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September 30th, 2013, 06:13 AM  #1 
Member Joined: Oct 2011 Posts: 81 Thanks: 0  Coin toss probability question
Question A coin has a probability p of showing head when tossed. It is tossed n times. Let p(n) denote the probability that no two or more consecutive heads occur. Prove that p(n)=(1p)*p(n1) + *(1p)*p(n2). Approach: My approach was by using Gap method. I filled all the n places with tails placed alternately and then worked out possibility of filling heads in those gaps because heads cannot occur consecutively. Also in that gap, i considered the possibility of filling a tail as well. But i couldn't express it in the general form given in the question. Am i applying the right approach? Please suggest a method or mistakes in my method. Any help would be appreciated. PS: I have a brief knowledge in Mathematical induction and would like to know if induction can be used in this question. Thanks. 
September 30th, 2013, 06:41 AM  #2 
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Re: Coin toss probability question
It's a sort of induction. The key point is: If you toss a coin n times and there are no two or more consecutive heads, then you must have tossed the coin n2 times and then n1 times with no consecutive heads. This allows you to calcuate p(n) in terms of p(n1) and p(n2). The other key point is whether such a sequence has a probability p of ending in a head or whether it's more likely to end in a tail. I'd take p = 0.5 and take a look at sequences to determine this. 
September 30th, 2013, 08:03 AM  #3 
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Re: Coin toss probability question
I did it like this. Let p(nT) = probability that there is no sequence of heads in n tosses and the sequence ends in a tail. And, p(nH) = the same but the sequence end in a head. First note that: This is because: The sequence up to n has no consecutive heads and ends in a tail if the sequence up to n1 has no consecutive heads and a tail is then tossed. The sequence up to n has no consecutive heads and ends in a head if the sequence up to n1 has no consecutive heads, ends in a tail and then a head is tossed. So, we have: 
October 6th, 2013, 02:20 PM  #4  
Member Joined: Oct 2011 Posts: 81 Thanks: 0  Re: Coin toss probability question Quote:
 

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