# probability of a run of k heads or more in N tosses of a fair coin N>>k

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### Accepted Answer

Image Analyst
on 6 Nov 2013

Edited: Image Analyst
on 7 Nov 2013

It's pretty trivial if you have the Image Processing Toolbox:

numberOfTosses = 1000000;

% Throw coin numberOfTosses times with tails defined as 0, heads as 1.

tosses = randi(2, 1, numberOfTosses)-1;

% Use the Image Processing Toolbox to find stretches of N.

N = 20; % Look for 20 in a row.

measurements = regionprops(logical(tosses), 'area');

% Extract the lengths of all stretches of heads in a row.

allRuns = [measurements.Area];

% Count the number of times it's N or more

numberOfLongStretches = sum(allRuns >= N)

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### More Answers (2)

Roger Stafford
on 7 Nov 2013

Here is an iterative method of finding the probability. Let k and N be the quantities you described. Then perform this matlab code:

p = zeros(N,1);

p(k) = 1/2^k;

if N > k, p(k+1) = p(k) + 1/2^(k+1); end

for n = k+2:N

p(n) = p(n-1) + (1-p(n-k-1))/2^(k+1);

end

Then p(n) is the probability for k consecutive heads out of n tosses for each of the values of n in 1<=n<=N. Note: This is not a Monte Carlo method; it is an exact computation.

This is based on the notion that if p is already evaluated from p(1) to p(n-1), then the probability p(n) event occurs in two mutually exclusive ways. Either the first n-1 of the n tosses contains such a contiguous sequence of k heads or else the last k of the n tosses are all heads, the toss immediately prior to them is a tails, and the remaining initial n-k-1 of them contains no such sequence of k consecutive heads. There is no other way this event can occur with the n tosses. The first of these ways clearly has probability p(n-1) and the second has probability equal to the product

1/2^(k+1)*(1-p(n-k-1))

and hence p(n) is their sum.

I ran this for two cases and arrived at the following results:

k = 3;

p(10) = 0.5078125 = 520/1024

k = 5;

p(50) = 0.5518753229536166 = 621356374702220/1125899906842624

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Ben Petschel
on 7 Nov 2013

The method in the link you gave can be modified to avoid the need for arbitrary-precision integer arithmetic in calculating the generalized Fibonacci numbers. Basically the solution to the recurrence relation can be expressed in terms powers of the roots of the characteristic polynomial. Then the solution can be written as:

r=@(k)roots([1,-ones(1,k)]); % roots of the characteristic polynomial

a=@(k)[1,zeros(1,k-1)]/vander(r(k)); % coefficients are solution to a Vandermonde system

% Have Fk(n) = a(k)*r(k).^(n+k-1) with Fk(0)=1

p=@(n,k)1-(2^k)*a(k)*(r(k)/2).^(n+k);

Some particular cases:

p(4,2) = 0.5000

p(10,3) = 0.5078 + 0.0000i

p(50,5) = 0.5519 - 0.0000i

p(1e6,20) = 0.3793 + 0.0000i

The small imaginary part in the solution is due to roundoff but it is about O(eps) in all cases indicating that the result is reasonably accurate.

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