Hello I'm trying to solve the following problem : a coin is tossed with probability 1/2 of giving head but if it lands head the probability of head rise to .9. If it gives tail again or if the number of head is bigger equal 5 in a row then it fails down to .5.(adsbygoogle = window.adsbygoogle || []).push({});

I wrote a code running for 100000000 trials and found the probability of head

were near .703Code (C):

#include<stdio.h>

#include<time.h>

#include<stdlib.h>

double rnd()

{

int maxv=1<<24;

int val=rand()%maxv;

return((double)val/(double)maxv);

}

int main(void)

{

int iter=100000000;

int numpile=0;

double p=.5;

int pileserie=0;

srand(time(0));

for(int i=0;i<iter;i++)

{

if(rnd()<p) // choose head

{

numpile++;

pileserie++;

if(pileserie>=5)

p=.5;

else

p=.9;

}

else // else tail

{

pileserie=0;

p=.5;

}

}

printf("%lf\n",(double)numpile/(double)iter);

}

I would like to calculate this probability exactly I started with an empirical equation $$p (h)=\underbrace {(.9-(.9-.5)e^{-5a})}_{p_1-(p_1-p_0)e^{-an}}p (h)+p_0*p (t) $$ ??*

With $$p (h)+p (t)=1$$.

I don't understand how to find the exact formula * for the probabilities given the problem since the parameter a above should be fitted with numerical results.

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# I Coin toss with changing probability

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