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of the points 0
represent points an
appropriately small distance to the left and right of x respectively. The argument
used for the particular case now show that choosing these points with probabilities
10
,
a
+ ,
b
+ , (
1
b
) −, (
1
a
) −,
1where x
+
and x
/
36 respectively is an optimal Infiltrator strategy.
One might have hoped that the expression for v
36
,
3
/
36
,
5
/
36
,
5
/
36
,
3
/
36
,
10
/
(
a
,
b
)
in Theorem 4 also covers
the case b
1
/
2
.
However this is not the case because Baston and Bostock [ 2 ]
(
/
,
/
)=
/
=
/
=
/
,
showed that v
1
3
1
2
1
4 whereas, for a
1
3and b
1
2
Theorem 3 gives
(
/
,
/
)=
{ (
+ λ
) / (
) , (
+ λ
) / (
) } =
/
.
v
1
3
1
2
1
max
1
2
λ
2
2
λ
1
6
1
1
0
0
Γ (
,
)
/
<
/
Theorem 4 enables us to express the value of
a
b
when 1
3
b
1
2ina
simpler form than Lee [ 8 ].
Theorem 5. Let a
b
,
1
/
3
b
<
1
/
2 and
λ i defined by ( 9.2 ).
(i) If
λ 2
2
,
v
(
a
,
b
)=(
2 m
6
) /
3 m where m
=
min
{
2
λ 0 ,
3
λ 1 ,
6
λ 2 }.
(ii) If
λ 2 =
1
,
v
(
a
,
b
)=(
m
1
) /
2 m where m
=
min
{ λ 0 ,
2
λ 1
1
}.
Λ + = {
Proof.
(i) Let
λ 2
2
,
then
3
}
and
λ 3 =
0 so, by Theorem 4 ,
3
+ λ 0
3
2
+ λ 1
3
1
+ λ 2
3
2
3
2
2
2
3
2
v
(
a
,
b
)=
1
max
{
λ 0 ,
λ 1 ,
λ 2 } =
max
{
λ 0 ,
λ 1 ,
λ 2 }
6
and (i) follows.
(ii) Let
Λ + = {
λ 2 =
1
,
then
2
,
3
}
so, by Theorem 4 ,
3
+ λ 0
3
2
+ λ 1
3
1
+ λ 0
2
λ 1
v
(
a
,
b
)=
1
max
{
λ 0 ,
λ 1 ,
λ 0 ,
1 }
2
λ 1
Now
(
3
+ λ 0 ) / (
3
λ 0 ) (
1
+ λ 0 ) / (
2
λ 0 )
because
λ 0
3and
( λ 1 ) / (
2
λ 1
1
)
(
2
+ λ 1 ) / (
3
λ 1 )
because
λ 1
2
.
Thus
1
1
v
(
a
,
b
)=
1
/
2
max
{
0 ,
) }
2
λ
2
(
2
λ
1
1
and (ii) follows.
We now give examples to show that every case in the theorem arises:
=
/
,
=
/
<
<
,
a
1
8
b
3
8gives
λ
2and6
λ
3
λ
2
λ
2
2
1
0
a
=
3
/
39
,
b
=
13
/
39 gives
λ 2
2and2
λ 0 <
3
λ 1 <
6
λ 2 ,
a
=
2
/
15
,
b
=
5
/
15 gives
λ 2
2and3
λ 1 <
2
λ 0 <
6
λ 2 ,
a
=
3
/
10
,
b
=
4
/
10 gives
λ 2 =
1and2
λ 1
1
< λ 0 ,
a
=
2
/
20
,
b
=
9
/
20 gives
λ 2 =
1and
λ 0 <
2
λ 1
1
.
Acknowledgements
The first author was supported, in part, by NATO Grant PST.CLG.976391.
 
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