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Bingo game at the nearby nursing home

  1. Jan 21, 2006 #1
    I run a bingo game at the nearby nursing home. The other day a lady covered all of the 25 [5x5] spaces on her card with 17 numbers yet to call. If there are 75 numbers to call at the beginning of the game, and the numbers are distributed in columns randomly among 1-15 under "B," 16-30 under "I," 31-45 under "N," 46-60 under "G," and 61-75 under "O" on a given card (with a "free" space in the central square), what are the odds that she could cover her card with at least 17 numbers to call?
     
  2. jcsd
  3. Feb 15, 2006 #2
    Give this another try, mathematicians; approximate if necessary. The lady at the nursing home for whom these odds occurred (and I, too) are quite curious as to what they might actually be. My guess is 1:5,000 to 1:10,000 games have such a coincidence.
     
  4. Feb 16, 2006 #3

    honestrosewater

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    Meh, I'll stick out my neck to try to get something going. Maybe it would be easiest to start with all the permutations of your set of 75 numbers since you presumably want the 58th number to be one of the 24 on your card. A game is just a permutation of those 75 numbers. There are 75! games. I can see at least two questions that you might be asking (which, for all I khow, might be the same question). For a given game, how many cards would be filled on the 58th turn? For a given card, how many games would fill it on the 58th turn?

    Can the same number appear more than once on a card?
     
    Last edited: Feb 16, 2006
  5. Feb 16, 2006 #4

    Tide

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    I'd like to know what the probability is of covering all 25 spaces WITHOUT shouting BINGO??
     
  6. Feb 16, 2006 #5
    honestrosewater,

    A given number can appear only once on a card (in the appropriate column, too), as your factorial seems to indicate.

    My question is for a given card.

    Thanks.
     
  7. Feb 16, 2006 #6

    JasonRox

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    Well, if you get a line, and there is still 50 calls left, you don't shout BINGO!

    17 calls left is still quite a bit.
     
  8. Feb 16, 2006 #7

    Tide

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    I thought the object was just to fill a line - shows how much I know about BINGO! :)
     
  9. Feb 16, 2006 #8

    JasonRox

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    If you get a line early on during the game, you wait.

    You have to be quick though. So that when someone calls BINGO!, you call it quickly afterwards.

    That way it was a tie, and you also took the chance to try and win more.

    Note: I don't play BINGO!
     
  10. Feb 16, 2006 #9

    shmoe

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    There's usually a free space, so 24 numbers on the card and 51 not on it right? With 58 numbers called you'd have C(51,34) ways of having a black out, the total number of ways the 34 of the 51 numbers off the card can be called. There are C(75,58) ways to choose 58 numbers, so the probability of a blackout after 58 numbers would be C(51,34)/C(75,58)=.0004977832...

    This is including the event that you had a blackout before call 58 as well.

    (here C(n,r)=n!/(r!(n-r)!) is the usual "n choose r")
     
  11. Feb 16, 2006 #10
    Thanks for wresting an answer, shmoe! I do believe it's approximate, though. Shouldn't each column (B, I, G, O) be treated individually as 5 chances out of 15, excepting column N, 4 chances out of 15?
     
  12. Feb 16, 2006 #11

    shmoe

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    No, it's exact. With respect to a blackout, your bingo card is just a subset of 24 numbers from the 75, the columns don't matter (nor do the positions within the columns). Of course not all subsets of these 24 numbers can form a valid bingo card due to the column restrictions, but that's not relevant here.

    If you'd like a second opinion, see:

    http://wizardofodds.com/bingo
     
  13. Feb 16, 2006 #12
    That's quite a second opinion! Thanks times two, shmoe.
     
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