James Sandefur is currently a Professor in the Department of Mathem...
The way to think about this for a specific scenario, say 3-to-2 is ...
There are 2 ways of splitting the next 2 points after a 3-to-3 tie:...
For each pair of splitting points there are always 2 ways of arran...
For $|x|<1$, we know that the sum of the geometric series $\sum_{n=...
CLASSROOM CAPSULES
EDITOR
Michael K. Kinyon
Indiana University South Bend
South Bend, IN 46634
Classroom Capsules consists primarily of short notes (1–3 pages) that convey new mathematical in-
sights and effective teaching strategies for college mathematics instruction. Please submit manuscripts
prepared according to the guidelines on the inside front cover to the Editor, Michael K. Kinyon, Indiana
University South Bend, South Bend, IN 46634.
AGeometricSeriesfromTennis
James Sandefur (sandefur@georgetown.edu), Georgetown Uni versity, Washington,
DC 20057
During the Wimbledon tennis finals, the commentator mentioned that one of the
players was winning 60% of his points on serve. I began wondering what fraction of
the games a person should win if the probability of winning any particular point was p.
In answering this question, I used some basic probability and summed a geometric
series. Others might want to share this with their students.
As a reminder, the winner of a game is the first person to score 4 points, unless the
game reaches a 3-to-3 tie. Then it continues until someone goes ahead by 2. (The four
points are called 15, 30, 40 and game, by the way.)
Assume that the probability that player A wins a point is p.Theprobabilitythat
player A wins the game within the first 6 points is the probability that A leads by a
score of 3-to-0, 3-to-1, or 3-to-2, and then wins the next point, which is
!"
3
0
#
p
3
+
"
4
1
#
p
3
(1 − p) +
"
5
2
#
p
3
(1 − p)
2
$
p = p
4
(15 − 24 p + 10 p
2
),
after simplification. If neither player has won by the 6th point, then the score must be
tied at 3-to-3. The probability of a 3-to-3 tie is
"
6
3
#
p
3
(1 − p)
3
= 20 p
3
(1 − p)
3
.
After that, the game must be won after an even number of points. The probability of
winning on the 8th point is the probability of a 3-to-3 tie, followed by winning the next
two points, which is
20 p
3
(1 − p)
3
p
2
.
The probability of winning on the 10th point is the probability of a 3-to-3 tie, splitting
the next two points, and then winning the 9th and 10th points, which is
20 p
3
(1 − p)
3
%
2p(1 − p)
&
p
2
.
224
c
" THE MATHEMATICAL ASSOCIATION OF AMERICA
More generally, the probability of winning on the (2n + 6)th point is the probability of
a3-to-3tie,20p
3
(1 − p)
3
,splittingeachpairofthenext2n − 2points[2p(1 − p)]
n−1
,
and then winning the last 2 points p
2
,whichis
20 p
3
(1 − p)
3
%
2p(1 − p)
&
n−1
p
2
.
Thus, the probability of A’s winning the game is
f ( p) = p
4
(15 − 24 p + 10 p
2
) + 20 p
3
(1 − p)
3
p
2
∞
'
n=0
%
2p(1 − p)
&
n
.
Summing the geometric series gives the function
f ( p) = p
4
(15 − 24 p + 10 p
2
) +
20 p
5
(1 − p)
3
1 − 2 p(1 − p)
,
whose graph is shown in Figure 1.
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
Figure 1. Graph of probability function f of winning a game as function of the probability p
of winning a point.
The answer to my original question is that players who win 60% of their points on
serve will win f (0.6) ≈ 0.74 or about 74% of their service games.
The graph suggests that f has rotational symmetry about the point (0.5, 0.5),al-
though this is not apparent from the form of f .Wecanverifytherotationalsym-
metry by noting that if q is the probability that the second player wins a point, then
f (q) gives the probability that the second player wins the game. But p + q = 1, so
f (q) = f (1 − p).Sinceoneofthetwoplayersmustwin, f ( p) + f (1 − p) = 1. If
we take p = 0.5 + x and g(x) = f (0.5 + x) − 0.5, we find that
g(x) + g(−x) = f (0.5 + x) + f (0.5 − x) − 1 = f ( p) + f (1 − p) − 1 = 0,
so g(−x) =−g(x).Thus,g is an odd function, and hence f has rotational symmetry.
From the graph, it also appears that there is a point of inflection at p = 0.5. It would
be tedious to check that f
%%
(0.5) = 0. Ho wever, if we remember the derivation
f
%%
(x) = lim
h→0
f (x + h) + f ( x − h) − 2 f (x)
h
2
VOL. 36, NO. 3, MA Y 2005 THE COLLEGE MATHEMA TICS JOURNAL 225
and apply that to g(x),weseethat
f
%%
(0.5) = g
%%
(0) = lim
h→0
g(h) + g(−h) − g(0)
h
2
= lim
h→0
0
h
2
= 0,
which shows that p = 0.5givesapointofinflection.Thismeansthatplayersgainthe
most in the number of games they win by increasing p when p ≈ 0.5. In other words,
if you are a weak or a strong player relative to your opponent ( p small or large), then a
small improvement in your serve (increasing p)doesn’tresultinasmuchimprovement
in the number of games you win as an improvement against a comparable opponent
(p about 0.5) does. A simple computation shows that f
%
(0.5) = 2.5, so a 1% gain in
p will result in about a 2.5% increase in your likelihood of winning a game.
◦
On Sums of Cubes
Hajrudin Fejzi
´
c(hfejzic@csusb.edu),DanRinne(drinne@csusb.edu),andBobStein
(bstein@csusb.edu), Department of Mathematics, California State University, San
Bernardino, CA 92407
Awell-knownidentityforthesumofthefirstn cubes is
1
3
+ 2
3
+···+n
3
= (1 + 2 +···+n)
2
. (1)
Some of our students noticed that, curiously, equality still holds if n − 1isreplaced
by 2, that is
1
3
+ 2
3
+···+(n − 2)
3
+ 2
3
+ n
3
=
(
1 + 2 +···+(n − 2) + 2 + n
)
2
.
This observation led us to ask whether other such switches are possible. In this note,
we investig ate those triples (k, m, n) for which
k−1
'
j=1
j
3
+ m
3
+
n
'
j=k+1
j
3
=
(
k−1
'
j=1
j + m +
n
'
j=k+1
j
)
2
. (2)
Clearly (2) holds (because of (1)) if m = k, so in what follows we assume m (= k.
Furthermore, as our students pointed out, (n − 1, 2, n) is a solution for every n ≥ 2.
Do other solutions exist, and if so, do they fit nice patterns? Our inquiry led us to some
interesting and unexpected answers, and ultimately to a connection with an unsolved
problem in number theory.
Our first result gives a necessary and sufficient condition for a triple (k, m, n) to be
asolutionof(2).(Weassumethatk and m are positive integers not exceeding n.)
Theorem 1. Atriple(k, m, n) with m (= ksatisfies(2) if and only if either
(a) k = n − 1 and m = 2,or
(b) there exists an integer p ≥ 2 and a positive divisor s of 3p( p − 1) for which
(k, m, n) =
"
2(p − 1) +
3(p − 1) p
s
, 2p + s, m + k − p
#
. (3)
Furthermore, different pairs ( p, s) yield different solutions.
226
c
" THE MATHEMATICAL ASSOCIATION OF AMERICA
There are 2 ways of splitting the next 2 points after a 3-to-3 tie: you either win the first point and lose the second or you lose the first point and win the second: $p(1-p)+(1-p)p=2p(1-p)$
For $|x|<1$, we know that the sum of the geometric series $\sum_{n=0}^{\infty}x^n = \frac{1}{1-x}$. Since $|2p(1-p)|<1$ we get
$$
\sum_{n=0}^{\infty}(2p(1-p))^n = \frac{1}{1-2p(1-p)}
$$
James Sandefur is currently a Professor in the Department of Mathematics and Statistics at Georgetown University. He got his PhD in Mathematics from Tulane University and in August 2006, he was awarded the MAA George Polya Award for my expository paper for this paper.
The way to think about this for a specific scenario, say 3-to-2 is to calculate the total number of possibilities of distributing the 2 points from player B in 5 different slots. You can do this using the binomial coefficient $\binom{5}{2}= 10$.
For each pair of splitting points there are always 2 ways of arranging them (depending on who wins the first point) so the probability is $(2p(1-p))^{n-1}$ and $p^2$ for the last 2 winning points.
