Calculating Probability for Bingo Game: 4 Corners, Row, and 2 Rows

[haplo3]

Hey all,
have a problem to solve for my new game game. really appreciate if someone can help me solve this issue:

* i need to calculate the probability for a given combination


1. Total number of balls in the box = 80
2. Total number of chosen balls 45
3. for simplicity we have only 1 card 5x5 (25 numbers in total)


question
what is the probability that i will get
a. 4 corners of the card?
b. any row ?
c. any 2 rows?

if someone can give me a start i will be happy

 

[alan2]

Here's a start. I'm assuming you are holding the card and the 4 corners are known. There are C(80,45) possible combinations of 45 drawn from 80. There are C(76,41) combinations that contain your 4 corners so the probability of getting all 4 corners is C(76,41)/C(80,45). Any row is similar except that there are 5 of them, each containing 5 numbers. Give it a try.

 

[haplo3]

Here's a start. I'm assuming you are holding the card and the 4 corners are known. There are C(80,45) possible combinations of 45 drawn from 80. There are C(76,41) combinations that contain your 4 corners so the probability of getting all 4 corners is C(76,41)/C(80,45). Any row is similar except that there are 5 of them, each containing 5 numbers. Give it a try.

thank you very much. for the rows because i can get 5 different combination

(c(76,41)/c（80,45))*5

is that correct?

i assume not cause it includes the probability of getting 2 rows together or 3 rows or 4 rows, right?

 

[haplo3]

is this the correct answer for 'any row'

(c(76,41)/c(80,45))*5 - ( (c(76,41)/c（80,45))^5 + (c(76,41)/c（80,45))^4 + (c(76,41)/c（80,45))^3 + (c(76,41)/c（80,45))^2 )

so change of getting a row multiply by 5 = this will give me the chance of geting 1st OR 2nd OR 3th OR 4th OR 5th

but will also give me AND 5,4,3,2 rows together - there for i deduct those