# Thread: Math Question: Bingo Blackout odds.

1. 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.

2.

3. 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

4. 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%.

5. 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....

6. 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?

7. 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

8. 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:

9. cool

10. 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.

11. Not to be rude, but what was the point in ressurrecting a three-year-old thread for a non-answer?

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