I was hoping there was a math whizz here that would be able to calculate some odds for me.

I'm working on a project for an insurance company where a bingo hall is wanting to insure a blackout bingo in x number of balls or less.

What i'm looking for is the following:

1. If 5,000 cards were sold what are the chances a full card would be called in 45 numbers or less?

2. If 10,000 cards were sold what are the chances a full card would be called in 45 numbers or less?

Any chance someone can give me the math equation or answer?

Quote:odiousgambitI thought insurance outfits were crawling with actuaries.

I'm with an Insurance Brokerage who is making a proposal to an insurance company. Doing some ground work

For example, I asked it for 1 winner with 9 drawn of each number on a "full card".

It said,

The probability of a single card winning is 0.00000028608740961552.

The most likely number of cards in play is 6,990,870.

This doesn't fully answer your question, as it lists only one combination of draws that can win.

But I'm guessing you're not going to have a lot of trouble safely insuring that action.

But the question is if there are 10,000 cards sold, that number would significantly go up I would assume.

5000 or 10000 cards are in roughly the same percentage area when you're talking about 10's of millions of cards to yield a single winner. But the chances are additive, considering every permutation that includes at least 5 of each number, except the N, which needs 4. Which means that all kinds of draws will not provide even the minimum number of picks in each column. But also means you would have to figure all combos up to and including 2 groups of all 15 numbers and the minimum in the other 3. And everything in between.

There are guys here who could set up the problem on a spreadsheet. I'm not one of them. Perhaps one or two will chime in.

A close approximation (expressed as an Excel formula) is:

`=1-BINOMDIST(0,cards,COMBIN(45,24)/COMBIN(75,24),FALSE)`

The Actual Probability

1) Identify the 45-number distributions which support a coverall. There are 7,931 such distributions.

2) For each distribution, compute the probability of a single card having a coverall:

`=HYPGEOMDIST(5,5,B,15)`

*HYPGEOMDIST(5,5,I,15)

*HYPGEOMDIST(4,4,N,15)

*HYPGEOMDIST(5,5,G,15)

*HYPGEOMDIST(5,5,O,15)

3) Calculate the probability of at least one coverall among the desired number of cards:

`=1-BINOMDIST(0,cards,p1,FALSE)`

4) Calculate the probability of this particular distribution:

`=COMBIN(15,B)`

*COMBIN(15,I)

*COMBIN(15,N)

*COMBIN(15,G)

*COMBIN(15,O)

/COMBIN(75,45)

5) Multiply p2 by p3; call this result p4.

6) Sum p4 for all 7,931 distributions.

The Results

For 5000 cards:

the approximation yields 0.000731665171869067 (1 in 1366.745)

the actual probability is_ 0.000731582811567932 (1 in 1366.899)

For 10000 cards:

the approximation yields 0.00146279500981428 (1 in 683.623)

the actual probability is_ 0.00146246608710499 (1 in 683.777)

The Disclaimer

I'm sure someone will chime in within 24 hours if they disagree with my math.

Quote:beachbumbabsYou might fiddle with this page.

For example, I asked it for 1 winner with 9 drawn of each number on a "full card".

Is it a full card? Center square is often free, and doesn't require a matching ball draw for a daub.

Going to full card -1 says we expect a coverall hit in 45 balls drawn (assuming 9/9/9/9/9) if there are about 67,000 cards in play (if I read it right).

That's still a very large bingo hall, even if everyone is playing 10 cards.

Quote:DieterQuote:beachbumbabsYou might fiddle with this page.

For example, I asked it for 1 winner with 9 drawn of each number on a "full card".

Is it a full card? Center square is often free, and doesn't require a matching ball draw for a daub.

Going to full card -1 says we expect a coverall hit in 45 balls drawn (assuming 9/9/9/9/9) if there are about 67,000 cards in play (if I read it right).

That's still a very large bingo hall, even if everyone is playing 10 cards.

I discussed the N only needing 4 matches later in the post. I was just using the Wiz's calculator and pulling a single example of 45 balls.

He was saying a distribution of 5000 cards, or 10000 cards, not 67000, if I read it correctly.

Quote:JBThe Approximation

A close approximation (expressed as an Excel formula) is:Replace cards with the desired number of cards.`=1-BINOMDIST(0,cards,COMBIN(45,24)/COMBIN(75,24),FALSE)`

The Actual Probability

1) Identify the 45-number distributions which support a coverall. There are 7,931 such distributions.

2) For each distribution, compute the probability of a single card having a coverall:where B, I, N, G, and O are the number of calls under that letter. Call this result p1.`=HYPGEOMDIST(5,5,B,15)`

*HYPGEOMDIST(5,5,I,15)

*HYPGEOMDIST(4,4,N,15)

*HYPGEOMDIST(5,5,G,15)

*HYPGEOMDIST(5,5,O,15)

3) Calculate the probability of at least one coverall among the desired number of cards:Call this result p2.`=1-BINOMDIST(0,cards,p1,FALSE)`

4) Calculate the probability of this particular distribution:Call this result p3.`=COMBIN(15,B)`

*COMBIN(15,I)

*COMBIN(15,N)

*COMBIN(15,G)

*COMBIN(15,O)

/COMBIN(75,45)

5) Multiply p2 by p3; call this result p4.

6) Sum p4 for all 7,931 distributions.

The Results

For 5000 cards:

the approximation yields 0.000731665171869067 (1 in 1366.745)

the actual probability is_ 0.000731582811567932 (1 in 1366.899)

For 10000 cards:

the approximation yields 0.00146279500981428 (1 in 683.623)

the actual probability is_ 0.00146246608710499 (1 in 683.777)

The Disclaimer

I'm sure someone will chime in within 24 hours if they disagree with my math.

Very kind of you to run an approximation, JB.

So, if you had a bingo hall big enough to get all 10000 cards in play, and you played 1 game a week, it would take about 13 years (on average) to award the prize? And with 5000, about 26 years? That's how I understand your numbers.

Quote:beachbumbabs

So, if you had a bingo hall big enough to get all 10000 cards in play

I think this becomes more of a concern with electronic bingo. I'm not sure how many cards people typically buy, but electronic auto-daubers mitigate some of the former limits of the game (you only have so much space on the table, and you only have so many seconds to daub your cards before the next ball is called).

There are (75)C(30) ways to draw 30 balls from the 75, and (51)C(30) sets of 30 balls among the 51, so the probability is:

(51)C(30) / (75)C(30)

= (51! / (30! x 21!)) / (75! / (30! X 45!)) = (51! x 45!) / (75! x 21!) = (45 x 44 x ... x 22) / (75 x 74 x ... x 52) = about 1/6831227.1405.

The probability that none of N cards is filled in 45 or fewer is (1 - 1/6831227.1405)

^{N}, so the probability that at least one is filled is 1 - ((1 - 1/6831227.1405)

^{N}).

For N = 5000, this is 1/1366.7454.

For N = 10,000, this is 1/683.6228.

Quote:ThatDonGuyFirst, figure out what the probability is that one card will have its 24 numbers drawn in 45 numbers or less. Instead of using balls numbered 1-75, let's use 24 red balls (representing the numbers on your card) and 51 white ones (representing the other numbers). You draw 45 balls, and want all 24 red balls to be in the 45. This is equivalent to drawing 30 balls (representing the 30 balls not drawn) and having all 30 be white.

There are (75)C(30) ways to draw 30 balls from the 75, and (51)C(30) sets of 30 balls among the 51, so the probability is:

(51)C(30) / (75)C(30)

= (51! / (30! x 21!)) / (75! / (30! X 45!)) = (51! x 45!) / (75! x 21!) = (45 x 44 x ... x 22) / (75 x 74 x ... x 52) = about 1/6831227.1405.

The probability that none of N cards is filled in 45 or fewer is (1 - 1/6831227.1405)^{N}, so the probability that at least one is filled is 1 - ((1 - 1/6831227.1405)^{N}).

For N = 5000, this is 1/1366.7454.

For N = 10,000, this is 1/683.6228.

Thank you for confirming my math for the approximation.

The approximation treats the game like keno, where a ticket may hold any of the available numbers, whereas bingo cards must conform to specific rules. Taking into account the 45-call distributions which preclude a coverall results in the probabilities I posted on page 1, which are naturally slightly lower than the approximation.

Quote:ThatDonGuyFirst, figure out what the probability is that one card will have its 24 numbers drawn in 45 numbers or less. Instead of using balls numbered 1-75, let's use 24 red balls (representing the numbers on your card) and 51 white ones (representing the other numbers). You draw 45 balls, and want all 24 red balls to be in the 45. This is equivalent to drawing 30 balls (representing the 30 balls not drawn) and having all 30 be white.

There are (75)C(30) ways to draw 30 balls from the 75, and (51)C(30) sets of 30 balls among the 51, so the probability is:

(51)C(30) / (75)C(30)

= (51! / (30! x 21!)) / (75! / (30! X 45!)) = (51! x 45!) / (75! x 21!) = (45 x 44 x ... x 22) / (75 x 74 x ... x 52) = about 1/6831227.1405.

The probability that none of N cards is filled in 45 or fewer is (1 - 1/6831227.1405)^{N}, so the probability that at least one is filled is 1 - ((1 - 1/6831227.1405)^{N}).

For N = 5000, this is 1/1366.7454.

For N = 10,000, this is 1/683.6228.

Hey! I understood this! Thanks, DonGuy!