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Theorem 5.
In the OEIG with n
≥
3
and
0
<
s
<
2
m, if
2
m
3
[
s
/
2
]
≤
is s
at
isfied, then an optimal strategy for player II is the uniform distribution on the
set B
0
,whereB
0
is the set defined by
(
3.26
) and the value of the game is equal to H
defined by (
3.27
). If, on the contrary,
2
m
3
,
[
s
/
2
]
>
(3.30)
then the value of the game is
min
H
H
=
,
v
with H defined by (
3.27
) and H
defined by (
3.29
).
I
f v
=
H then an optimal strategy
for player II is the uniform distribution on the set B
0
,whereB
0
is the set defined by
(
3.26
),
if
v
H
then an optimal strategy for player II is the uniform distribution on
the set B
0
,whereB
0
is the set defined by (
3.28
).
=
Proof.
As we know, the OEIG satisfies hypothesis of Theorem
2
, therefore, an op-
timal strategy for player I is the uniform distribution on
X
=
F
and the value of the
game is given by
1
m
n
min
A
∈F
v
=
M
(
A
,
B
)
which in this case is written as
n
i
=
1
s
i
1
m
n
min
f
1
m
n
min
j
=
i
(
m
−
s
j
)
.
(
s
1
,
s
2
,...,
s
n
)=
2
m
3
is satisfied, then from Corollary
1
it follows that the minimum of
f
is
equal to
K
defined by (
3.23
), therefore
[
/
]
≤
If
s
2
v
=
H
and an optimal strategy for player II is the uniform distribution on the set
B
0
,where
B
0
is the set defined by (
3.26
).
If, on the contrary, (
3.30
) is satisfied, then, Theorem
4
proves that the minimum
for
f
is the minimum of the values
K
,definedby(
3.23
), and
K
defined by (
3.24
).
If this minimum is
K
, then the value of the OEIG is equal to
H
defined
b
y (
3.27
)
and an optimal strategy for player II is the uniform distribution on the set
B
0
,where
B
0
is the set defined by (
3.26
). If this minimum is
K
, then the value of the OEIG
is equal to
H
defined b
y
(
3.29
) and an optimal strategy for player II is the uniform
distributionontheset
B
0
,where
B
0
is the set defined by (
3.28
). This proves the
theorem.
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