Probability and Tossing a CoinDate: 10/25/2004 at 13:21:47 From: Kevin Subject: Coin Toss... Probability Question What is the probability for HEAD face up 2 times in a row before TAIL face up a total of 10 times (or X times)? For example, T T T T H T T H H = Good T T T H T H T T T T H T H T = No Good I thought of a long way to calculate this, like adding HH, then THH, then HTHH, HTTHH, but I'm hoping there is a formula for this problem. Date: 10/25/2004 at 16:54:14 From: Doctor Vogler Subject: Re: Coin Toss... Probability Question Hi Kevin, Thanks for writing to Dr. Math. This is a counting problem. You will always end after, at most, 20 coin tosses. So the thing to do is count how many ways there are for a string of n H's and T's to have two H's in a row before getting to 10 T's. Then the probability of any single such string is 2^(-n), so you multiply the probability times the count, add those up for n from 1 to 20, and viola! Now, in order not to count a string twice, we should only count the ones that END in HH, and (except for the 2-long string HH) they should end in THH (or they would have ended sooner). So let's suppose our string has length n and ends with THH, so there are n-3 letters before the THH, (n-3 letters)THH and those n-3 letters contain no more than 8 T's (so that there are fewer than 10 after including the THH) and no double H. Well, how many H's could there be? There must be at least n - 3 - 8 = n - 11 H's (if n > 11, or there could be any number if n <= 11). So let's suppose that there are m H's in those first n-3 letters. Then how many ways are there to place them in those n-3 places? You might be tempted to say n-3 ( ), m also called "n-3 choose m," but you need to account for the fact that no two H's can be next to one another. So we might group each H with the T following it. Then we have to include the T from the THH, and then there are n-2 letters, but only n-2-m T's, and we just want to know which T's have an H before them. So the answer for a particular n is that there are n-m-2 sum ( ), m m where the sum is take from m = max(0,n-11) up to the biggest m with n-m-2 > m, which is floor((n-2)/2). Do this sum for each n. (Why do you get 0 when n > 20?) Multiply each result by 2^(-n), and add them all together. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
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