April 19th, 2018 at 3:06:39 PM
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I want to calculate the probability of hitting a certain bingo, such as an "L" or "4 corners" and it's just me playing. What is the math to calculate? I was figuring that you use the 15 numbers in the "B" column and the single numbers in the others when making a formula for the "L" and single numbers for the 4 corners.

April 19th, 2018 at 6:01:30 PM
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The first question is, since it's just you playing, what is the maximum number of balls that can be drawn? Obviously, every conceivable result can be obtained if all 75 balls are drawn.

The second question is, how many cards are you playing?

If it is just one card, then keep in mind that each possible bingo is a specific set of numbers, and the balls don't care what the letters are.

For example, if you have a card with B11-15, I21-25, N31/32/free/34/35, G51-55, and O71-75 (each in order from top to bottom), then "four corners" is 11, 15, 71, 75; the letters are irrelevant.

I have a feeling what you are really asking is, how many different cards are there for a particular game?

For four corners, that requires two numbers in 1-15 and two in 61-75

There are (15)C(2) = 105 pairs of numbers in 1-15, and for each pair, there are 105 pairs in 61-75, for a total of 105 x 105 = 11,025 different sets of numbers. (Note that there are actually four times as many cards, but a card with 1 in the top left, 5 in the bottom left, 71 in the top right, and 75 in the bottom right is the same as 5 in the top left, 1 in the bottom left, 75 in the top right, and 71 in the bottom right, as both win when those four numbers are called.)

For an L, there are (15)C(5) = 3003 sets of 5 B numbers, 15 possible numbers for the bottom of the I column, 15 for the N, 15 for the G, and 15 for the O, for a total of 152,026,875 different sets of winning numbers. Again, note that I am considering the card with 1, 2, 3, 4, 5 in the B column from top to bottom, and the rest of the bottom row I-30, N-45, G-60, and O-75 the same as if the B column was 2, 4, 5, 1, 3 from top to bottom and the other four numbers were the same (30, 45, 60, 75), as the same set of nine numbers is needed to win.

The second question is, how many cards are you playing?

If it is just one card, then keep in mind that each possible bingo is a specific set of numbers, and the balls don't care what the letters are.

For example, if you have a card with B11-15, I21-25, N31/32/free/34/35, G51-55, and O71-75 (each in order from top to bottom), then "four corners" is 11, 15, 71, 75; the letters are irrelevant.

I have a feeling what you are really asking is, how many different cards are there for a particular game?

For four corners, that requires two numbers in 1-15 and two in 61-75

There are (15)C(2) = 105 pairs of numbers in 1-15, and for each pair, there are 105 pairs in 61-75, for a total of 105 x 105 = 11,025 different sets of numbers. (Note that there are actually four times as many cards, but a card with 1 in the top left, 5 in the bottom left, 71 in the top right, and 75 in the bottom right is the same as 5 in the top left, 1 in the bottom left, 75 in the top right, and 71 in the bottom right, as both win when those four numbers are called.)

For an L, there are (15)C(5) = 3003 sets of 5 B numbers, 15 possible numbers for the bottom of the I column, 15 for the N, 15 for the G, and 15 for the O, for a total of 152,026,875 different sets of winning numbers. Again, note that I am considering the card with 1, 2, 3, 4, 5 in the B column from top to bottom, and the rest of the bottom row I-30, N-45, G-60, and O-75 the same as if the B column was 2, 4, 5, 1, 3 from top to bottom and the other four numbers were the same (30, 45, 60, 75), as the same set of nine numbers is needed to win.

April 20th, 2018 at 11:56:04 AM
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Here is a Bingo math problem I came up with that has always intrigued me.

Given two (random) bingo cards in a game, what is the probability that both cards will make Bingo on the same called number?

I believe the answer is dependent on the pattern required for bingo.

Easier cases: 1) Four corners 2) Blackout

Harder cases 3) Standard Bingo - any vertical, horizontal or diagonal line

4) Any block of nine (3x3)

I think this problem is more easily stated than solved. ; )

Edit: Some internet research reveals that people calculate bingo probabilities using simulation models, including (apparently) the WOO site. I have always been intrigued by using combination math to do this. I think that for a simpler problem, such as 4 corners, it should be possible.

Given two (random) bingo cards in a game, what is the probability that both cards will make Bingo on the same called number?

I believe the answer is dependent on the pattern required for bingo.

Easier cases: 1) Four corners 2) Blackout

Harder cases 3) Standard Bingo - any vertical, horizontal or diagonal line

4) Any block of nine (3x3)

I think this problem is more easily stated than solved. ; )

Edit: Some internet research reveals that people calculate bingo probabilities using simulation models, including (apparently) the WOO site. I have always been intrigued by using combination math to do this. I think that for a simpler problem, such as 4 corners, it should be possible.

Last edited by: gordonm888 on Apr 20, 2018

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

April 20th, 2018 at 2:44:30 PM
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Okay, I realize I'm talking to myself but I will solve the problem for the 4 corners pattern

Probability of 2 bingo cards tieing for Bingo: 4 corners pattern.

In order for two bingo cards to get 4 Corners Bingo simultaneously:

- The cards must have in common one or more numbers in their 4 Corners.

- Of all the numbers in the 4 Corners of the two cards, the last one to be called must be a common number

Zero numbers in common:

This will occur with a probability of (c(13,2)/C(15,2))^2 = 0.551836735.

In this scenario, the two cards will never tie for bingo.

One number in common:

This will occur with a probability of 2*c(13,2)*c(13,1)*c(2,1)/C(15,2)^2 = 0.367891156

The two cards will tie for bingo 1/7 of the time, because the common number must be the last of 7 different numbers in the 8 corners to be called out.

Two numbers in common:

Probability of either 2 B numbers in common or 2 O numbers in common = 2*c(13,2)*c(2,2)/C(15,2)^2 = 0.01414966

Probability of one B number and one O number in common =(c(13,1)*c(2,1)/C(15,2))^2 = 0.061315193

The two cards will tie for bingo 2/6 of the time.

Three numbers in common:

This will occur with a probability of 2*c(2,2)*c(13,1)*c(2,1)/C(15,2)^2 = 0.004716553

The two cards will tie for bingo 3/5 of the time.

Four numbers in common:

Probability of occurrence= (c(2,2)/C(15,2))^2 = 0.000907029

The two cards will always tie for bingo.

Combining all of that, two cards will get a 4 Corners Bingo on the same called-out number 8.0631% of the time.

Probability of 2 bingo cards tieing for Bingo: 4 corners pattern.

In order for two bingo cards to get 4 Corners Bingo simultaneously:

- The cards must have in common one or more numbers in their 4 Corners.

- Of all the numbers in the 4 Corners of the two cards, the last one to be called must be a common number

Zero numbers in common:

This will occur with a probability of (c(13,2)/C(15,2))^2 = 0.551836735.

In this scenario, the two cards will never tie for bingo.

One number in common:

This will occur with a probability of 2*c(13,2)*c(13,1)*c(2,1)/C(15,2)^2 = 0.367891156

The two cards will tie for bingo 1/7 of the time, because the common number must be the last of 7 different numbers in the 8 corners to be called out.

Two numbers in common:

Probability of either 2 B numbers in common or 2 O numbers in common = 2*c(13,2)*c(2,2)/C(15,2)^2 = 0.01414966

Probability of one B number and one O number in common =(c(13,1)*c(2,1)/C(15,2))^2 = 0.061315193

The two cards will tie for bingo 2/6 of the time.

Three numbers in common:

This will occur with a probability of 2*c(2,2)*c(13,1)*c(2,1)/C(15,2)^2 = 0.004716553

The two cards will tie for bingo 3/5 of the time.

Four numbers in common:

Probability of occurrence= (c(2,2)/C(15,2))^2 = 0.000907029

The two cards will always tie for bingo.

Combining all of that, two cards will get a 4 Corners Bingo on the same called-out number 8.0631% of the time.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

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