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Thread: Math Question: Bingo Blackout odds.

  1. #1 Math Question: Bingo Blackout odds. 
    Forum Isotope (In)Sanity's Avatar
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    Bingo, for those that don't know the game the cards look something like my example below. The X is a free space.

    Code:
     B    I   N   G   O  
     1   25  34  48  63
    11   24  33  57  65
     7   27  X   60  72
    15   30  35  52  73
    13   16  40  53  70
    Bingo is won normally by getting 5 in a row in any direction that is straight _ / | \

    There is also a special way to win known as Blackout (aka Coverall). To win this you need to cover all the spaces after so many balls (numbers) have been picked.

    The Balls that are picked are picked one at a time and range is 1-75.

    The B column can contain the first 15 numbers (1-15) the I the second 15 numbers (16-31) etc.

    Here's the question for you, what are the odd's of getting a blackout where all the numbers on your card have been picked in 48 picks?

    Would also love to know the formulla for this.


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  3. #2  
    Forum Ph.D.
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    I'm sure you would love to see the formula, lol.

    Is the question with a random scoreboard, or with the scoreboard shown in your post? Although it shouldn't matter, I have a funny way of calculating chances, and it often leaves me counting. Sadly, in the Math courses I did, chance was barely covered. The only thing I remember is a boxplot. Oh, and a permutation .

    Oh, and as my math teacher always told me to fill in something, I'd say it's 0.42%, as I like the number 42

    Mr U


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  4. #3  
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    Quote Originally Posted by HomoUniversalis
    I'm sure you would love to see the formula, lol.

    Is the question with a random scoreboard, or with the scoreboard shown in your post? Although it shouldn't matter, I have a funny way of calculating chances, and it often leaves me counting. Sadly, in the Math courses I did, chance was barely covered. The only thing I remember is a boxplot. Oh, and a permutation .

    Oh, and as my math teacher always told me to fill in something, I'd say it's 0.42%, as I like the number 42

    Mr U
    LOL, Well the odds of getting the first number would be 32% 24/75 the odds of getting the second ball assuming you got the first ball would be 31.08% 23/74. This is where it gets confusing, im trying to figure out the odds in 48 picks. Those odds should be the same with any Bingo card. I know it will be far less then 1%.
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  5. #4  
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    I think it would be something like !24/!75 . I should take a statistics class one of these days, maybe then i'd learn. You could also probably write that in a sum of a series

    Code:
    24
    ----
    \  (24-i)/(75-i)
    /  
    ----
    i=0
    I don't know how exactly that would work, but those are my two guess. meh....
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  6. #5  
    Forum Isotope (In)Sanity's Avatar
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    Well I found this formula

    combin(74-24,54-24)/combin(54,54) = 0.000054

    54 being the number of balls called, however I have no clue what combin does. Anyone have a clue?
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  7. #6  
    Forum Isotope (In)Sanity's Avatar
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    Well I found out that combin is in Excel, so I decided to plug a formulla in and run a range of 24-75 here is what I got

    Code:
    25,778,699,578,994,600,000	24
    1,031,147,983,159,780,000	25
    79,319,075,627,675,600	26
    8,813,230,625,297,290	27
    1,259,032,946,471,040	28
    217,074,645,943,283	29
    43,414,929,188,657	30
    9,803,371,107,116	31
    2,450,842,776,779	32
    668,411,666,394	33
    196,591,666,587	34
    61,785,952,356	35
    20,595,317,452	36
    7,236,192,618	37
    2,665,965,701	38
    1,025,371,424	39
    410,148,569	40
    170,061,602	41
    72,883,544	42
    32,204,357	43
    14,638,344	44
    6,831,227	45
    3,267,109	46
    1,598,798	47
    799,399	48
    407,857	49
    212,085	50
    112,281	51
    60,459	52
    33,081	53
    18,378	54
    10,359	55
    5,919	56
    3,427	57
    2,009	58
    1,192	59
    715	60
    434	61
    266	62
    165	63
    103	64
    65	65
    41	66
    26	67
    17	68
    11	69
    7	70
    5	71
    3	72
    2	73
    1	74
    1	75
    This is based on

    1/(combin(24,x)*combin(51,y-x)/combin(75,y))

    x being the number of places to fill in the card, y being the number of calls.

    See the following URL.

    http://wizardofodds.com/games/bingo/probbingo.html
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  8. #7  
    Forum Freshman Geodesic's Avatar
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    Combin is short for combination, it's the number of choices of n items from a total of m, regardless of order. Sometimes written <sup>m</sup>C<sub>n</sub>, it's given by m!/(m-n)!n!
    If you were interested in the possible permutations, ie. order is important, you would have <sup>m</sup>P<sub>n</sub> = m!/(m-n)! :wink:
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  9. #8  
    Forum Ph.D.
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    cool
    Stumble on through life.
    Feel free to correct any false information, which unknown to me, may be included in my posts. (also - let this be a disclaimer)
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  10. #9 Re: Math Question: Bingo Blackout odds. 
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    Quote Originally Posted by (In)Sanity
    Bingo, for those that don't know the game the cards look something like my example below. The X is a free space.

    Code:
     B    I   N   G   O  
     1   25  34  48  63
    11   24  33  57  65
     7   27  X   60  72
    15   30  35  52  73
    13   16  40  53  70
    Bingo is won normally by getting 5 in a row in any direction that is straight _ / | \

    There is also a special way to win known as Blackout (aka Coverall). To win this you need to cover all the spaces after so many balls (numbers) have been picked.

    The Balls that are picked are picked one at a time and range is 1-75.

    The B column can contain the first 15 numbers (1-15) the I the second 15 numbers (16-31) etc.

    Here's the question for you, what are the odd's of getting a blackout where all the numbers on your card have been picked in 48 picks?

    Would also love to know the formulla for this.
    The matrix can simulate by matlab or get an answer.I am not sure how exactly that would work.
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  11. #10  
    Forum Radioactive Isotope MagiMaster's Avatar
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    Not to be rude, but what was the point in ressurrecting a three-year-old thread for a non-answer?
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