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February 22nd, 2019 at 6:31:10 PM
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Hope this is right forum. If not let me know.

A bingo game with 75 balls. Only need three numbers.

What is the math to figure out the odds to get 3 numbers in the first 3 balls, the first 20 balls and the first 40 balls.

A bingo game with 75 balls. Only need three numbers.

What is the math to figure out the odds to get 3 numbers in the first 3 balls, the first 20 balls and the first 40 balls.

February 22nd, 2019 at 8:30:49 PM
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Use the hypergeometric distribution. In excel, it is =hypgeomdist(3,3,n,75) where n is the number of balls drawn (either 3, 20, or 40).

I heart Crystal Math.

February 22nd, 2019 at 10:48:35 PM
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I thought there was only one Free Space on a Bingo card and you need 4 to 5 numbers to win.

My advice is to only play when there are less than 75 players in the room. Going to a place with 1000 players and you won't win often.

My advice is to only play when there are less than 75 players in the room. Going to a place with 1000 players and you won't win often.

February 23rd, 2019 at 8:35:35 AM
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Quote:ChumpChangeI thought there was only one Free Space on a Bingo card and you need 4 to 5 numbers to win.

My advice is to only play when there are less than 75 players in the room. Going to a place with 1000 players and you won't win often.

It sounds like it's some sort of "speed game."

Question for the OP: by "need three numbers," does this mean that you need to get any 3 of the 24 numbers on the card called?

Assuming it does, it might be easier to figure out the probability of not getting at least 3 of your numbers drawn, and subtract that from 1.

Let's color your 24 numbers' balls red, and the other 51 white.

There are (75)C(N) ways to pull out the first N balls.

Of these, (51)C(N) have no red balls, (51)C(N-1) x (24)C(1) have one, and (51)C(N-2) x (24)C(2) have two.

Add these three up, divide by (75)C(N), and subtract that fraction from 1.

February 25th, 2019 at 12:05:25 PM
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Thank you all for the answers.

Let me give a little more detail. The card only has 3 numbers on it. It is Lightning Bingo. You put a credit on each number and as your number is called you get that credit back. Once a player has all 3 credits returned, the player wins all the other players left over credits. I filed a patent for it. I added progressives if you get all your numbers called within certain limitations. Will make it an online game or multiplayer, multi-unit game for land based casinos.

Let me give a little more detail. The card only has 3 numbers on it. It is Lightning Bingo. You put a credit on each number and as your number is called you get that credit back. Once a player has all 3 credits returned, the player wins all the other players left over credits. I filed a patent for it. I added progressives if you get all your numbers called within certain limitations. Will make it an online game or multiplayer, multi-unit game for land based casinos.

February 25th, 2019 at 2:41:54 PM
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Quote:jrblackieThank you all for the answers.

Let me give a little more detail. The card only has 3 numbers on it. It is Lightning Bingo. You put a credit on each number and as your number is called you get that credit back. Once a player has all 3 credits returned, the player wins all the other players left over credits. I filed a patent for it. I added progressives if you get all your numbers called within certain limitations. Will make it an online game or multiplayer, multi-unit game for land based casinos.

You just now filed a patent for this? I’ve played a version of this with only 60 numbers instead of all 75. It’s very fun.

http://bjs-bingo.com/speed-bingo/ .

“Man Babes” #AxelFabulous

February 26th, 2019 at 7:00:48 AM
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I hope my calculations is correct :

1) To get 3 number correct in first 3 balls, odds = 1/C(75,3) = 0.0000148093

2) To get 3 number correct in first 20 balls, odds = C(72,17)/C(75,20) = 0.016882636

3) To get 3 number correct in first 40 balls, odds = C(72,37)/C(75,40) = 0.146316179

There is a formula for that, Bet x numbers, get exact y correct numbers from the first n numbers pick from N numbers, where y<=x, y<=n and n<=N.

odds to get exact y numbers correct= C(x,y) * C( N-x, n-y) / C(N, n).

For jrblackie question, x=y=3, odds = C(N-3. n-3) / C(N, n).

1) To get 3 number correct in first 3 balls, odds = 1/C(75,3) = 0.0000148093

2) To get 3 number correct in first 20 balls, odds = C(72,17)/C(75,20) = 0.016882636

3) To get 3 number correct in first 40 balls, odds = C(72,37)/C(75,40) = 0.146316179

There is a formula for that, Bet x numbers, get exact y correct numbers from the first n numbers pick from N numbers, where y<=x, y<=n and n<=N.

odds to get exact y numbers correct= C(x,y) * C( N-x, n-y) / C(N, n).

For jrblackie question, x=y=3, odds = C(N-3. n-3) / C(N, n).

Last edited by: ssho88 on Feb 26, 2019