Question

MATALB MATALAB

Getting 4-of-a-kind

Consider the following experiment:

oYou draw 5 cards from a deck of 52 cards. This is considered a single experiment. If you get 4-of-a-kind the experiment is considered a "success".

oYou repeat the experiment N=100,000 times, keeping track of the"successes".

oAfter the N experiments are completed count the total successes, andcalculate the probability of getting4-of-a-kin

Answer 1

Hello,
       Please find the answer attached below. If the answer has helped you please give a thumbs up rating. Thank you and have a nice day!

********** Matlab Code **********

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Four-of-a-kind experiment in Matlab

%%%%%%%%%% Define the parameters %%%%%%%%%%
numOfTrials = 100000          % persons playing cards.
numCardsPerTrial = 5         % cards each person holds
four_Of_A_Kind = 0;

% Perform Monte Carlo simulations for the random distribution game:
for h = 1 : numOfTrials
   % Get the cards. Each card is numbered from 1 - 52
   cards_pick = randperm(52, numCardsPerTrial);                    % picking the cards randomly
  
   % Check for four of a kind
   sameSuit = (min(cards_pick) >= 1 && max(cards_pick) <= 13) || ...     % cards are numbered from 1 till 52
   (min(cards_pick) >= 14 && max(cards_pick) <= 26) || ...              % a gap of 13 in the card numbering indicates a different kind
   (min(cards_pick) >= 27 && max(cards_pick) <= 39) || ...
   (min(cards_pick) >= 40 && max(cards_pick) <= 52);
   if sameSuit
       four_Of_A_Kind = four_Of_A_Kind + 1;                            % counting the successes
    end
end

success = four_Of_A_Kind;               % total number of successes
prob_success = success/numOfTrials     % probability of the success

************ End of Code **********

Output:

numOfTrials =

      100000

numCardsPerTrial =

     5

prob_success =

    0.0019

Final answer: The code gives the probability of success as 0.0019 or 0.19%

NOTE 1: Since this is a random experiment, the answer that you obtain might be slightly different from the answer that is given here. The answers can be made quite close by repeating the experiment for a large number of times.

NOTE2: The complete program has been commented in a manner that it is easy to re-code and understand the code.