How can I calculate 5 lions in a row?

Question:A fair coin is tossed 10 times. What is the probability of getting exactly 5 lions in a row? ( 6 times in a row Don't count the number of taps or more.)

my code:

clear all

close all

clc

for i=0:1023

x=fliplr(de2bi(i));

coin(i+1,:)=[ zeros(1,10-length(x)) x ];

end

% =============================================

z=0;

yek=zeros(1,1024);

for i=1:1024

x=coin(i,:);

for j=1:10

if(x(:,j)==1)

z=z+1;

yek(1,i)=z;

end

end

z=0;

end

yek=yek';

y=zeros (253,1);

m=0;

for i=1:1024

if yek(i,:)>=5

y(m+1,1)=i;

m=m+1;

end

end

c=0;

q=0;

for i=y'

x=coin(i,:);

c=0;

n=0;

% ===============================

for j=1:9

if x(1,j)==1 && x(1,1+j)==1

c=c+1;

else c=0;

end

if c==4

n=4;

end

end

if n==4

q=q+1;

end

w(i,1)=c;

e(i,:)=x ;

end

my question:How can I extract 5 lions in a row through generated states?

Answer 1

[Torsten]

n = 1024;
S = 0:(n-1); % 2^10 possible sequences of coin tosses.
A = de2bi(S);
A = [zeros(n,1),A,zeros(n,1)];
nn = 0;
for i = 1:size(A,1)
  idx = strfind(A(i,:),[0 1 1 1 1 1 0]);
  if ~isempty(idx)
   nn = nn + 1;
  end
end
format long
fraction = nn/n

fraction = 
   0.062500000000000

[John D'Errico]

But there is no need to do a large simulation though, since the entire problem space contains only 1024 possible sequences, each of which is equally probable.

[Torsten]

Correct. I borrowed some code of yours :-)

Answer 2

[John D'Errico]

I'm so used to coins having heads or tails though. ;-) I guess I can deal with lions.

Anyway, your code gets at the question at hand, but you can make it simpler.

S = 0:1023; % 2^10 possible sequences of coin tosses.
sequences = de2bi(S)

sequences = 1024×10

0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0

I dropped the semicolon so we could see the first few sequences. All you need now is to count the number of sequences where there were at least 5 tosses in a row.

Probably it is easiest here to just use a loop, despite the fact that it COULD be vectorized.

count5 = 0;
for i = 1:2^10
    count5 = count5 + ~isempty(strfind(sequences(i,:),ones(1,5)));
end
count5

count5 = 112

count5/2^10

ans = 0.1094

Look carefully at the inner line of code there. It uses strfind to identify sequences that have a sub-sequence of length 5, that matches the string of interest. Since all you care about is that it finds a sub-sequence of length at least 5, this is adequate. Then the ~isempty call retiuurns true, when such a sequence was found.

[Torsten]

@John D'Errico

What is the probability of getting exactly 5 lions in a row?

[John D'Errico]

Oops. I missed the word exactly. But that is easy enough to fix. Appending a 0 at each end, and then searching for the subsequence [0 1 1 1 1 1 0] is the solution that I would use. Since I see Torsten does exactly that, he is correct.

Answer 3

[Image Analyst]

Yet another way (Monte Carlo)

numExperiments = 100000
numTosses = 10
tosses = randi([0, 1], numExperiments, numTosses);
maxLength = zeros(numExperiments, 1);
% Find the lengths of each run of 1's
for k = 1 : numExperiments
    if nnz(tosses(k, :)) == 0
        continue;
    else
        props = regionprops(logical(tosses(k, :)), 'Area');
        maxLength(k) = max([props.Area]);
    end
end
fraction5 = sum(maxLength == 5) / numExperiments

fraction5 =

0.06296