Annotated output from the automatic solitaire playing program.
This uses a pretty simple strategy, which seems to win about one time 
in thirty, and gets partial wins sufficient to lose only about $1.68
per game over the long haul.

[Output from various runs of many games. "exp value" is the
average number of cards raised up (i.e., the ones you get $5 for). If the
average were ever 10.40, we would break even for that set of runs.
Unfortunately, as you can see, averages over fairly large sets of games
tend to be about  10.06 +- 0.15 so at least with the present strategy
it looks unlikely that one could even break even in a long set of games.]

For this run, numgames and exp value are  32768 	 10.129150390625 
For this run, numgames and exp value are  32768 	 10.155731201171875 
For this run, numgames and exp value are  32768 	 10.046051025390625 
For this run, numgames and exp value are  32768 	 10.07061767578125 
[100 repetitive, boring lines deleted]
For this run, numgames and exp value are  32768 	 10.0517578125 
For this run, numgames and exp value are  32768 	 10.061248779296875 
For this run, numgames and exp value are  32768 	 10.093719482421875 
For this run, numgames and exp value are  32768 	 10.1175537109375 
For this run, numgames and exp value are  32768 	 10.097442626953125 
For this run, numgames and exp value are  32768 	 10.058441162109375 
For this run, numgames and exp value are  32768 	 10.0125732421875 
For this run, numgames and exp value are  32768 	  9.897796630859375 
For this run, numgames and exp value are  32768 	 10.047515869140625 
For this run, numgames and exp value are  32768 	 10.073883056640625 

total games played:  3618364 

Average number of cards raised up is  10.0644, that is, this strategy tends to 
lose  $1.67732793  per game.


In the following table, the number in the second column is the number
of games among those 3,618,364 in which I was able to raise n cards
for the $5, where n is the number in the first column.  Thus for
example in 20,568 of the games (0.6%) I never found an Ace.
Observe that in about half the games, I was only able to raise  7  or
fewer cards up.

 0 	 20568 
 1 	 71967 
 2 	 158243 
 3 	 246657 
 4 	 314696 
 5 	 350063 
 6 	 344655 
 7 	 321322 
 8 	 283372 
 9  	 243280 
 10 	 202712 
 11 	 168035 
 12 	 136813 
 13 	 112636 
 14 	 90555 
 15 	 73704 
 16 	 59979 
 17 	 49245 
 18 	 39939 
 19 	 33204 
 20 	 26807 
 21 	 22369 
 22 	 18486 
 23 	 15138 
 24 	 12439 
 25 	 10660 
 26 	 8973 
 27 	 7257 
 28 	 6133 
 29 	 5380 
 30 	 4643 
 31 	 4070 
 32 	 3524 
 33 	 3235 
 34 	 2710 
 35 	 2297 
 36 	 1990 
 37 	 1776 
 38 	 1482 
 39 	 1310 
 40 	 1103 
 41 	 1025 
 42 	 838 
 43 	 790 
 44 	 543 
 45 	 618 
 46 	 322 
 47 	 455 
 48 	 219 
 49 	 298 

There are no entries next to 50 or 51 since it is impossible to be left
with precisely  1  or  2  cards to be raised up. (I'm sure that virtually
all the cases counted in the last few lines could have been won by a
human willing to move cards around. But this would only increase the
average number of cards raised by a few  0.001's -- possibly decreasing 
average loss by about a penny.

Next to  '52'  would be the total of all the remaining games -- those in
which all cards were raised to their suit pile, and the maximum $260 
is won. Stupidly, I did not record this number but if I did the subtractions
correctly, this happened in  129829  of the  3618364  games (which
is 3.58806% of the games: one out of every 27.87 games).

Could I break even with some cleverness? Let's suppose I managed to
completely win all the games in which the computer got more than half
of the cards raised. That would mean winning about half-again more games
(181847 total, out of 3618364 games: just over 5% (one in 20) fully won).
In that case I would still lose  $1,015,918, i.e. about 28 cents per game.
If I could also bring the 8973 26-cards-moved-up games to the full-win
column, I finally clear a profit ($150,572 -- about 4 cents per game).
Of course it would be far more helpful to move up just one or two
additional cards whenever the computer manages to move a dozen or so,
but in my experience there are not many opportunities for human
intervention in those cases.

I conclude it is possible but perhaps unlikely to design a program
which will play the game with an expected net profit, and that even a
very good program cannot expect to win as many as one game in ten.
But these are simply estimates!