September 12th, 2018 at 3:33:25 PM
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I saw a game played at Indian Casinos in Florida. It involved a card with only 3 numbers and used 75 balls. So, I was wondering about the math.

I know it is 75x74x73 to get Bingo in the first 3 balls. How about the math to get bingo in the rest of the calculation. Not asking to put a table here but give me the math equation. Such as getting bingo in 5 or 10 or 20 balls.

What is the equation in doing this probability?

I know it is 75x74x73 to get Bingo in the first 3 balls. How about the math to get bingo in the rest of the calculation. Not asking to put a table here but give me the math equation. Such as getting bingo in 5 or 10 or 20 balls.

What is the equation in doing this probability?

September 12th, 2018 at 4:29:00 PM
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Actually, it is 75 x 74 x 73 / 6, since the order does not matter. If the card's numbers are 1, 2, 3, then it doesn't matter if the balls come out in the order 1, 2, 3; or, 2, 3, 1; or, 3, 1, 2; or any of the other three orders.

For calculating it in N balls or less:

Instead of numbering the balls, assume there are 3 red balls (the ones with your three numbers on it) and 72 white ones.

There are (75)C(N) ways to select N balls.

Of these, there are (72)C(N-3) ways to choose the (N-3) white balls in a winning combination.

The probability is (72)C(N-3) / (75)C(N)

= ( 72! / ( (N-3)! (72-(N-3))! ) ) / ( 75! / (N! (75-N)!) )

= 72! / ( (N-3)! (75-N)! ) x (N! (75-N)!) / 75!

= (72! N! (75-N)!) / ( (N-3)! (75-N)! 75!)

= (72! (N-3)! x (N-2) x (N-1) x N) / ( (N-3)! 72! x 73 x 74 x 75)

= ((N-2) x (N-1) x N) / (73 x 74 x 75)

For N = 3, this is 6 / (73 x 74 x 75), as already shown

For N = 5, this is 60 / (73 x 74 x 75)

Remember, this is 5 balls or fewer; to calculate the number for exactly 5 balls, take the value for 5 balls or fewer and subtract the value for 4 balls or fewer

For calculating it in N balls or less:

Instead of numbering the balls, assume there are 3 red balls (the ones with your three numbers on it) and 72 white ones.

There are (75)C(N) ways to select N balls.

Of these, there are (72)C(N-3) ways to choose the (N-3) white balls in a winning combination.

The probability is (72)C(N-3) / (75)C(N)

= ( 72! / ( (N-3)! (72-(N-3))! ) ) / ( 75! / (N! (75-N)!) )

= 72! / ( (N-3)! (75-N)! ) x (N! (75-N)!) / 75!

= (72! N! (75-N)!) / ( (N-3)! (75-N)! 75!)

= (72! (N-3)! x (N-2) x (N-1) x N) / ( (N-3)! 72! x 73 x 74 x 75)

= ((N-2) x (N-1) x N) / (73 x 74 x 75)

For N = 3, this is 6 / (73 x 74 x 75), as already shown

For N = 5, this is 60 / (73 x 74 x 75)

Remember, this is 5 balls or fewer; to calculate the number for exactly 5 balls, take the value for 5 balls or fewer and subtract the value for 4 balls or fewer

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