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We estimate E
[ μ n ]
using ideas from [ 11 ]. Let F be the distribution function of
X v ,
v
V . For any real number t ,wehavethat
+ v ( X v t ) + ,
μ n
t
n t
which gives that E
[ μ n ]
h
(
t
)
:
=
t
+
[
1
F
(
x
)]
dx ,forany t
R . Differentiating
n t n
1
n )
F 1
h
( · )
one finds that its minimum is at t n :
=
(
1
and so E
[ μ n ]
t n +
[
1
e λ x it follows that E
1
F
(
x
)]
dx .Since1
F
(
x
)=
[ μ n ]
λ (
1
+
log n
)
. Hence
2
λ (
E
[ τ ]
2
·
E
[ μ n ]
1
+
log n
) .
We summarize the preceding results into a Theorem which is an improvement of
the Main Theorem from [ 2 ].
Theorem 3. Let G be a graph on n vertices and maximum degree
Δ
. If the number
of available colors is at least
2 and if all players adopt the simple strategy, then
for any starting assignment of colors, the network coloring game reaches a proper
coloring at time
Δ +
τ
that is stochastically smaller than a random variable T , such that
2
,where 2
[
]
λ (
+
)
λ
,
E
T
1
log n
1
023 .
4.5 Proof of Theorem 2
This section is devoted to the proof of Theorem 2 . We want to show that the median
of X d is
3 d + 2
4 ,where X d is the number of different colors after a toss of d coins
that are colored using d colors. Before proving this theorem we need some notation
and remarks.
Suppose that we have d coins that are colored with d colors. Let G be the depen-
dency graph corresponding to this set of coins. We are going to orient G as follows.
Toss all the coins and orient each edge towards the vertex (color) that came up in
the toss. Thus a toss of the coins gives rise to an orientation on the edges of G .Asa
consequence, X d =
j corresponds to the fact that j vertices have positive in-degree,
which means that d
j vertices must have in-degree 0. Also note that none of the
vertices of zero in-degree can be adjacent.
We denote the in-degree of a vertex v by deg (
v
)
and by Z d the number of ver-
tices of zero in-degree. Thus X d =
Z d .
It turns out that the median of X d can be estimated through the median of E d ,
the number of even in-degree vertices, whose distribution is easier to determine. We
will need the following two graph-theoretic results.
d
Lemma 6. Suppose that G is a (possibly disconnected) graph on d vertices and m
edges. Fix some orientation on the edges and let O d , m ,
E d , m be the number of odd
and even in-degree vertices respectively. Then the parity of E d , m equals the parity of
m
d.
 
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