What is the probability of getting seven duplicates between two bingo cards?

Question

Assume two standard 5x5 bingo cards. On both, there are 25 total squares, and the middle square is a bonus square, and will not have a number in it. All 24 other squares will have a number on it as determined by the following rule:

For each card, the first column of 5 squares can have any number between 1 and 16 (inclusive). There many not be duplicate numbers on the same card. The second column of 5 squares can have any number between 17 and 32, and so on with each row having a random number drawn from a set of incrementing 16, up to the fifth column having any numbers between 65 and 80.

What is the probability that there is at least one number that appears on both cards? The location of the number is not important, just that there is a duplicate number (in the same column) between both cards.

Then, what is the probability that there are exactly seven duplicate numbers between the two cards? Assume the highest chance to do so, which I think is having 2 duplicates in 2 columns that are not the middle column (as the bonus square cannot be a duplicate number), and then one duplicate number each in the other three columns.

Answer 1

For the probability of getting exactly one match, we calculate 1 - P(no match)

The number of ways to have no matches when choosing 5 from 16 would mean you matched 5 of the remaining 11 = C(11,5). For the middle row, there would be C(12,4), The number of possible selections of 5 from 16 is C(16,5) and the number of selections of 4 from 16 is C(16,4)

C(16,5)⁴C(16,4) = 6.62524444 * 10¹⁷

Then, the number of ways to have no selections is

C(11,5)⁴C(12,4) = 22551378862320

Thus, the probability of no selections is 1 - C(11,5)⁴C(12,4)/(C(16,5)⁴C(16,4)) = 1 - (11/104)^4*99/364 =

1 - 1449459/42582851584 = 42581402125/42582851584 = 0.999965961438793

For the probability of 7 matches, I came up with two ways. One is to find out all ways to add up to 7 (looking at the middle row separately), find the number of ways to get a pattern in one order (for example, 5 from the first row, 2 from the second row, 0 from the third through 5th), determining the number of ways to get 5, 2, 0, 0, and 0 = 1*C(11,3)*Combin(5,2)*Combin(12,4)*combin(11,5)*combin(11,5), and then determining that there are 4*3 = 12 ways to get 5, 2, 0, 0, 0 in different orders (0 in row 3)

It turns out that there are 31 different patterns.

Excel files don't paste well here.

The other way is to write a program where we loop through the number of matches for the five rows and add when the sum is 7.

The program is below.

We get 134966548526116128 for 7 and the total is 662524444712632320 (matching the total above)

134966548526116128/662524444712632320 = .2037

Interestingly, 7 is the most likely result.

The program is below (I actually wrote out counts for 0 through 24 matches; you can check the program by seeing that count2 is the denominator listed previously- there are some mathematical precision issues that have to be taken into account - if you only want seven change the loop to 7:7 from 0:24.

The program is in R, but the coding is simple, so you can easily change into other languages.

choose is R's function for combinations. If you can't find the function, you can use the factorial function.

C(n,r) = n!/(r!(n-r)!)

count2 <- 0

for(z in 0:24){

counts <- 0

for(a in 0:5){

for(b in 0:5){

for(c in 0:4){

for(d in 0:5){

for(e in 0:5){

if((a+b+c+d+e) == z){

counts <- counts + choose(11,5-a)*choose(5,a)*

choose(11,5-b)*choose(5,b)*

choose(11,5-d)*choose(5,d)*

choose(11,5-e)*choose(5,e)*

choose(12,4-c)*choose(4,c)}

}}}}}

print(paste(z,counts),sep=',')

count2 <- count2 + counts}

print(paste("Total: ",count2,sep=""))