Solve Probability Question: Tossing 3 Heads in a Row out of 64 Tosses

AI Thread Summary
The discussion focuses on calculating the probability of tossing three consecutive heads in 64 coin tosses. The initial interpretation clarifies whether the question pertains to getting exactly three heads in a row or at least once. It is established that there are 62 groups of three consecutive flips in 64 tosses, with a 1 in 8 chance for each group to yield three heads. The calculations involve multiplying the probabilities of not achieving three heads in previous groups to determine the overall likelihood. The total probability approaches 1, indicating a high chance of getting at least one occurrence of three consecutive heads.
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hi, i posted this question in the probability forum here but no one is even browsing that one.

my brother needs to know this but I'm in exam block right now and can only think of the physics I am studying this term. i can't remember how to figure this out anymore i just know i'll probably have to use nCr and nPr.
heres the question:
how do i find the probability of getting tossing 3 heads in a row out of 64 tosses?

thanks, much appreciated.
 
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Do you mean getting 3 heads and 61 tails or 3 heads in a row?
 
My interpretation was getting 3 heads in a row (at least once or exactly once? Exactly three or would 4 heads in a row also count?) anywhere in the 61 flips. Looks to me like a very hard problem with a lot of calculation involved.
 
I will assume the problem is having at least 3 consecutive heads at least once.
In 3 flips you have the following possibilities:
TTT
TTH
THT
THH
HTT
HTH
HHT
HHH
So you have a 1 in 8 probability to get threee heads in a row with 3 flips.
You have 62 groups of three consecutive flips in 64 tosses:
1 2 3 - 2 3 4 - ... - 62 63 64.
You have 1/8 probability of getting 3 heads in the first group and 7/8 of not getting it.
The probability of getting the 3 heads in the second group given that you did not get it in the first is
7/8 * 1/8 = 7/64
So you have a 57/64 chance of not getting 3 heads in the first 2 groups.
The probability of getting the 3 heads in the third group given that you did not get it in the first two is:
57/64 * 1/8 = 57/512
and so on...
The total probability is the sum 1/8 + 7/64 + 57/512 + ... very near 1.
 
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