Blackjack how to define cards as their value

Hey everybody, I am trying to write a very simple blackjack code where the player either hits 21 and wins, or busts and loses. So far i have this,

function card = dealcard;

%This program will allow someone to draw a card at random from a full deck
and see what that card would be. you can go through the whole deck

persistent dec; %persistent makes it so matlab stores this variable

if length(dec)<53

      dec=[1:52]; %one deck of cards

      dec=dec(randperm(length(dec))); %this shuffles the cards
  end;
  card=dec(end);

I do not know how to define the cards as being an ace, 2,3,...queen, and king. If anyone could help me with this that would be great. Thank you.

Answer 1

[Image Analyst]

You don't need to make it persistent. Just pass the dec in. And to keep track of both the card number, and the card's value, you'll need a 2 D array.

cardNumbers = [1:52];
cardValues = [1:13, 1:13, 1:13, 1:13];
dec = [cardNumbers', cardValues']
% Now shuffle
shufflingOrder = randperm(size(dec, 1));
dec = dec(shufflingOrder, :)
% Prepare your output
cardNumber = dec(end, 1);
cardValue = dec(end, 2);

[Ryan B]

So keep most of what I have except for the persistent? Then add your code after it? thanks

[Image Analyst]

I would do this:

function [cardNumber , cardValue, dec] = dealcard(dec)
cardNumbers = [1:52];
cardValues = [1:13, 1:13, 1:13, 1:13];
dec = [cardNumbers', cardValues']
% Now shuffle
shufflingOrder = randperm(size(dec, 1));
dec = dec(shufflingOrder, :)
% Prepare your output
cardNumber = dec(end, 1);
cardValue = dec(end, 2);
% Remove that card from the dec so next time it's not there
dec = dec(end-1,:);

Of course I would make it better by using try catch, and things like that, but this is a good start. Be sure to make the first two lines be able to handle an incoming "dec" that has less than 52 cards in it (because some were dealt/removed). Actually I'd probably use an array of structures (to make it easier to remove cards from the dec and send it back in), but I was just trying to stick as close as possible to the simplistic way you were doing it.

[Ryan B]

Thank you very much for your help so far. I was wondering if you could help me out further by pointing me in the right direction of how to define the cards so there are 4 2's, 4 3's,.....4 9's, 16 10's (grouping the 10's, jacks, queens, and kings together). Or what the next step would be in making this code. Sorry im so bad at matlab. Thanks for your time

[Image Analyst]

After you assign cardValues, just set everything greater than 10 equal to 10. So add this line:

cardValues(cardValues>10) = 10

Alternatively you could do

cardValues = min(cardValues, 10);

[Ryan B]

after putting that in it is still giving me 47 and other large numbers. this is where i put it...

function [cardNumber , cardValue, dec] = Dealcards(dec)

cardNumbers = [1:52];

cardValues = [1:13, 1:13, 1:13, 1:13];

cardValues(cardValues>10) = 10;

dec = [cardNumbers', cardValues'];

%now shuffle

shufflingOrder = randperm(size(dec, 1));

dec = dec(shufflingOrder, :);

%prepare output

cardNumber = dec(end,1);

cardValue = dec(end,2);

%now removing card from deck

dec = dec(end-1,:);

[Image Analyst]

The cardValue can/will never be greater than 10. Obviously the cardNumber might be greater than 10 - it can go from 1-52 because there are 52 cards. What exactly is giving you 47 that you think should not?

[Ryan B]

Well i was trying to get it so it would come out with an answer of 2-11 simulating two through ace. I was hoping that i could get 4 of the cards to show an answer of 2, answer of 3 ect. instead of having 1 2, 1 3, 1 4... ect. I hope im making sense. Thanks

[Image Analyst]

If you want, just manually set up card values giving all 52 values. Then you can assign any value to any card number.

cardValues = [...
11, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 10,...
11, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 10,...
11, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 10,...
11, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 10];

(Assumes Aces are card numbers 1, 14, 27, and 40. Or set up whatever values you want for each of the 52 elements

[Ryan B]

Oh wow. I cant believe i didnt think of this. Thank you for all your help!