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different
n
-sequences. A
particular n
-sequence will be called a
-number,
i.e.:
n
Â
£
(
)
=
()
(
i
-
1
)
n
0
mn
,
jim
£
m
(33)
-
1
i
=
1
where 0 £
j
(
i
) £ (
m
- 1) represents the outcome labeled
j
at the
i
th trial.
The probability of a
particular n
-sequence (represented by a
-number)
is then
n
'
1
()
=
(
[]
p
p
j i
(34)
n
i
i
=
where
p
i
[
j
(
i
)] gives the probability of the color labeled
j
to occur at the
i
th
trial in accordance with the specific
-number as defined in Eq. (33).
Since after each trial with a 'don't return” outcome all probabilities are
changed, the probability of an event at the
n
th trial is said to depend on
the “path,” i.e., on the past history of events, that led to this event. Since
there are
m
n
-1
possible paths that may precede the drawing of
j
at the
n
th
trial, we have for the probability of this event:
m
n
-
Â
2
-
1
()
=
(
(
)
)
pj
pjm
◊
n
-
1
+
n
-
1
,
m
n
n
=
0
where
j
◊
m
n
-1
+
(
n
- 1,
m
) represent a
(
n
,
m
)-number which begins with
j
.
From this a useful recursion can be derived. Let
j
* be the colors of balls
which when drawn are
not
replaced, and
j
the others. Let
n
j
*
and
n
j
be the
number of preceding trials on which
j
* and
j
came up respectively (S
n
j
*
+
S
n
j
=
n
- 1), then the probability for drawing
j
(or
j
*) at the
n
th trial with
a path of S
n
j
*
withdrawals is
N
Nn
p
Â
j
(
)
()
=
pj
◊
n
(35a)
n
Â
n
-
1
j
*
-
j
*
and
Nn
Nn
p
-
Â
j
*
j
*
(
)
()
=
pj
*
◊
n
(35b)
n
Â
n
-1
j
*
-
j
*
where
N
=S
N
j
+S
N
j
*
is the initial number of balls, and
N
j
and
N
j
*
the initial
number of balls with colors
j
and
j
* respectively.
Let there be
N
balls to begin with in an urn,
N
w
of which are white, and
(
N
-
N
w
) are black. When a white ball is drawn, it is returned; a black ball,
however, is removed. With “white” ∫ 0, and “black” ∫ 1, a particular
n
-
sequence (
n
= 3) may be
(3, 2) = 1 0 1
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