The B i n g o

Paradox
Arthur Benjamin, Joseph Kisenwether,
and Ben Weiss

B
elieve it or not, when a large number of people can sink our teeth
play bingo, it is much more likely that the B I N G O into. For a randomly
winning card has a completed horizontal row 3 27 34 50 67 generated sequence of
than a completed vertical column. How can bingo numbers, what
this be? If you randomly mark off numbers 6 19 35 48 65 is the probability that
on your own bingo card, you are just as likely to get a all five letters appear
FREE
horizontal bingo as a vertical bingo. Why should the 7 25 SPACE 54 61 before any single letter
winning card be any different? appears five times?
A computer program was written that generated 13 26 36 55 64
1,000 random bingo cards and played the game 100,000 Analyzing the
times. To our surprise, horizontal winners were almost 10 30 42 49 70 Paradox
twice as likely as a vertical winner. Figure 1. A typical bingo card. For simplicity, let’s
To better understand this paradox, let’s review the assume we are
rules of this popular game. A typical bingo card, like playing with so many bingo cards that as soon as all
the one in figure 1, has five columns, labeled B, I, N, G, five letters appear, we have a horizontal winner and
and O, and each column has five numbers underneath as soon as one letter appears five times, we have a
it. Column B has five numbers from 1 through 15, in a vertical winner.
random order. Similarly columns I, N, G, and O have We define a horizontal sequence to be an arrangement
five random numbers from 16 to 30, 31 to 45, 46 to 60, of the 75 bingo numbers so that all five letters appear
and 61 to 75, respectively. Some bingo cards have a free before any letter appears five times. Otherwise, it
space in the middle of column N. is called a vertical sequence. We say a sequence has
The caller calls out numbers—such as “B11!”—pulled property Hn if it becomes a horizontal sequence on
randomly from a container. Players place markers on the nth number and has property Vn if it becomes a
the corresponding spaces on their card if they have vertical sequence on the nth number. A sequence that
them. The first person to complete a row, column, or begins with the eight numbers given above would have
diagonal yells “bingo!” and wins a prize. In the analysis property H8. Note that it is impossible for a sequence to
that follows, we will initially ignore the effect of the free be both horizontal and vertical since a number like O75
space, but we will consider it later. could not simultaneously be the first time that the letter
Suppose the first eight numbers were drawn in this O appears and the fifth time that the letter O appears.
order: B11, I23, G58, B13, I21, N34, G55, and O75. The probability of a horizontal sequence in five
With these numbers, each letter has appeared, so it is draws (the minimum number possible) is
possible for there to be a bingo card with a horizontal
or diagonal win. But since no letter has appeared five
times, it is impossible to have a vertical bingo at this since after the first number is drawn, 60 of the
point. This suggests a mathematical question that we remaining 74 numbers will produce a new letter, then

18 September 2017 : : Math Horizons : : www.maa.org/mathhorizons

45 of the remaining 73 numbers will produce a third a collection of 15 items. For example, the number of
letter, and so on. Similarly, the chance of a vertical ways to assign numbers to BBBBIIING is
sequence in five draws is

Say we choose the numbers B1, B2, B3, B4, I16, I17,
So a horizontal sequence is about 50 times more likely I18, N31, G46. Then these nine numbers can be ar-
than a vertical sequence on the fifth draw. ranged, like Scrabble tiles in a rack, in 9! ways. Finally,
What about after the fifth draw? Here, the counting the remaining 65 numbers (appearing after O75) can
gets a little more complicated, but we can do it. Let’s be arranged in 65! ways. Putting this all together, the
find the probability of achieving a horizontal sequence probability that a bingo sequence becomes horizontal
in exactly 10 draws. Note that there are 75! equally on the 10th draw with shape 4311 is
likely sequences of bingo numbers. How many of them
result in a horizontal sequence on the 10th draw? Let’s
choose the 10th number first. There are 75 possibili-
ties, so let’s mentally choose the 10th to be O75. Similarly, the probabilities of becoming horizontal
The nine previous numbers can have four possible with the other possible shapes are
shapes based on how many of each of the letters B, I,
N, and G appear: 4311, 4221, 3321, or 3222. For ex-
ample, the sequence N31, N41, G59, I26, B5, N35, B8,
B9, B7 (inspired by the digits of pi) has shape 4311,
since a letter appears four times (B), another letter ap-
pears three times (N), and the other letters (I and G) and
each appear once.
The horizontal shapes are partitions of the integer
9 into four positive parts, where all parts have size at Altogether, the probability of achieving a horizontal
most 4. Thus, a shape like 5211 is not allowed since it
sequence on the 10th draw is
contains five of the same letter and would therefore be
a vertical sequence.
The letters B, I, N, and G can be given a shape of
4311 in 12 different ways (BBBBIIING, BBBBINNNG, We perform similar calculations to find P(V10), the
and so on). Likewise, there are 12 ways to give them probability of a vertical sequence on the 10th draw.
a shape of 4221, 12 ways to give them a shape of Again, there are 75 possibilities for the 10th draw,
3321, but only four ways to give them a shape of 3222 which we will assume is O75. The previous nine num-
(we have four choices for which letter appears three bers must include exactly four Os, which can be chosen
times, and the other letters must all appear twice). In in ways. The remaining five letters can have one
general, a four-digit shape can be assigned to B, I, N, of four possible shapes: 4100, 3200, 3110, or 2210. We
and G in 1, 4, 6, 12, or 24 ways depending on whether exclude the shapes 5000 and 2111, since the first shape
it consists of all the same digit, a tripled digit, two would create an earlier vertical sequence and the sec-
pairs of digits, one pair of digits, or all different digits, ond shape (combined with the other Os) would create
respectively. an earlier horizontal sequence. Each of the shapes has
Once we have determined how many of each letter one digit that appears twice and can therefore be as-
is to be used, we can count the ways to assign them signed letters in 12 ways. Therefore,
numbers using binomial coefficients. Recall that the
binomial coefficient

is the number of ways we can choose k items out of

www.maa.org/mathhorizons : : Math Horizons : : September 2017 19

 n Shapes P(Hn) P(Vn) Ratio Sum Cumulative
5 1 0.04400 0.00087 50.57 0.0449 0.0449
6 1 0.08800 0.00373 23.60 0.0917 0.1366
7 2 0.11350 0.00956 11.87 0.1231 0.2597 when it happens after the 13th draw,
which is only about 7 percent of the
8 3 0.12052 0.01903 6.33 0.1396 0.3992
time, then the vertical sequences have
9 4 0.11220 0.02902 3.87 0.1412 0.5404 the edge.
10 4 0.09415 0.03652 2.58 0.1307 0.6711 When all cards begin with a free space
11 5 0.07191 0.03968 1.81 0.1116 0.7827 in the middle, the chance of a vertical
sequence increases slightly, since column
12 4 0.04972 0.03748 1.33 0.0872 0.8699
N now has 15 numbers instead of 14
13 4 0.03075 0.03085 1.00 0.0616 0.9315 numbers to cover the remaining four
14 3 0.01658 0.02178 0.76 0.0384 0.9698 spaces. Joe Kisenwether and Dick Hess
15 2 0.00742 0.01260 0.59 0.0200 0.9898 independently discovered that when
the free space is used, the chance of
16 1 0.00254 0.00558 0.45 0.0081 0.9980
a horizontal win is 73.73 percent (see
17 1 0.00052 0.00151 0.34 0.0020 1.0000 Dick Hess’s The Population Explosion
Total 35 0.752 0.248 1.000 and Other Mathematical Puzzles, World
Table 1. The probabilities of a vertical or a horizontal bingo. Scientific, 2016).

The Numbers of Shapes

Although we have answered our original question, more
interesting mathematics is lurking behind the analysis.
and
When we enumerated sequences with properties H10
and V10, we had to analyze—in both cases—exactly
four shapes. This is not a coincidence. For each n, the
So, the probability of achieving a vertical sequence on sequences that are horizontal and vertical on the nth
the 10th draw is draw yield the same number of shapes. This is not
obvious. The number of shapes for Hn is the number of
partitions of by four positive integers less than
5, and the number of shapes for Vn is the number of
Comparing P(V10) to P(H10), we see that even on the partitions of by four nonnegative integers less
10th draw, horizontal sequences are more than twice as than 5, at least one of which is 0. What’s going on?
likely to appear as vertical sequences. We will illustrate the one-to-one correspondence
Every sequence will become horizontal or vertical between the shapes for Hn and the shapes for Vn in the
within 17 draws: After 16 draws, we can have a case but the reasoning is the same for any n.
sequence of four Bs, Is, Ns, and Gs, say, but the next The shapes used for the enumeration of H10 consist
number will be either an O (creating a horizontal of four positive numbers that add to 9, where all
sequence) or a B, I, N, or G (creating a vertical numbers are less than or equal to 4. These partitions,
sequence). We summarize our findings in table 1. 4311, 4221, 3321, and 3222, are displayed pictorially
The upshot is that the probability of a horizontal in figure 2. Such representations are called Ferrers
sequence is 75.2 percent, which is about three times diagrams. Note that they fit in a 4-by-4 box.
more likely than a vertical sequence. A sequence If we subtract 1 from each value, or equivalently,
becomes horizontal or vertical by the 12th draw about delete the first columns of dots in the Ferrers diagrams,
87 percent of the time. In all
these cases, horizontal sequences
are much more likely than vertical
sequences. When it happens on
the 13th draw (about 6 percent
of the time) the sequences have Figure 2. Ferrers diagrams for the partitions 4311, 4221, 3321, and 3222 (left to
almost the same probability, and right) fit into a 4-by-4 box.

20 September 2017 : : Math Horizons : : www.maa.org/mathhorizons

 Figure 3. Partitions of 5 into nonnegative values less
than or equal to 3 fit into a 4-by-3 box.

Figure 4. The conjugate

partitions of those in figure 3 fit
into a 3-by-4 box.

as in figure 3, we get partitions

of 5 into four nonnegative parts,
where all numbers are less than or
Then a binomial coefficient like
equal to 3. The partitions now fit
inside a 4-by-3 box.
Next, interchange the rows and Figure 5. The
columns of dots to create the lattice path
has a corresponding q-binomial coefficient
corresponding to
conjugate partitions in figure 4.
the partition 3200.
These Ferrers diagrams fit into
a 3-by-4 box, or equivalently a 4-by-4 box with an
empty last row. They correspond to partitions of 5 by
four nonnegative integers less than or equal to 4, at
least one of which is 0. In particular, these are the V10
shapes 2210, 3110, 3200, and 4100. Believe it or not, after simplifying this rational
Thus, this technique gives a bijection between the function, we obtain a 12th-degree polynomial in
n
shapes for H10 and the shapes for V10; and the same which
e the coefficient of the qqn term is the number of
technique works for other values of n. partitions of the integer n that fit in a 4-by-3 box (see
In fact, we can get more information from these Integer Partitions by George E. Andrews and Kimmo
diagrams. In the Vn case, each partition carves out a Ericsson, Cambridge University Press, 2004, for a
lattice path from the point (0,0) to (4,3) in the 3-by- justification). In other words, it’s the shape-counting
4 box. For example, the partition 3200 creates the polynomial
lattice path in figure 5. Consequently, the number of
shapes needed to compute all of the vertical sequences
(and hence the number needed to compute all of the as seen in the second column of table 1. n
horizontal sequences) corresponds to the number of
lattice paths from (0,0) to (4,3). Since each lattice path Arthur Benjamin teaches at Harvey Mudd College
takes, in some order, four steps to the right and three and is a past editor of Math Horizons. He thanks
steps up, the total number of shapes is Jay Cordes, Adam Busis, and Bob Koca for valuable
assistance.

Joseph Kisenwether is a mathematics and game design

as seen at the bottom of the second column of table 1. consultant to the casino industry and founder of
Finally, there is a slick way to generate the number Craftsman Gaming. He describes his job as “the reason
of shapes for each n (the rest of the entries in the you can’t win.”
second column of table 1) using q-binomial coefficients,
Ben Weiss is the developer of the Frax app that allows
which are polynomial generalizations of binomial
users to navigate fractal images in real time. He has
coefficients.
competed as a free diver for the USA in international
First we replace the integer m with the polynomial
competitions and has held his breath for over seven
minutes. He works for Google in Southern California.
http://dx.doi.org/10.4169/mathhorizons.25.1.18
For example,

www.maa.org/mathhorizons : : Math Horizons : : September 2017 21