How to similuate a coin flip with probablility p

How do I simulate getting a result, either 0 or 1, with probability p. So if p=0.5 I should get an output of 0 half of the time, and 1 half of the time.

[ali fadi]

probability and statics the probability function flipping 3 coins in matlab

Answer 1

[Wayne King]

100 tosses with p=0.5.

x = round(rand(100,1));

If you want a probability other than p=0.5, then realize that rand() is uniform random number generator between [0,1], so you can assign the output of rand() accordingly. For example, for p=0.25:

y = zeros(100,1);
x = rand(100,1);
y(x<0.25) = 1;

Or if you just want to simulate the number of 0's or 1's in a certain number of trials. Say 100 for example. Here is a simulation of ten such experiments. Requires Statistics Toolbox.

R = binornd(100,0.5,10,1);

Answer 2

[Nick]

Alternatively you could use the randi function in MATLAB which generates random integers.

100 tosses with 0=heads, 1=tails

coin=randi([0:1], [100,1])

It should more or less give you 50 0's and 50 1's.

If there is more than 2 possible outcomes and they all occur with the same probability then just increase the integer range of the randi function.

[Evangelia Lo]

What if i want a program whick calculate how many times it will give 0000000000 (10×0)like that in a row ...How can i do it ?

[Evangelia Lo]

Which *

[Image Analyst]

@Evangelia Lo, if you have the Image Processing Toolbox you can use bwareafilt() to extract runs of only the length you want and then bwlabel() or regionprops() to count them. If you want to find the index where each run starts, use regionprops():

coins = logical(randi([0, 1], 1, 30000)); % 30,000 coins
patternLength = 10; % Whatever you want
% Throw out all regions that don't match the length we want using bwareafilt().  Need to invert.
coins2 = bwareafilt(~coins, [patternLength, patternLength]);
% Count them
props = regionprops(coins2, 'PixelIdxList');
numPatternOccurrences = numel(props)
% For fun, find out where they occur.
indexes = vertcat(props.PixelIdxList); % All 10 indexes in a run
startingIndex = indexes(1 : patternLength : end) % Just the starting index.

[Evangelia Lo]

How can i do it without the Image Processing Toolbox ?

Answer 3

[Cory London]

This code is easily minupulated and works pretty well

Ntoss = 100;
x = rand(1, Ntoss); % what is the resulting size of x?
toss = (x < 0.5);

[Umar Abdulhamid]

Can you briefly explain this for me please?

Answer 4

[SARATHRAJ V]

Ntoss = 100; x = rand(1, Ntoss); % what is the resulting size of x? toss = (x < 0.5);