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have been expected that the justification of a RED optimal strategy would prove
more troublesome than a BLUE one as it involves triplets of intervals played with
certain probabilities whereas a BLUE strategy is simply a probability distribution
on [0,1]. However the reverse was true because a RED optimal strategy could be
easily verified by showing that each point of [0,1] meets its intervals with a cer-
tain probability. In many cases the RED optimal strategies could be derived using
three appropriate inequalities of the form
x
i
1
+
+
≥
=
,
,
3where
the
x
ij
are non-negative integers; to be appropriate, a necessary condition is that
there are positive integers
p
α
x
i
2
α
x
i
3
α
1
i
1
2
1
2
3
,
,
+
+
=
q
r
satisfying
px
i
1
qx
i
2
rx
i
3
W
for each
i
result-
ing in the value of the game being
Note that the inequalities represent
three different coverings of the unit interval by segments of lengths
(
p
+
q
+
r
)
/
W
.
α
3
.
This follows a similar pattern to that found in the Two Intervals Game where the
RED optimal strategies can mostly be derived from two different coverings of the
unit interval.
α
1
,
α
2
and
3 might be treated in
general, there are sufficient exceptions to suggest that further ideas are necessary if
substantial progress is to be made. For example most have a BLUE optimal strategy
which uses each of 0 and 1 with a probability equal to the game value. However no
such BLUE strategy was found for the case
Although the examples give indications of how the case
k
=
Its
value is 13/56 but the BLUE optimal strategy obtained used each of 0 and 1 with
probability 10/56. This case is interesting in another way as the Red optimal strategy
was derived from four covering inequalities rather than the usual three; in addition
to the obvious 3
α
1
=
1
/
3
,
α
2
=
1
/
4and
α
3
=
1
/
5
.
1 is needed. It seems
therefore that an interesting challenge on the way to solving the case
k
α
1
≥
1
,
4
α
2
≥
1and5
α
3
≥
1
,
α
1
+
2
α
2
+
α
3
≥
=
3 would
be to answer the following open question.
Question 1.
What is the value of the Three Intervals Game when the lengths of the
intervals are of the form
=
/
,
α
=
/
=
/
,
α
1
u
1
v
and
α
1
w
when
u
v
and
w
are
1
2
3
≤
≤
≤
positive integers satisfying 3
u
v
w
?
To introduce a note of optimism, Woodward in his work on the Three Intervals
Game has taken the length of the largest interval to be at least a third and it could be
that the case when the longest interval has length less than a third will have a more
unified treatment. After all, in the Two Intervals Game, there is a single expression
for the value of the game when the length of the longest interval is less than a half
but this expression does not always hold when it is greater than or equal to a half.
Woodward has also obtained some minor results for the
n
interval case. In partic-
ular he has shown that, when
α
1
>
1
/
2
,
the value of the game is one half of the value
of the game with
n
−
1 intervals with lengths having values
α
2
/
(
1
−
α
1
)
,...,
α
n
/
(
1
−
α
1
)
.
Also if there are
n
intervals all of the same length 1
/
k
where
k
>
n
,
the
value of the game is 1
−
k
/
n
.
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