Yahoo Answers: Answers and Comments for Coin Probability (Pattern) Question? [Mathematics]
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Thu, 20 Mar 2008 19:28:22 +0000
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Yahoo Answers: Answers and Comments for Coin Probability (Pattern) Question? [Mathematics]
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https://answers.yahoo.com/question/index?qid=20080320192822AAEG6iq
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From Low Key Lyesmith: Let x, y, and z be the expected number of toss...
https://answers.yahoo.com/question/index?qid=20080320192822AAEG6iq
https://answers.yahoo.com/question/index?qid=20080320192822AAEG6iq
Thu, 20 Mar 2008 20:38:58 +0000
Let x, y, and z be the expected number of tosses until one sees H, HH, and HHH, respectively.
x = 2 because the probability of heads is 1/2. (In general, if the probability of heads is p, then the expected number of tosses until heads appears is 1/p.)
Before we can get HH, we must wait for an H. Once we get an H, we have a 1/2 chance of getting HH on the next toss; otherwise we must start from the beginning.
So y = x + .5 + .5(y+1), hence y = 6.
Before we can get HHH, we must wait for HH. Once we get HH, we have a 1/2 chance of getting HHH on the next toss; otherwise we must start from the beginning.
So z = y + .5 + .5(z+1), hence z = 14.
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We analyze the pattern HTH in a similar fashion. Let x, y, and z be the expected number of tosses until one sees H, HT, and HTH respectively.
As before, x = 2.
In order to get HT, we must first wait for the first H, and then we must wait for the first T. Each stage requires 2 tosses on average, so y = 4.
Once we get HT, we have a .5 chance of getting HTH on the next toss; otherwise we must start from the beginning.
So z = y + .5 + .5(z + 1), hence z = 10.
EDIT: One could also find x = 2 by solving x = .5 + .5(x+1).