5 people participate in a custom game. They are given blank cards, in which they have to fill numbers from 1-25 in a 5x5 table. Each card must contain all the numbers from 1-25 without repetition. The host of the game, then calls out the numbers randomly (between 1-25, without repetition). After each call from the host, the players have to scratch the called out number from their respective cards. The first person to complete a row of scratched out numbers is the winner of the game, and is awarded 100 dollars.

However there is a special clause to the game. If somebody wins(completes a row) within the first 15 calls, they are awarded 300 dollars. The 5 players mutually agree that its more profitable to win the 300 dollars and split it 5 ways, rather than play individually for the 100 dollars.

How should they arrange the numbers on their cards, in such a way that at least one of them would get a scratched out row within the first 15 calls?

  • 1
    $\begingroup$ You use the word "row" specifically. Should we take this to mean that we are interested only in horizontal rows and unlike traditional bingo we are not interested in columns or diagonals? $\endgroup$ – JMoravitz May 24 '16 at 15:01
  • $\begingroup$ That's correct. Only horizontal rows. $\endgroup$ – Kash May 24 '16 at 15:07
  • $\begingroup$ I've never heard of a bingo variant where columns or diagonals mattered. $\endgroup$ – Henrik May 24 '16 at 15:08
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    $\begingroup$ I suspect you mean that they have to fill the numbers from $1$ to $25$ into their table? As the question is currently phrased, a trivial solution is to fill each of the $25$ rows with one of the $25$ numbers. $\endgroup$ – joriki May 25 '16 at 11:42
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    $\begingroup$ There are 25 rows total, since there are 5 cards with 5 rows/card. @joriki's trivial solution was to fill the rows with 5 repetitions of each number, so ex. player 1's card might be "1 1 1 1 1 / 2 2 2 2 2 / 3 3 3 3 3 / 4 4 4 4 4." But, to be clear, every number from 1-25 must be present on every player's card? If so, it would be worth editing this into the OP. This is an important difference. $\endgroup$ – Hungry May 27 '16 at 0:57

This is not an answer, but input for inspiration on the question.

I've looked at bingo puzzles with the same rules, but with smaller cards. That is, I've looked at cards with the sizes $N*N$ where $N = 2$ (with the numbers $1-4$), $N=3$ (with the numbers $1 - 9$) and $N=4$ (with the numbers $1 - 16$). As a guideline for how many calls were made in each case, I had the rule that at least $\frac{15}{25} = 0.6$ of the numbers were called. The result is as follows:

enter image description here

"Ca" is the number of calls divided by the total set of numbers and "TotCo" is the total number of combinations of called numbers this gives (i.e. for $N=4$ the number of called numbers is $10$ out of $16$ which $= 8008$). Each card is outlined by a square and to the right of each row in each card the number of combinations it removes from the total number of combinations, is given. For each card the subtotal of combinations removed is given and at the bottom of the subtotals, is the total number of combinations removed.

The case for $N=2$ and $N=3$ were derived manually, while the case for $N=4$ was done by an exhaustive search algorithm.

Now for the analysis. For $N=2$, only $1$ player is needed to exhaust the possible combinations. For $N=3$, $3$ players are needed to exhaust the possible combinations. For $N=4$, $5$ players are needed to exhaust the possible combinations. If this pattern holds for $N=5$, there is no solution to the posed question.


The scenario in the OP is not possible. The following (informal) proof has nothing to do with my previous "answer" so I thought I'd make a new one.

The proof relies on the following 2 facts:

  1. The minimum number of Rows Needed to ensure that at least one row has a winning combination depends on the number of Calls, and increases monotonically as Calls decreases from 25 to 5.
  2. Already at Calls = 20, Rows Needed = 30 (= 6 cards). As the OP suggested that it was possible with Calls = 15 to have Rows Needed = 25, the OP is disproved.

Re. 1

If Calls = 25, all numbers have been called and just 1 number is needed to assure a winning combination. As the minimum unit we are working with is rows, just 1 row is needed. If Calls = 24, 1 number has not been called. This 1 number could be in our first row, so we need at least 2 rows of unique numbers. If Calls = 23, 2 numbers have not been called and we need at least 3 rows of unique numbers. For Calls = 22, 4 rows are needed. For Calls = 21, 5 rows are needed.

For Calls = 20 we can no longer introduce a new row of unique numbers, as all 25 unique numbers have been used in the first 5 rows (= the first card). However, it is clear that more rows are needed as 5 numbers have not been called and there could be one such number in each of the 5 rows we already have. It should also be clear that for Calls < 20, even more rows are needed. The maximum Rows Needed occurs when Calls = 5, at which point we need all possible combinations of 5 numbers selected from 25 numbers or $\frac{25!}{(20!)(5!)}= 53,130$ rows.

Re. 2

At Calls = 20, there are 5 numbers which have not been called. It is therefore not enough to have 5 rows of unique numbers as there could be one uncalled number in each row. So how many rows are needed to ensure a winning combination?

We can guarantee a winning combination if we can show that at least one row in a card has 2 uncalled numbers. If one row has 2 uncalled numbers, there are only 3 uncalled numbers left, but we have 4 rows left on the card. That means at least one row has only called numbers.

So how many cards are needed to show that at least one row will have 2 uncalled numbers?

Suppose the number 1 is uncalled. If we have enough cards such that we can be sure that 1 has been together with every other number, then we know that at least in one of those cards, 1 will be together with another uncalled number. This is possible with 6 cards (= 30 rows). In fact, it is possible with 30 rows, no matter what the initial number (1 in this case). With 30 rows (= 6 cards) every number can be in a row with every other number and therefore at least one row in a card has a row containing only called numbers. Below is an example:

enter image description here

Edited to add

One needs at least 30 rows because there are 25 numbers. There are therefore 24 other numbers for any given number. A given number can only occur in one row in a card. It can therefore only be together with 4 new numbers in every card. With 4 new numbers per card, one needs at least $\frac{25-1}{4}=6$ cards.


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