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September 30th, 2013, 06:13 AM   #1
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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.
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September 30th, 2013, 06:41 AM   #2
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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.
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September 30th, 2013, 08:03 AM   #3
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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 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:

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October 6th, 2013, 02:20 PM   #4
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Re: Coin toss probability question

Quote:
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:





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:

Well, thanks so much for that.
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