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For 1
/
2
<
x
≤
1
−
(
2
−
β
)
/
9
,
it is
2
−
β
x
(
1
−
(
1
+
β
)(
1
−
F
(
1
−
x
))) =
x
(
1
−
β
+
)
)
9
(
1
−
x
Rout
ine calculations
show that this expression has a minimum at
x
=
1
−
)
(
(
]
,
its minimum occurs at 1/2. The payoff is continuous on the right in [0,1] so, for
x
1
/
3
2
−
β
)
/
(
1
−
β
)
≤
1
/
2when
β
≥
1
/
5; it is convex so, in
[
1
/
2
,
(
2
−
β
)
/
9
>
1
/
2
,
it is at least that of the interval [0,1/2] which is at least the payoff
(
2
−
β
)
/
9
of [0,1/2).
(ii) For
(
2
−
β
)
/
9
≤
x
<
1
/
2
,
the payoff of the interval
[
x
,
1
/
2
)
is
(
/
−
)(
−
(
+
β
)(
(
/
−
)
−
(
))) = (
/
−
)(
−
(
−
β
)(
−
)
/
)
.
1
2
x
1
1
F
1
2
F
x
1
2
x
1
2
1
2
x
9
x
The two (main) brackets are decreasing functions of
x
so the maximum occurs
at
x
=(
−
β
)
/
(
−
β
)
/
.
2
9 and is less than
2
9
(iii) Now
F
((
1
+
x
)
/
2
)
−
F
((
1
−
x
)
/
2
)=
1
−
2
F
((
1
−
x
)
/
2
)
so the expected payoff
of
[(
1
−
x
)
/
2
,
(
1
+
x
)
/
2
]
is
x
(
1
−
(
1
+
β
)(
1
−
2
F
((
1
−
x
)
/
2
))) =
x
(
2
−
β
)(
1
−
(
4
/
(
9
(
1
−
x
)))
.
This expression is concave and has a maximum of
(
2
−
β
)
/
9when
x
=
1
/
3
.
Lemma
1
tells us that RED can ensure an expected payoff of at least
(
2
−
β
)
/
9so
we have established the following theorem.
Theorem 1.
The value of the point greedy interval game is
(
−
β
)
/
/
≤
2
9
when
1
5
β
≤
.
This leads to the following conjecture.
Conjecture 1.
The point greedy interval game has value max
5
/
7
n
2
{
(
n
−
1
−
β
)
/
}
where
the maximum is taken over all positive integers.
Of course there are a legion of further challenges with an obvious one being the
case when BLUE is allowed to select more than one point. An alternative approach
would be to investigate other forms of payoff. We have introduced the idea of a
cost to RED of being discovered so a natural extension would be to allow BLUE
to choose the number of points to play but levy a cost on BLUE for them. BLUE
would then have to make a judgement about the amount of resource to employ, thus
creating a doubly greedy game. This might entail a movement away from zero-sum
games but, on the basis that what is good for me is bad for my enemy and vice-
versa, might realistically still remain in the zero-sum environment. As mentioned
above Ruckle did frame his game in terms of BLUE having at most
k
points but, in
the absence of a penalty for using more points, BLUE does not in fact have to make
a judgement call. For the record, in the modified length greedy game Ruckle did
introduce a modified payoff to RED of
a
is the length of RED's in-
terval,
n
is the number of points in RED's interval and
a
and
b
are positive constants
so this may also be setting off point for allied games.
α
−
bn
where
α
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