# Re: Studying Blackjack (#151) **URL:** \ **Category:** Ruby\ **Created:** [January 7, 2008, 4:41am UTC](https://www.ruby-forum.com/t/re-studying-blackjack-151/125214 "2008-01-07T04:41:26Z")\ **Posts on this page:** 1\ **Page:** 1
**Author:** ![jameskelly](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/j/3ab097/32.png) [@jameskelly](https://www.ruby-forum.com/u/jameskelly)\ **Post date:** [January 7, 2008, 4:41am UTC](https://www.ruby-forum.com/t/re-studying-blackjack-151/125214/1 "2008-01-07T04:41:26Z")
My solution works by expanding a nested “probability hash.” Each card is a key in this probability hash which contains another probability hash equivalent to what would happen if the dealer were to next gain that card. For example, the information on the situation in which the dealer starts with a 2 up and an ace down and then hits a jack would be contained in $probs[2][:A][:J]. In addition, each hash contains the following: 1)The hand of the dealer . E.g.: $probs[1][:A][:K][:hand]==[1,:A,:K] 2)The remaining deck, as represented by a card=\>number of instances of that card remaining hash. 3)The probability of this situation occurring. This is easily calculated as the probability of the parent situation occurring times the probability of a the new card being drawn. The program starts with the base case of full decks, empty hand, and probability of 1, and then expands down the sub-situations down to the point of 0-probability, bust, or dealer’s hand being above 17. The probabilities of all the sub-situations of an upcard are then summed and outputted. Interestingly, while most of my table is within rounding error of Dennis’s, the results for the aces are remarkably different. ``` 17 18 19 20 21 BUST ``` 2 13.94% 13.33% 13.07% 12.40% 11.93% 35.33% 3 13.28% 13.07% 12.46% 12.18% 11.54% 37.48% 4 13.07% 12.02% 12.10% 11.64% 11.31% 39.85% 5 12.10% 12.28% 11.73% 10.90% 10.73% 42.25% 6 16.62% 10.62% 10.67% 10.12% 9.75% 42.21% 7 37.05% 13.82% 7.80% 7.88% 7.34% 26.11% 8 12.97% 36.12% 12.90% 6.89% 6.96% 24.16% 9 12.09% 11.20% 35.41% 12.11% 6.10% 23.09% 10 11.29% 11.22% 11.30% 33.56% 11.31% 21.32% J 11.29% 11.22% 11.30% 33.56% 11.31% 21.32% Q 11.29% 11.22% 11.30% 33.56% 11.31% 21.32% K 11.29% 11.22% 11.30% 33.56% 11.31% 21.32% A 12.85% 13.09% 13.02% 13.12% 36.34% 11.59% Here’s my program: CARDS = (2…10).inject({}){|h,n|h[n]=n;h}.merge( {:J=\>10, :Q=\>10, :K=\>10, :A=\>nil}) MAX\_HAND = 21 HIT\_THRESHOLD = 17 NUM\_DECKS = 2 HIGH\_ACE = 11 LOW\_ACE = 1 CARD\_REPS = 4 class Array def count(obj) select{|el|el==obj}.size end def sum inject(0){|s,n|s+n} end end def sum\_hand(hand) sum = hand.map{|c|CARDS[c]}.compact.sum sum += hand.count(:A)\*HIGH\_ACE hand.count(:A).times{sum -= (HIGH\_ACE-LOW\_ACE) if sum \> MAX\_HAND} sum \<= MAX\_HAND ? sum : nil end def expand\_prob\_hash(prob\_hash) return if 0 == prob\_hash[:prob] or nil == sum\_hand(prob\_hash[:hand]) or HIT\_THRESHOLD\<=sum\_hand(prob\_hash[:hand]) CARDS.keys.each do |c| mod\_deck = prob\_hash[:deck].clone mod\_deck[c] -= 1 prob\_hash[c] = {:hand=\>prob\_hash[:hand]+[c], :deck=\>mod\_deck, :prob=\>prob\_hash[:prob]\* prob\_hash[:deck][c]/prob\_hash[:deck].values.sum} expand\_prob\_hash(prob\_hash[c]) end end def sum\_probs(prob\_hash) probs=((HIT\_THRESHOLD…MAX\_HAND).to\_a+[nil]).inject({}){|h,n|h[n]=0.0;h} if prob\_hash.has\_key? :A CARDS.keys.each do |c| prob\_part = sum\_probs(prob\_hash[c]) prob\_part.each\_pair {|k,v| probs[k] += v} end elsif 0 == prob\_hash[:prob] #do nothing else probs[sum\_hand(prob\_hash[:hand])] += prob\_hash[:prob] end probs end $probs = {:hand=\>[], :deck=\>CARDS.keys.inject({}){|h,c|h[c]=CARD\_REPS\*NUM\_DECKS;h}, :prob=\>1.0} expand\_prob\_hash($probs) puts " "+((HIT\_THRESHOLD…MAX\_HAND).to\_a+[“BUST”]).map{|el| “%8s”%[el]}.join [2,3,4,5,6,7,8,9,10,:J,:Q,:K,:A].each do |c| p = sum\_probs($probs[c]) printf "%2s ",c ((HIT\_THRESHOLD…MAX\_HAND).to\_a+[nil]).each{|n|printf "%6.2f%% ",p[n]_100_CARDS.size} puts end ----- Original Message ---- From: Ruby Q. [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) To: ruby-talk ML [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) Sent: Friday, January 4, 2008 7:04:45 AM Subject: [QUIZ] Studying Blackjack (#151) The three rules of Ruby Q.: 1. Please do not post any solutions or spoiler discussion for this quiz until 48 hours have passed from the time on this message. 2. Support Ruby Q. by submitting ideas as often as you can: [http://www.rubyquiz.com/](http://www.rubyquiz.com/) 1. Enjoy! Suggestion: A [QUIZ] in the subject of emails about the problem helps everyone on Ruby T. follow the discussion. Please reply to the original quiz message, if you can. -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= The majority of the strategy in Blackjack hinges around the dealer’s hand. The reasons are likely obvious to most of you: that’s the hand you have to beat and the dealer plays by fixed rules we can predict. For those unfamiliar with Blackjack, you only need to know a tiny bit about the game for the purposes of this exercise. The goal for both the player and the dealer is to draw cards to make a hand with the highest total possible, without going over 21. Going over 21 is called “busting” and it means you lose the hand. Face cards count for ten, aces are one or eleven (whichever is better for the hand), and all other cards count for their face value. You start with two cards and, if they happen to be a ten valued card and an ace (a count of 21), the hand is called a “natural.” A natural is an automatic win in most cases. The dealer begins with one of his two cards face up and one face down. We call the former the “upcard.” The dealer will “hit” or take more cards until he reaches a count of 17 or higher. After that he will “stand” or leave the hand where it is. That tells us that there are only seven possible outcomes for the dealer: get dealt a natural, bust, or hit to a total of 17, 18, 19, 20, or 21. We start every hand knowing half of what the dealer holds thanks to the upcard. Believe it or not, you can make pretty reliable guesses about how the hand will go with just that knowledge. Write a Ruby program that shows the percent chance of a dealer reaching each possible outcome based on the upcard showing. I’ll give you some hints to verify your results. Basic Blackjack strategy teaches that we should assume the dealer “has a ten in the hole” (as the face down card). It’s not always true, of course, but 17 is a common outcome for a dealer with an upcard of seven. Finally, we call five and six “the dealer’s bust cards” for reasons that will become obvious if you are outputting correct percentages. In the casinos Blackjack is often played with more than one deck shuffled together. One, two, six, and eight deck games are common. You may want to offer the option to adjust the deck size your program uses. Either way, let’s default to two decks as an average of what a player will face. ``` ____________________________________________________________________________________ ``` Never miss a thing. Make Yahoo your home page. [http://www.yahoo.com/r/hs](http://www.yahoo.com/r/hs)