THE MATHEMATICAL GAZETTE
278
76.5 Snap
The environmental card game was proving too complicated for my
three-year old son, so we decided to play snap instead. There were 50
cards in 25 pairs. To our disappointment it took three goes to get even
one snap, so we soon put the cards away in disgust and played dominoes
instead. Later I decided to have a mathematical look at snap, and came
up with what I thought was an interesting result.
At first sight, snap might appear to be the famous derangement
problem (see for example [1] and [3]) in disguise. However, there are
two important differences: in snap the players do not in general start with
the same cards, one of each type; and for a snap to take place a card has
to agree with the card just played or with the next card. The interesting
result is that the probability of getting no snap with n pairs of cards tends
to 1/e as n tends to «., the same limit as occurs in the derangement
problem.
The proof is a nice application of the inclusion-exclusion principle
(see [1] and [3]). A game of snap can be thought of as a permutation of
the cards, namely the order in which they are played. If we think of the
pack as consisting of n pairs xlt x2 where 1 < x < n, then we want to find
out how many of the (2n)\ permutations have the property that for each x
= u, v, w,... the cards * j , x2 are not adjacent.
Let N(u, ...,v) denote the number of permutations in which, for each
x = u, ...,v, the cards xlt x2 do appear in adjacent positions. Then, by the
inclusion-exclusion principle, the number of permutations with no
adjacent pair is equal to
(2n)! - £
N(«) + £ N ( u , v ) -
U
U , V
£
N(M,V,W) + ... .
(1)
U , V , »V
Consider, for example, N(«, v, w) for one of the „C3 choices of u, v, w.
Here we can think of the pairs Uj, u2; Vj, v2; w1( w2 as three objects
which, along with the 2n-6 remaining cards, give 2n-3 objects to be
permuted. Since the cards in each pair can be ordered in 2 ways, we must
have N(«,v,w) = 2 3 (2n-3)!. Thus, from (1), the number of permutations
of all In cards with no pair an adjacent pair is
(2«)l - , £ , 2.(2/i-l)! +„C 2 2 2 .(2n-2)! - ... .
The required probability is now obtained by dividing by (2n)!: thus the
probability of no snap occurring is
Y>
(-2) r „c r
2*l2n(2n-l)
r=0
... (2n-r+l)
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279
NOTES
-z
*- (-2)r n(n-l) ... (n-r + l)
_
r! 2n(2n-l) ... (2/I-/-+1)
r=0
X ^ ( - 1 ) 2n(2w-2) ... (2«-2r+2)
2*1
r\ 2n(2n-l) ... ( 2 n - r + l )
r=0
- Z^C-sbiX-i^-O-Ef^u)-
<»
r=0
Note that this last expression is remarkably like
r =0
which is the probability occurring in the derangement problem, and
which tends to 1/e as n tends to «>.
The difference between the rth terms in (2) and (3) is
M'-O-shX'-fihM'-E^n,)}
(4)
This is clearly less than 1/r! ; but further, if r < nl/i, we can use the fact
that 1 - (1-XJ)(1-JC 2 ) ... (l-xr) < xl + JC2 + ... + xr to estimate (4) as less
than
Il_L-+_2_+
r!l2«-l
2«-2
*"
i
r-\
1
2n-(r-l)i
r!(2n-r+l)
r! ' n '
2
rlnl/3
So the difference between (2) and (3) is at most
r<ntn
r>nm
r=0
<
-fn
1/3
n
+
r>nl/3
—1/3
h~
[« ]!
( s e e [2D
-» ° as n -> oo .
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280
THE MATHEMATICAL GAZETTE
Thus the probability of getting no swap tends to 1/e as n tends to infinity.
If the same analysis is carried out for An cards (n sets of 4), a similar, but
messier, calculation shows that the probability of no snap tends to 1/e3 =
0.0498 ... as n tends to infinity. This value is very different from the
probability 0.01623 ... obtained in the case n - 13 of a similar but
essentially different matching problem posed as Problem 73.F in the
October 1989 Gazette.
References
1. Anstice, J., "The worst postman problem". Math. Gazette 72 (1986) 226-228.
2. Clifford, R, "e on a micro". Math. Gazette 74 (1990) 289-291.
3. Dennett, J.R., "Cricket and derangements". Math. Gazette 74 (1990) 2-5.
IAN ANDERSON
Department of Mathematics, University of Glasgow G12 8QW
76.6 On the scalar triple product and determinantal products
The purpose of this note is to comment on F. Gerrish's note, published with the same title in the Gazette of December 1988. A confirmation of the multiplication rule for third order determinants was obtained
by extending a prior note by H.B. Davies in the Gazette of December
1987. Gerrish's use of a Cartesian base to effect a second expansion
could leave the reader with the impression that the resultant rule depended in an essential way on the use of such a base. Any interested reader
can verify that the rule is more general, and that it follows immediately
from effecting Gerrish's second expansion using any set of linearly
independent vectors.
J.R. GOSSELIN
Dept of Physics, Royal Military College of Canada, Kingston, Ontario
Canada K7K5LO.
76.7 A drawbridge in balance
Recently in the Gazette of June 1990, 124-127, Martyn Cundy described the curve that a counterbalance weight for a drawbridge must
follow in order to be in equilibrium in all possible positions. The curve
is a cardioid. The photograph below shows one such bridge. It was
taken in Australia in 1976. My notes on the photograph only say
"Kyalite River", and as I remember it was near the point where South
Australia, New South Wales and Victoria meet.
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