Playing breaking bingo. The last round is a jackpot round where the caller calls seven balls containing at least a B, I, G, and O. (The freespace provides the N if no N is called). To win the Jackpot, you must get bingo on the first call. So I wonder, given the above, What are the odds of winning the jack pot, i.e. getting bingo in those first seven calls.


The seven balls are a subset of size $7$ taken from $\{1,2,3,\dots,75\}$ with the property that at least one ball is in each of the columns $B$, $I$, $G$, $O$. We assume all such subsets are equally likely. Our first task is to determine how many such subsets exist. We will use the Principle of Inclusion and Exclusion (PIE) for this.

Without any restriction on the balls in the columns, there are $N=\binom{75}{7}$ possible subsets. Number the columns of the card from 1 to 5 and say that a set of seven balls has "Property $i$" if none of the balls are in column $i$, for $i=1,2,4,5$. Let $S_j$ be the number of subsets with $j$ of the properties, for $j=1,2,3,4$. Then $$\begin{align} S_1 &= \binom{4}{1} \binom{60}{7} \\ S_2 &= \binom{4}{2} \binom{45}{7} \\ S_3 &= \binom{4}{3} \binom{30}{7} \\ S_4 &= \binom{4}{4} \binom{15}{7} \end{align}$$ By PIE, the number of subsets with none of the properties, i.e. the number of subsets with at least one ball in each of columns 1,2,4, and 5, is $$N_0=N-S_1+S_2-S_3+S_4 = 704,143,125$$

Now we need to count the number of subsets which result in bingo. A bingo could occur in any one of rows 1 through 5 or in either diagonal. No bingo is possible in a column because with at least one ball in each of columns 1,2,4, and 5, any one of these columns can contain at most 4 balls, which is not enough for bingo, and column 3 can contain at most 3 balls, also not enough for bingo.

To have a bingo in row 1, the 7-member subset must contain each of the 5 numbers in row 1, with 2 numbers left over which can be drawn from any of the remaining 70 numbers. So there are $\binom{70}{2}$ subsets that result in a bingo in row 1. Rows 2, 4, and 5 are similar.

By essentially the same argument, the number of subsets that result in bingo in row 3, which contains the free cell, is $\binom{71}{3}$. The two diagonals are similar.

So in all, there are $$n=4 \binom{70}{2} + 3 \binom{71}{3} = 181,125$$ draws of seven balls that result in an immediate bingo, and the associated probability is $$\frac{n}{N_0} = \boxed{2.5723 \times 10^{-4}}$$


Let us assume there are $75$ balls like in a normal BINGO game and there are the same $\frac15$ probability of getting any given letter at the start. We need at least $1$ B, $1$ I, $1$ G, and $1$ O to get the jackpot. We can split this into $2$ cases, the first one being when a ball is chosen it is returned and the second being when a ball is not replaced.

Case 1: There are $\binom74$ ways to make $7$ balls where there is at least $1$ B, I, G, or O, as shown by the first variant of stars and bars here. (However, since we can include another star since there are no restrictions on the amount of N balls) There are also $\binom{11}7$ ways to choose these $7$ balls without restrictions. So, the probability of getting at least $1$ B, I, G, or O is $\frac{\binom74}{\binom{11}7}=\frac7{66}$

Case 2: It is unfeasible to go and find the probability of all of these ways happening, so the only way is by using computational assistance. I used the following Python 3 code:

import random
for i in range(20000000):
    for i in range(7):
       if chosenBalls[i]<16:
       elif chosenBalls[i]<31:
       elif chosenBalls[i]<46:
       elif chosenBalls[i]<61:
    if BBalls>0 and IBalls>0 and GBalls>0 and OBalls>0:

The code runs 20 million simulations, where $7$ numbers are picked from the numbers $1$ and $75$, assigns a "letter" based on their size (e.g. if one of the numbers is $57$ then since it would be an N in Bingo it is assigned the letter N), and counts the number of simulations where the above requirements are met. The output I recieved was


i.e. there are $7094837$ simulations where the above conditions are met. So, the estimated probability of the requirements being met in Case 2 is $35.47\%$.

  • $\begingroup$ Why the large disparity between the two meathods? $\endgroup$ – Chris Dec 19 '18 at 14:17
  • $\begingroup$ I’m not quite sure, actually. But remember these are 2 separate cases, and different results should be expected. $\endgroup$ – Kyky Dec 19 '18 at 14:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.