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Assume "standard" bingo (75 numbers) with the columns ranging the following inclusive "semi-random" values B: 1 to 15, I: 16 to 30, N: 31 to 45, G: 46 to 60, O: 61 to 75. By semi-random I mean restricted to a small range (15 at a time). There is a free space in the middle of the 5x5 playing board. Numbers (from 1 to 75 inclusive) are randomly drawn one a time without replacement (without any repeats in each game) and with equal probability of being drawn. A win (Bingo) is defined as a completed line segment of 5 adjacent squares made only from the drawn numbers but which may include the free space (and must include it if it is beneficial). A bingo card has 25 of these board squares arranged in a 5x5 matrix.

So my question is if there are 20 players, each with a unique playing card (randomly generated by computer out of I think 552 septillion possible playing cards), what are the chances/probability that 2 or more players will get Bingo on the same drawn number?. For example, someone could win bingo with as few as 4 drawn numbers but likely it would take much more. So I am asking if balls are drawn until at least one person wins, what are the chances that at least 2 people will win at the same time? The game is considered finished / decided when there is at least 1 winner for that game. You can assume that all players are good enough not to make any mistakes (not true in real bingo but assume here).

I am not sure how to set this up mathematically and because there are so many possible bingo cards, computer simulation of all of them is not a good idea. Perhaps what can be done with simulation is to first simulate 20 legitimate bingo cards (out of 552 septillion), and then have the computer draw one random number at a time until we have at least 1 winner. Do this for maybe 1 million trials and count how many have simultaneous multiwinners. For example, after 11 balls drawn there are no winners yet for that game but on the 12 drawn ball, there are 2 or more winners. I would like to know how often the multiwinner situation occurs.

I could probably do the simulation with a fair amount of work but wanted to know if this problem can be done mathematically or if is too difficult to set up.

One concern I see is that if one card has for its first column (the B column), from top to bottom, 1, 4, 7, 10, and 15 and some other card has for column B (also from top to bottom), 15, 10, 7, 4, and 1 in that order. Problem there is even though the cards have different order for the B column numbers, if those 5 numbers are drawn, both players may win at the same time. So my point is it makes a difference if we say (or not) that multiple bingo cards cannot have the same exact 5 numbers in a row, column, or diagonal but just in a different order. That might be an interesting problem in itself to figure out how many fewer "legit" bingo cards there are with that constraint so someone could comment about it but the actual question is about the multiwinner probability. I think the answer to the no permutation bingo card restriction is about 111 quadrillion legit bingo cards.

  • Programming this is a project in one of my computational probability classes. Unrealistic assumptions about 'semi-random' nature of bingo cards are necessary. Free-space a complication. Suspect combinatorial approach may be intricate. Will watch Answers with interest. – BruceET Aug 19 '15 at 16:31
  • Reason I ask is because at work we actually have this scenario or very close to it. The first person to get bingo wins a monetary prize and I suspect there are about 20 players all competing. I calculated that the # of different possible "standard" bingo cards is [(151413*12) ^ 5] * (11 ^ 4) which is about 552 septillion (5.52 * 10^26). – user263122 Aug 19 '15 at 19:29
  • It seems very unlikely to get an answer by explicitly counting, but straightforward programming to get an answer by simulation. – Michael Lugo Aug 19 '15 at 20:30
  • Yes I agree this should be able to be simulated and a good approximation obtained, however, I wonder if bingo card manufacturers just plop randoms numbers on there or they make sure no 2 cards are identical or they even go as far as to make sure no row, column, or diagonal has an exact permutation of the numbers of any other card in the set for that same row, column, or diagonal (taking into consideration the free space as well). – user263122 Aug 19 '15 at 20:32
  • Bruce, why would free space be a complication for simulation? It seems like it would make it slightly easier and faster cuz there is one less thing to check since that block is "always covered". For example, you could number the 25 blocks 1 thru 25 and not have to check block 13. – user263122 Aug 20 '15 at 04:33
  • Anyone gunna simulate this? I am trying to get my account finished signed in so I can hopefully award points for answers and good comments. I will try to simulate it too but not until September cuz I am moving so others have time to try. – user263122 Aug 21 '15 at 15:19

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