Bingo math

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  • Thread starter thetexan
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  • #1
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Main Question or Discussion Point

I am collecting bottle caps off 2 liter and 20 oz soda bottles because the folks at the nursing home uses them to cover bingo numbers instead of sliding the little window on the bingo card because that's difficult for them to manipulate.

Which got me to thinking. Considering the rules of bingo...

1. played on a grid of 5x5 numbers
2. must cover 5 in a row to win
3. the center position is a "FREE" number

what is the maximum number of bottle caps one would need to ensure there are enough bottle caps to play a single card game of bingo?

I say 15. In other words I could cover 15 numbers and still not have bingo. But as soon as I get the 16th (remembering the center position is free) number covered I must win.

Am I right?

tex
 

Answers and Replies

  • #2
34,058
9,930
You can arrange 5 "still empty" spots in such a way that every row and every column has exactly one of those spots. That means you need 25-5-1=19 caps. Or 20 if you want to mark the bingo.
 
  • #3
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I can fill the bottom left 4x4 with 16 caps including covering the center free spot. This means that the next number will create the first 5 streak.

Doesn't that mean that 17 is the minimum needed. Is there a more correct number?
 
  • #4
jim mcnamara
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@mfb The rules of bingo are 5 in a row with row usually defined as:
diagonally
vertically
horizontally

At least in church sponsored games. Casinos have a lot of variations on these rules.
 
  • #5
34,058
9,930
I can fill the bottom left 4x4 with 16 caps including covering the center free spot. This means that the next number will create the first 5 streak.

Doesn't that mean that 17 is the minimum needed. Is there a more correct number?
There are other arrangements where you need more. Example, O=empty, X=cap, F=free:

Code:
OXXXX
XXOXX
XXFXO
XOXXX
XXXOX
19 "X", no bingo (also not on the diagonals).
 
  • #6
238
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ok. So far 19 is the maximum minimum. What other possible configuration would require more?
 
  • #7
34,058
9,930
There is no possible configuration that needs more (but many others that need 19 as well). In order to avoid a bingo, every row needs an empty spot, so we can fill at most 25-5=20 spots, one of them is the free spot in the middle. And, as demonstrated, that maximum is possible.
 

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