwise, your opponent (who is guaranteed to have that number somewhere on their board) places the marker on their corresponding square.

The game ends when one of you has “bingo,” meaning five markers in a row going across, down, or diagonally somewhere on the board. Who is more likely to win this game: you (with the 5×5 board) or your opponent (with the 8×8 board)?

At first, I didn't see that "Otherwise" where I added the emphasis, and said to myself, "Self, this is clearly a losing proposition, the 8x8 board has so many possible winning bingo configurations, $96,$ to be precise, while I and my 5x5 board only have the standard 12." I even offered this version of puzzle in the car while driving to my in-laws house on Sunday morning, and the entire cadre of kids ages 7 through 13 answered that they'd prefer the 8x8 board (though some of their logical reasoning had some gaps). However, then the old adage "Reading comprehension is the silent killer" came back with a vengeance, since then the ``otherwise'' and the blurb at the top saying "you get priority" turned thought process upside-down.

In this case, despite the random arrangement, let's assume that the numbers on the 8x8 board are in lexicographical order, so that the squares along the top row are numbered $1, 2, \dots, 8$ from left to right, then the second row is $9, 10, \dots, 16,$ and so on. Since I am the only player that can mark down any of the numbers on my board, we can effectively remove any of the 25 random numbers on my board from the 8x8 board. For instance, if my board is the numbers say 1 through 25, then we see that my opponent would have a board with 26 through 64 remaining, which would have a total of 22 available bingos in it. On the other hand if my board has the following values $$B = \{1,4,7,10,13,14,16,19,22,24,25,28,29,34,39,40,44,45,48,50,53,54,59,62,64\}, $$ then there is no possible way for your son to win because there are 0 available bingos left on the big board.

Now unlucky for us there are $\binom{64}{25} = 4.01 \times 10^{17}$ different combinations of how to remove 25 squares from a total of 64, so simple enumeration is not going to cut it in this case. However, we can hope that numpy.random.permutation does a relatively good job and hope to simulate our way out of this mess. Using $N=1,000$ samples, first calculate our 5x5 board and then calculate the winning bingos in your 5x5 board and all those remaining bingos that do not contain any of the in the 8x8 board. get the following that the average number of remaining bingos available to your opponent on the 8x8 board is only about 7.3 compared to your guaranteed 12 possible bingos. Given that again we are going to rely on the intuitive sense that the player with more available winning combinations is more likely to win, then you are more likely to win on the 5x5 board than your opponent on the 8x8 board. Going further, though with much less confidence, let's hope that by running a bunch of additional numpy.ranom.permutations and checking whether the 5x5 or 8x8 board wins, we get a win probability that is roughly about $65.56\%$ using these $N=1,000$ samples.

import numpy as np

bingos = [

    [ 1, 2, 3, 4, 5], [ 2, 3, 4, 5, 6], [ 3, 4, 5, 6, 7], [ 4, 5, 6, 7, 8],

    [ 9,10,11,12,13], [10,11,12,13,14], [11,12,13,14,15], [12,13,14,15,16],

    [17,18,19,20,21], [18,19,20,21,22], [19,20,21,22,23], [20,21,22,23,24],

    [25,26,27,28,29], [26,27,28,29,30], [27,28,29,30,31], [28,29,30,31,32],

    [33,34,35,36,37], [34,35,36,37,38], [35,36,37,38,39], [36,37,38,39,40],

    [41,42,43,44,45], [42,43,44,45,46], [43,44,45,46,47], [44,45,46,47,48],

    [49,50,51,52,53], [50,51,52,53,54], [51,52,53,54,55], [52,53,54,55,56],

    [57,58,59,60,61], [58,59,60,61,62], [59,60,61,62,63], [60,61,62,63,64],

    [ 1, 9,17,25,33], [ 2,10,18,26,34], [ 3,11,19,27,35], [ 4,12,20,28,36],

    [ 5,13,21,29,37], [ 6,14,22,30,38], [ 7,15,23,31,39], [ 8,16,24,32,40],

    [ 9,17,25,33,41], [10,18,26,34,42], [11,19,27,35,43], [12,20,28,36,44],

    [13,21,29,37,45], [14,22,30,38,46], [15,23,31,39,47], [16,24,32,40,48],

    [17,25,33,41,49], [18,26,34,42,50], [19,27,35,43,51], [20,28,36,44,52],

    [21,29,37,45,53], [22,30,38,46,54], [23,31,39,47,55], [24,32,40,48,56],

    [25,33,41,49,57], [26,34,42,50,58], [27,35,43,51,59], [28,36,44,52,60],

    [29,37,45,53,61], [30,38,46,54,62], [31,39,47,55,63], [32,40,48,56,64],

    [ 1,10,19,28,37], [ 2,11,20,29,38], [ 3,12,21,30,39], [ 4,13,22,31,40],

    [ 9,18,27,36,45], [10,19,28,37,46], [11,20,29,38,47], [12,21,30,39,48],

    [17,26,35,44,53], [18,27,36,45,54], [19,28,37,46,55], [20,29,38,47,56],

    [25,34,43,52,61], [26,35,44,53,62], [27,36,45,54,63], [28,37,46,55,64],

    [ 5,12,19,26,33], [ 6,13,20,27,34], [ 7,14,21,28,35], [ 8,15,22,29,36],

    [13,20,27,34,41], [14,21,28,35,42], [15,22,29,36,43], [16,23,30,37,44],

    [21,28,35,42,49], [22,29,36,43,50], [23,30,37,44,51], [24,31,38,45,52],

    [29,36,43,50,57], [30,37,44,51,58], [31,38,45,52,59], [32,39,46,53,60],

]

​

def asymmetric_bingo(wins, losses, N=1000):

    hits = np.zeros(N)

    for r in range(N):

        perm = np.random.permutation(np.arange(64))

        end = False

        k = 5

        while (not end):

            test = set(perm[:k])

            for w in wins:

                if w <= test:

                    hits[r] = 1

                    end = True

            for l in losses:

                if l <= test:

                    end = True

            k += 1

    return hits.sum() / N

smpl = 1000

remaining = np.zeros(smpl)

winprob = np.zeros(smpl)

for i in range(smpl):

    board = (np.random.permutation(np.arange(64))[:25]+1)

    b_ = board.reshape(5,5)

    wins = [ set(b_[i,:]) for i in range(5)] + [ set(b_[:,i]) for i in range(5) ] + [{b_[i,i] for i in range(5)}, {b_[i,4-i] for i in range(5)}]

    losses = [set(b) for b in bingos if len(set(board).intersection(b)) == 0]

    remaining[i] = len(losses)

    winprob[i] = asymmetric_bingo(wins, losses)

remaining.mean(), winprob.mean()