Studying Blackjack (#151)

The solutions took two totally different approaches this week. Some
exhaustively searched the possible hands a dealer can have. The search
isn’t too big in this case so that’s a viable approach that gives exact

Others chose to solve the problem with a simulation. This approach is
simple: deal a ton of hands, play them out by the dealer rules, and
keep track
of the results. If you simulate enough hands, this approach should zero
in on
the actual numbers.

Let’s see if the two approaches agree. Here are the expected results
from Eric
I.'s exhaustive search:

upcard bust 17 18 19 20 21

   2 |   35.33%   13.94%   13.33%   13.07%   12.40%   11.93% 

3 | 37.48% 13.28% 13.07% 12.46% 12.18% 11.54%
4 | 39.85% 13.07% 12.02% 12.10% 11.64% 11.31%
5 | 42.25% 12.10% 12.28% 11.73% 10.90% 10.73%
6 | 42.21% 16.62% 10.62% 10.67% 10.12% 9.75%
7 | 26.11% 37.05% 13.82% 7.80% 7.88% 7.34%
8 | 24.16% 12.97% 36.12% 12.90% 6.89% 6.96%
9 | 23.09% 12.09% 11.20% 35.41% 12.11% 6.10%
10 | 21.32% 11.29% 11.22% 11.30% 33.56% 3.55%
ace | 11.59% 12.85% 13.09% 13.02% 13.12% 5.28%

Now here are the same results from Chris L.'s simulation code (with
number of games increased to 100,000):

Upcard Bust 17 18 19 20 21 Natural
c2 35.02% 14.06% 13.36% 13.12% 12.53% 11.90% 0.00%
c3 37.24% 13.47% 12.94% 12.67% 12.27% 11.42% 0.00%
c4 39.29% 13.25% 12.58% 12.16% 11.59% 11.13% 0.00%
c5 41.78% 12.29% 12.24% 11.78% 11.20% 10.71% 0.00%
c6 42.05% 16.36% 10.75% 10.69% 10.14% 10.02% 0.00%
c7 25.97% 37.00% 13.89% 7.83% 7.75% 7.57% 0.00%
c8 24.52% 12.75% 35.93% 12.75% 6.94% 7.10% 0.00%
c9 22.86% 12.00% 11.97% 35.02% 12.08% 6.07% 0.00%
ct 21.43% 11.04% 11.11% 11.05% 34.37% 3.36% 7.63%
cj 21.39% 11.11% 11.15% 11.17% 34.07% 3.44% 7.67%
cq 21.16% 10.97% 11.30% 11.13% 34.19% 3.47% 7.79%
ck 21.23% 11.27% 11.13% 11.24% 34.07% 3.49% 7.58%
ca 13.13% 12.88% 12.86% 12.71% 12.71% 4.95% 30.77%

As you can see, they are very close to each other.

There are advantages to both approaches and I had a hard time picking
what to
talk about in this summary. If accuracy is your biggest concern, you’re
probably better off sticking with the exhaustive search. However,
gets very close results, is probably a little easier to code up, and
would be an
option even if the search space was much larger (though the results may
be less
accurate for that).

Given that, I’m going to show a simulation solution here. If you are
interested in the exhaustive search, Eric I.'s code is well commented
and easy
to read.

The solution I want to look at below is Sander L.'s first submission.
not perfect and I’ll try to point out the problems as they come up, but
the code
is a very straight-forward simulation with a few good tricks in it. I
that makes it worth a read through.

Sander’s solution is written for Ruby 1.9, so it doesn’t run without
modification on earlier versions. I’ll point out the new features as we
go and
that will give us a chance to see how some of the new stuff gets used.

Here’s the start of the code:

class Array
def score
sort.inject(0){|s,c| s+c > 21 && c==11 ? s+1 : s+c }

This is Sander’s code for valuing Blackjack hands. Sander’s notion of a
hand is
simply an Array of 2 through 11 Integers. Blackjack isn’t affected by
suits, so
this code doesn’t bother to track them.

This code counts drops an ace an to 1 if counting it as 11 would bust
the hand.
Note that the hand is first sorted to push aces to the end and make sure
are counted last. Everything else gets the normal card value added on
to a
running total.

While, this code gets very close, it doesn’t work in all cases.
Consider a hand
containing a ten, an ace, and another ace. The code will count the ten,
eleven to it, because it wouldn’t bust the hand, then add one more.
This gives
a final total of twenty two, instead of the correct count of twelve. At
three solutions made the same error.

One way to fix the code is:

class Array
def score
s+c > 21 - (size - (i + 1)) && c==11 ? s+1 : s+c

This does the same thing, but reduces the hand limit for each card we
have left
to count once we reach the aces. They are sorted to the end so we know
we only
have aces left and we could choose to count them as one each. Note that
I used
a 1.9 feature here of calling each_with_index() without a block to get
Enumerator object. That allows me to inject() over the values with
their index.

Here’s the next bit of Sander’s code:

unless ARGV[0]
(2…11).each{|n| puts ruby1.9 #{__FILE__} #{n}}

I thought this was a great trick. As you are about to see, the rest of
the code
is written such that it only worries about a single upcard. That
simplified the
code, but it doesn’t let us see all of the results at once. To fix
that, Sander
just calls his own program once for each upcard, when one isn’t
provided. It’s
a recursive process call, so to speak.

The downside here is that this code didn’t run for me as written. I
call my
ruby 1.9 install ruby_dev, so the hardcoded name bit me. A more
portable way to
get the name would be:

unless ARGV[0]
require “rbconfig”
puts #{Config::CONFIG["ruby_install_name"]} #{__FILE__} #{n}

Let’s more on. Here’s the code that reads the upcard and number of
decks to

puts “upcard: #{upcard = ARGV[0].to_i}”
NDECKS = (ARGV[1]||2).to_i
CARDS = (((2…11).to_a+[10]*3)4NDECKS).tap{|c|
c.delete_at c.index(upcard)

We know we have an upcard if we made it this far, since the code before
calls us
with one when the user doesn’t pass one in. That makes it safe to read
in at
this point. The number of decks is also read if available, or the
default of
two is assigned when it’s not.

The last bit of this code builds a full deck. It does that by
generating the
values 2 through 11, adding on three additional 10 values for the face
multiplying the whole set by 4 for the suits, and multiplying that
standard deck
by the count of decks we want to play with.

This gives us the number of requested full decks, but we need to remove
already used upcard. That’s what the tap() call does and it’s new in
Ruby 1.9.
delete_at() returns the element deleted, but we need the resulting deck
instead, to store it for future use. That’s what tap() is for. You can
into a chain of calls to run some code, but you still get the object
itself back
as the result of the tap() call.

We’re now ready to run the simulation:

score_count = [0]*27
cards = []
cards = CARDS.dup.shuffle if cards.size < 17
dealer = [upcard]
dealer << cards.pop while dealer.score < 17
score_count[dealer.score] += 1

Believe it or not, that’s the entire simulation code.

It begins by creating an Array to hold the tally of scores, indexed by
total. It’s interesting to look at the size of this Array, initialized
twenty seven 0’s. That makes the highest Array index twenty six, which
is the
largest hand a dealer will ever bust with. That is, they can draw a
face card
when hitting a sixteen (a ace would count as one in this case). Indices
through sixteen won’t be touched, since a dealer won’t stop on these

The rest of this code creates an Array to hold the cards left, and
decides how
many rounds to run the simulation for based on user input or a default.
simulation loop then starts, doing just four things: deal and shuffle()
in Ruby 1.9) a fresh deck if we may not have enough cards to deal
another hand
(seventeen aces at minimum), initialize a hand with the upcard, deal
cards until we crest a total of seventeen, and record the final hand

The good point about this code is that it reuses the deck for as long as
it can.
This simulation gets fairly accurate results by trying many possible
hands with
each trip through the deck.

One downside is that it doesn’t distinguish between a total of twenty
one and a
natural. You could fix this by dedicating some slot in the score Array
for that
(or switching to a score Hash of and adding a check for
special hand after the hitting code.

With the results tallied, printing is all that remains:

puts %w[17 18 19 20 21 bust].join(’ ')
puts (score_count[17…21] << score_count[22…-1].inject(:+)).
map{|x| '%-4.1f%% ’ % (100.0*x / N )}.join

The only real point of interest in this code is another 1.9 idiom.
inject() has
be enhanced so that it can now take a Symbol argument for the method to
call in
the block. That greatly simplifies the common summing case, as we see

Again the other solutions and approaches to this problem were all
Do take the time to look through them.

My thanks to all the card players that decided to ante up for this quiz.

Tomorrow we will continue the Blackjack theme…