poorboy
poorboy
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August 19th, 2022 at 6:13:53 AM permalink
Hi,

I am definitely not a math person, but had a question about Bingo.

I know there is a lot of information available about getting a full card on a bingo card, in whatever number of ordinals, but my question is this:

Is there an simple way, using those odds, to factor in if the card bein played on is a Double Action type card? Two numbers in each square, either of which is need to 'mark' that spot. Or is there a whole different calculation?

Thanks
ThatDonGuy
ThatDonGuy
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August 19th, 2022 at 1:45:37 PM permalink
Different calculation, I think.

For the normal game, think of the game as having 24 red balls (your 24 numbers) and 51 white balls (the remaining numbers); it's pretty straightforward to determine the probability of drawing N balls and having all 24 red balls in the draw.

For the Double Action game, now there are 2 balls of each of 24 colors, plus 27 white balls, and you need to draw at least one ball of each of the 24 colors. This involves counting (a) exactly one of each color, (b) both of one color and one of each of the 11 others, (c) both of two colors and one of each of the other 10, (d) both of three colors and one of each of the other 9, and so on through (m) both balls of all 12 colors.
DJTeddyBear
DJTeddyBear
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August 19th, 2022 at 5:29:24 PM permalink
Quote: poorboy

Two numbers in each square, either of which is need to 'mark' that spot. Or is there a whole different calculation?link to original post

That’s a whole different calculation. After all, two numbers in one spot could be on two different spots on a different card.

As if bingo math wasn’t hard enough….
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ ————————————————————————————————————— Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
avianrandy
avianrandy
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August 19th, 2022 at 6:04:09 PM permalink
I have played this before.their are 2 numbers in each square. Day one spot had16 and 30 init. If either 16 or 30 is called,that spot is considered covered.just wanted to clear that up if their is any confusion
Vegasrider
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August 19th, 2022 at 8:13:47 PM permalink
Quote: avianrandy

I have played this before.their are 2 numbers in each square. Day one spot had16 and 30 init. If either 16 or 30 is called,that spot is considered covered.just wanted to clear that up if their is any confusion
link to original post



I won’t play unless the progressive is at 38 numbers
ThatDonGuy
ThatDonGuy
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poorboy
August 20th, 2022 at 8:16:21 AM permalink
Quote: ThatDonGuy

Different calculation, I think.

For the normal game, think of the game as having 24 red balls (your 24 numbers) and 51 white balls (the remaining numbers); it's pretty straightforward to determine the probability of drawing N balls and having all 24 red balls in the draw.

For the Double Action game, now there are 2 balls of each of 24 colors, plus 27 white balls, and you need to draw at least one ball of each of the 24 colors. This involves counting (a) exactly one of each color, (b) both of one color and one of each of the 11 others, (c) both of two colors and one of each of the other 10, (d) both of three colors and one of each of the other 9, and so on through (m) both balls of all 12 colors.
link to original post


Actually, it's 25 things to count, not 13. If there are N balls drawn, there are C(75,N) ways to draw the balls, of which:
(1) 2^24 x C(27,N-24) have one number from each square
("How did you calculate that?" There are 2 possibilities for the number in square 1, 2 for the number in square 2, ..., 2 for the number in square 24, and since these are 24 of the numbers drawn, there are N-24 numbers remaining to be drawn from the 27 not in any of the squares.)
(2) C(24,1) x 2^23 x C(27,N-25) have both numbers from one square and one from each of the other 23
(3) C(24,2) x 2^22 x C(27,N-26) have both numbers from two squares and one from each of the other 22
(4) C(24,3) x 2^21 x C(27,N-27) have both numbers from 3 squares and one from each of the other 21
...
(24) C(24,23) x 2^1 x C(27,N-47) have both numbers from 23 squares and one from the other one
(25) C(24,24) x C(27,N-48) have both numbers from all 24 squares
Note that, in some cases, the value will be 0 because it is impossible; for example, it is impossible to draw both numbers from all 24 squares if only 47 numbers are drawn.

Here is the probability of filling the card for anywhere from 30 to 74 numbers drawn (note that each probability is for filling the card in that many numbers or fewer):

301 / 13,365,664
311 / 3,998,505
321 / 1,327,324
331 / 481,748
341 / 188,989
351 / 79,404
361 / 35,465
371 / 16,735
381 / 8299
391 / 4307
401 / 2330
411 / 1310
421 / 763
431 / 459
441 / 285
451 / 182
461 / 119
471 / 80.3
481 / 55.3
491 / 39
501 / 28.1
511 / 20.65
521 / 15.48
531 / 11.82
541 / 9.181
551 / 7.255
561 / 5.827
5721.042 %
5825.417 %
5930.265 %
6035.544 %
6141.193 %
6247.131 %
6353.261 %
6459.472 %
6565.641 %
6671.642 %
6777.345 %
6882.626 %
6987.367 %
7091.465 %
7194.834 %
7297.405 %
7399.135 %
74100 %

poorboy
poorboy
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August 21st, 2022 at 8:01:16 AM permalink
Thank you everyone for replying, I appreciate you taking the time.

The table really helps as well!
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