;
I wish
to have the probability that, by
a number x
of casts, I will bring about these
n faces in
sequence in
the order
.
> eqn12:=y(x)=y(x-1)-y(x-n)/n^n+1/n^n;
Example
1.
Laplace solves this in the case of
only. We must have
and
.
> g2:=rsolve({eqn12_2,y(1)=0,y(2)=1/4},y(x),'makeproc'):
When
do we have an even
wager that the sequence 12 will appear? When
.
> g3:=rsolve({eqn12_3,y(1)=0,y(2)=0,y(3)=1/27},y(x),'makeproc'):
Thus
the other two roots
can be expressed in terms of the first as
and
Moreover, we may rewrite
as
We
may then substitute this
new expression for
in the
solution given above.
An
alternative is to
compute the values recursively . We
do this
in general with the procedure
.
> G:=proc(n,x::nonnegint) options remember;
> if x<n then 0 elif x=n then 1/n^n else evalf(G(n,x-1)-G(n,x-3)/n^n+1/n^n); fi; end:
be the probability.
Laplace obtains the recurrence relation
subject to the initial conditions
that
for
and
.
. The initial
conditions are that
,
, and
. It is
for reason of difficulty, I suspect,
that Laplace stopped with
. 
is a root of
. Then
divides
and
the factorization of this polynomial
is 
. 
,
substituting them successively for
. Since the polynomial
is cubic, we can
find the roots exactly. These roots are