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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)=(1-p)*p(n-1) + *(1-p)*p(n-2). 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 n-2 times and then n-1 times with no consecutive heads. This allows you to calcuate p(n) in terms of p(n-1) and p(n-2). 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: $p(nT)= p(n-1)(1-p)$ $p(nH)= p((n-1)T)p$ This is because: The sequence up to n has no consecutive heads and ends in a tail if the sequence up to n-1 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 n-1 has no consecutive heads, ends in a tail and then a head is tossed. So, we have: $p(n)= p(nT) + p(nH) = p(n-1)(1-p) + p((n-1)T)p = p(n-1)(1-p) + p(n-2)(1-p)p$
October 6th, 2013, 02:20 PM   #4
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Re: Coin toss probability question

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 Originally Posted by Pero 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: $p(nT)= p(n-1)(1-p)$ $p(nH)= p((n-1)T)p$ This is because: The sequence up to n has no consecutive heads and ends in a tail if the sequence up to n-1 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 n-1 has no consecutive heads, ends in a tail and then a head is tossed. So, we have: $p(n)= p(nT) + p(nH) = p(n-1)(1-p) + p((n-1)T)p = p(n-1)(1-p) + p(n-2)(1-p)p$
Well, thanks so much for that.

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