To prove (d), we note that when the distance 
, there are two ways
that the next distance 
can be j. The first way occurs if
the dealer's secret card 
has face value m-1-j. This occurs with
probability 
. The second way occurs if the dealer's secret
card 
has face value m, thereby overshooting the player's closest
secret card 
by one, and the face value of 
is j+1, making
. This second event occurs with probability 
.
Thus we have shown (d) for all j < m-1. To finish (d), note that the
only way 
and 
is when both face values of
and 
are m. This occurs with probability 
.
The first row of M is the absorbing state, hence (c).
To prove (b) we consult Figures 1 and 2. We see that the crucial
difference between 
and 
is the penultimate card
to 
, call it star. the shaded boxes contain the only possible
occurences of the other 
. If we imagine an arc drawn between any
two adjacent 
, then we have two cases. The first case occurs when
the star card remains vacant -- no 
occupy star.
Then the same number of arcs which occured in the 
event now
occur in the 
event. In the second case, star is occupied
by some 
, and now the total number of arcs is 1 more than there
were in the 
event, giving us a probability of
. In summary
, thus proving (b).
Note that when i>j, the dealer's secret card 
could have a face value
of i-j, allowing 
. This event occurs with
probability 
and cannot happen when 
,
whence the different structure of the upper triangular portion of M. In
fact the same reasoning as in the proof of (b) allows us to subtract
this one event from those occuring in 
,
thus subtracting a probability of 
at the
end of the process. This proves (a).
