Baby Bingo

Consider a smaller version of the game with a 3-by-3 grid: a “Free” square surrounded by eight other squares with numbers. Suppose each of these eight squares is equally likely to be called, and without replacement (i.e., once a number is called, it doesn’t get called again).

On average, how many markers must you place until you get “bingo” in this 3-by-3 grid?

To do this, let's define some terms. Let's denote the squares, as follows: the top row moving left to right is $a, b, c$, then the left box on the middle row is $d$ and the right box on the middle row is $e$, and finally the bottom row from left to right is $f, g, h.$ Let $N$ be the random number of markers needed until getting bingo on this 3-by-3 grid. Since it is a random variable on the natural numbers, we can use the sum of tail probabilities to obtain the expected value, that is, $\mathbb{E}[N] = \sum_{n=0}^\infty \mathbb{P} \{ N \gt n \}.$

We see that since you need three in a row, the smallest number of markers you will need to get bingo with the help of the free center square is $2$. So we have $\mathbb{P} \{ N \gt 0 \} = \mathbb{P} \{ N \gt 1 \} = 1.$ Additionally, we see that there are only four configurations where $N=2,$ namely, $\{a,h\}$, $\{b,g\}$, $\{c,f\}$ and $\{d,e\}$. Since these can be selected in any order, we actually have $2! \cdot 4 = 8$ possible winning combinations out of a total possible of $8 \cdot 7 = 56$ combinations of $2$ squares, so $\mathbb{P} \{ N = 2 \} = \frac{8}{56} = \frac{1}{7},$ so therefore, $\mathbb{P} \{ N \gt 2 \} = \frac{6}{7}.$ Similarly, we see that there are 11 winning combinations with $N \leq 3,$ namely $\{a,b,c\},$ $\{a,b,g\},$ $\{a, b, h\}$, $\{a, c, f\},$ $\{a, c, h\}$, $\{a, d, e\},$ $\{a, d, f\},$ $\{a, d, h\}$, $\{a, e, h\},$ $\{a, f, h\},$ $\{a, g, h\},$ $\{b, c, f\},$ $\{b, c, g\},$ $\{b, d, e\},$ $\{b, d, g\},$ $\{b, e, g\},$ $\{b, f, g\},$ $\{b, g, h\},$ $\{c, d, e\},$ $\{c, d, f\},$ $\{c, e, f\},$ $\{c, e, h\},$ $\{c, f, g\},$ $\{c, f, h\},$ $\{d, e, f\},$ $\{d, e, g\},$ $\{d, e, h\},$ and $\{f, g, h\}.$ If we wanted to make sure that $N = 3,$ then we would need to make sure we know the right order to ensure that for instance we don't have $d$ then $e$ then $a$, which would win after $N=2.$ However, let's just focus on $N \leq 3$, where we don't care, so we have $3! \cdot 28 = 168$ possible winning combinations with $N \leq 3$ out of $8 \cdot 7 \cdot 6 = 336$ possible combinations of 3 squares. So we have $\mathbb{P} \{ N \leq 3 \} = \frac{168}{336} = \frac{1}{2},$ which leaves $\mathbb{P} \{ N \gt 3 \} = \frac{1}{2}.$

We are making good progress. One item to note is that we also have $\mathbb{P}\{ N \leq 5 \} = 1$ or conversely $\mathbb{P} \{ N \gt n \} = 0$ for all $n \geq 5.$ We can see this by appealing to a slightly modified pigeonhole principle. We see that there are 8 total sets of three, but since these sets contain overlapping squares, we will assume that whenever a square is selected a marker is added to each set that contains that square. From the pigeonhole principle, if we have 17 total markers, then at least one set must be complete. We start off with 4 markers from the free center square. We also note that the corner squares are each in three sets, while the middle squares are in two apiece. If we have 5 squares to play with and want to try not to have a set, then we could have at most one of the squares from the middle column and at most one of the squares from the middle row, which would mean that we have 3 corner squares (for a total of 9 markers) and 2 middle squares (for a total of 4 markers). Together with the free central square we see that 5 squares gives at least 9 + 4 + 4 = 17 markers, so we are guaranteed to have a full set with $N=5.$

So the only thing remaining is to determine $\mathbb{P} \{ N \gt 4 \}.$ We see that there are 8 configurations of four squares that do not yet win, namely $\{a, b, e, f\},$ $\{a, c, d, g\},$ $\{a, c, e, g \},$ $\{a, e, f, g \},$ $\{b, c, d, h \},$ $\{b, d, f, h\}$ $\{b, e, f, h\},$ and $\{c, d, g, h\}.$ Since the order of these non-winning configurations does not matter, we have a total of $4! \cdot 8 = 192$ not yet winning combinations out of a total of $8 \cdot 7 \cdot 6 \cdot 5 = 1680$ four square combinations, so we see that $\mathbb{P} \{ N \gt 4 \} = \frac{96}{1680} = \frac{4}{35}.$

Therefore putting everything together we get that the expected number of markers you must place until you get bingo on a 3-by-3 grid is $$\mathbb{E}[N] = \sum_{n=0}^\infty \mathbb{P} \{ N \gt n \} = 1 + 1 + \frac{6}{7} + \frac{1}{2} + \frac{4}{35} = \frac{243}{70} = 3.4\overline{714285}.$$