(This question is inspired by: If I have an event where the outcomes aren't uniformly distributed, how would I make the "fairest" bingo card out of the events? )

You have a $5 \times 5$ bingo card filled with $25$ distinct numbers, one per square. There is also a pot containing each number once and you draw them out one by one without replacement.

A line is any of the $5$ rows, $5$ columns, or $2$ main diagonals. A line is completed when its $5$ numbers have been drawn. A line wins if it is the first line to be completed.

Question: Do the $12$ lines have equal chance of winning? If not, which lines have higher chance of winning?

Why I ask: Define $T_l$ to be the time to completion for line $l$. It is obvious that all $T_l$ are equi-distributed, and thus all $E[T_l]$ are equal. However, because the lines overlap, the different $T_l$'s are dependent, and it is not clear to me that they have equal chances of being first.

In particular, imagine that the $5$-subsets are not arranged in rows, columns and diagonals, but are clustered in some non-uniform way. Then a subset that shares a lot of elements with other subsets might have a lower chance to win than a subset that does not share a lot of elements with other subsets. (I can provide a simple example if there is interest.)

On the bingo card, the $5$-subsets are pretty uniform but not exactly uniform, due to the diagonals. So my suspicion is that the line win chances are almost equal but not exactly equal. And I am curious as to which lines have higher chances.

I imagine (but would be happy to be proven wrong) that calculating the exact win prob for a line might be difficult/tedious, so qualitative arguments based on e.g. symmetry are also welcome.

Clarifications: A drawn number can complete multiple lines, so that needs some special handling. However what I am interested in is the question of equality, so I will accept any reasonable way to handle such "shared" wins, i.e. if $N>1$ lines are completed at the same draw (and no line has been completed before this draw), then you can treat this as if...

  • they all win (in which case the sum of the $12$ win probabilities exceed $1$, but that doesn't matter since I am interested in which are higher/lower), or,

  • the whole experiment is repeated from the beginning (i.e. we condition on such shared wins not happening), or,

  • you flip an $N$-sided die to determine the winner (i.e. this effectively counts as $1/N$ win for each involved line), etc.

  • 1
    $\begingroup$ After this, you can do BINGO cards with a "free" space in the middle. $\endgroup$ – GEdgar Mar 8 at 20:37
  • $\begingroup$ I don't see why you would think any particular line would have a higher chance of winning if there is an equal probability of each square being picked every time. A line requires 5 numbers being picked and the probability of each 5 number combinations are exactly the same. $\endgroup$ – Matthew Liu Mar 8 at 20:53
  • $\begingroup$ @MatthewLiu Because when one line gets a number, it also helps other lines. The diagonal are at a disadvantage. Look at my solution for $3\times3$ bingo. $\endgroup$ – saulspatz Mar 8 at 21:24
  • $\begingroup$ @MatthewLiu what you said is why all the $T_l$ are identically-distributed, but they are not independent. A line $L$ wins if $T_L < T_l \forall l \neq L$. $\endgroup$ – antkam Mar 8 at 22:15
  • $\begingroup$ Ok you guys are correct. The diagonal lines are at a strict disadvantage. $\endgroup$ – Matthew Liu Mar 11 at 17:14

I must admit, when I read the title of the question, I thought "Of course they all have an equal chance of winning. What a silly question." But then, I read the body, and was delighted to see that you had a valid point, and I was the silly one. I don't know how to solve for $5\times5$ bingo in any reasonable time, but for $3\times3$ bingo we have only $9!$ permutations of the numbers to consider. I wrote a python script to exhaustively test this.

from itertools import permutations
from fractions import Fraction

score = 9*[0]
card = 8*[0]

for p in permutations(range(9)):
    card[0] = {0,1,2}
    card[1] = {3,4,5}
    card[2] = {6,7,8}
    card[3] = {0,3,6}
    card[4] = {1,4,7}
    card[5] = {2,5,8}
    card[6] = {0,4,8}
    card[7] = {2,4,6}
    for number in p:
        for line in card:
            line -= {number}
        winners = len([line for line in card if not line])
        if winners:
            for idx, line in enumerate(card):
                if not line: score[idx] += Fraction(1, winners)

for idx in range(8):
    print(idx, score[idx])

This produced the output

0 47520
1 45600
2 47520
3 47520
4 45600
5 47520
6 40800
7 40800    

This means that the diagonals are worst, and the lines along the edges of the card are best, with the horizontal and vertical lines through the center in the middle.

I should mention that in the case of ties, I awarded an equal fraction of the game to each winner, so that the sum of the scores does work out to $9!.$

  • 1
    $\begingroup$ What does it produce if you give a full win to each line if 2/3/4 lines win at once? $\endgroup$ – Banbadle Mar 8 at 21:58
  • $\begingroup$ awesome! i had thought the diagonals are worst but all the rows and columns are equal, but you showed that the row/col passing through the center are penalized. interesting... and now i wonder if the 1st & 2nd rows are different in the 5x5. ooh, yeah, i would like to see the idea of @Banbadle tested too. would the multiple full wins cancel the effect of sharing? $\endgroup$ – antkam Mar 8 at 22:00
  • 1
    $\begingroup$ @Banbadle I just modified saulspatz's code to try other scoring modes. If a full win is given to each simultaneously winning line, the win counts are 54000 > 53290 > 46656 for the edges, center row/col, diagonals respectively. If we condition on no shared wins (i.e. those permutations award no wins to anybody), then the win counts are 41040 > 38880 > 35424, again for edge, center row/col, diagonals. So the order is the same in all 3 win-counting models. $\endgroup$ – antkam Mar 8 at 22:44

I simulated a standard bingo game, one million games with ten random cards for each game. I recorded the number of times each line triggered bingo, including bingo on multiple cards and multiple lines winning simultaneously on the same card.

The results:

  • Horizontal through center wins $16.2\%$ of games
  • Vertical through center wins $16.1\%$ of games
  • Diagonals each win $15.9\%$ of games
  • Other horizontal lines each win $5.8\%$ of games
  • Other vertical lines each win $5.6\%$ of games

Note that this totals over $100\%$, accounting for multiple wins in a single game (there were nearly $1.1$ million wins in one million games)

After ensuring that these results are indeed accurate and giving some thought to why the diagonals win less than the central row and column, it is clear that this is due to intersecting lines. Each of the squares on the diagonals, when drawn, has the potential to trigger a win on either of two intersecting lines, while each square in the central row or column can only trigger a win on a single intersecting line.

I have also run the simulation without the free center square:

  • Rows win $9.42\%$
  • Diagonals win $9.17\%$
  • Columns win $9.05\%$
  • $\begingroup$ when you say "standard bingo game" do you mean the center square is free (i.e. already marked off)? if so that must be why lines through center win much more coz they only require 4 cells. $\endgroup$ – antkam Mar 8 at 22:04
  • $\begingroup$ @antkam Yes, center square is free in my simulation. $\endgroup$ – Daniel Mathias Mar 8 at 22:05
  • $\begingroup$ Even in the free-center case it is open (at least, in my mind) whether the diagonals are at a disadvantage. If you found that diagonals win 10% vs the center row/col winning 20% each, that would be quite conclusive. But at 15.9.% vs 16.1% I am not sure this is a real effect or just within the error bar. $\endgroup$ – antkam Mar 8 at 22:18
  • 1
    $\begingroup$ @antkam A run of ten million games: $16.18\%$, $16.09\%$, $15.91\%$, $15.90\%$ for center row, center column, downward diagonal, upward diagonal. Of course, this only proves that my simulation produces consistent results. It is entirely possible that the inequality is a result of some unintended bias, though I have found no potential cause for bias. $\endgroup$ – Daniel Mathias Mar 8 at 23:31
  • 1
    $\begingroup$ @DanielMathias Thanks. I knew very little about bingo before reading this. Now I have upgraded my knowledge from "very little" to just "little". $\endgroup$ – badjohn Mar 9 at 14:54

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