Probability of at least 5 consecutive heads in 10 coin flipsby Biotic Tags: coin, consecutive, flips, heads, probability 

#1
May2112, 11:56 AM

P: 5

Everything is in the title, basically. We flip a coin 10 times. What is the probability of at least 5 consecutive heads?
I thought it was like this: Those 5 heads can start at spots 16 in 10 flips, so there are 6 possibilities. The rest of times, we can get anything, so there are 2^5 possibilities. That means 6*2^5 in total (favorable outcomes) Total outcomes: 2^10 so P=(6*2^5)/(2^10) However, this is supposedly not correct. Can someone tell me why and provide the solution? Thanks. 



#2
May2112, 12:02 PM

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P: 26,167

Hi Biotic!
(try using the X^{2} button just above the Reply box ) for example, you've counted 10 heads 6 times!Try again! 



#3
May2112, 12:03 PM

P: 5

1. The problem statement, all variables and given/known data
Everything is in the title, basically. We flip a coin 10 times. What is the probability of at least 5 consecutive heads? 2. Relevant equations 3. The attempt at a solution I thought it was like this: Those 5 heads can start at spots 16 in 10 flips, so there are 6 possibilities. The rest of times, we can get anything, so there are 2^5 possibilities. That means 6*2^5 in total (favorable outcomes) Total outcomes: 2^10 so P=(6*2^5)/(2^10) However, this is supposedly not correct. Can someone tell me why and provide the solution? Thanks. 



#4
May2112, 12:04 PM

P: 642

Probability of at least 5 consecutive heads in 10 coin flips
You already asked this: http://www.physicsforums.com/showthread.php?t=607785




#5
May2112, 12:21 PM

P: 5

Ok, so i did it manually counting separatelly for 5,6,7,8,9 and 10 consecutive heads.
For example, when 5 heads start from flip no. 1, 6th flip can only be tails and the rest can be anything  which is 2^{4} possibilities, the same when heads are at the end. But when they are in the middle, the first flip befpre and after have to be tails (to maintain only 5 in a row)  this means 2^{3} possibilities. In total, for 5 in a row, 64. Using the same method on other number of consecutive heads ther is a total of 112 favorable outcomes which is correct. However, is there a less kindergartenish way to do this? 



#6
May2112, 12:31 PM

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P: 4,672

The answer will be P(0,end) = Prob{>= 1 run of 5 or more} = 7/64. RGV 



#7
May2112, 01:00 PM

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P: 26,167

i don't think so
most probability questions don't have a neat solution 



#8
May2112, 04:35 PM

P: 325

The answer is related to the "Fibonacci nstep numbers", where n=5 in your case.
See http://mathworld.wolfram.com/FibonaccinStepNumber.html 



#9
May2212, 05:55 AM

P: 961

If all are head, only one event, but you're counting it 6 times.
At least 5. Isolated group of 5,6,7,8,9 &10 



#10
May2212, 06:42 AM

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P: 14,468





#11
May2212, 10:18 AM

P: 1

To get this in painful detail, you can follow my journey through these three posts on marknelson.us. The middle article details the path to the answer.
(Note: the board does not allow me to post links. If you go to marknelson.us and search on "heads", the three articles in question will pop up) Innumeracy, Revisited 20 Heads In a Row  What Are the Odds? A Big Problem That Doesn't Need a BigNum  Mark 



#12
May2312, 05:42 PM

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P: 9,205

For the present problem the number of tosses happens to be twice the run length of interest. As DH showed, that makes it much easier. That's because there's no room for two separated runs of length n in 2n tosses. For n heads in a row in 2n tosses we have chances of (n+2)/2^(n+1). With n=5 that's 7/64. 



#13
May2412, 06:33 AM

P: 1

>The calculation is wrong.
That's why there are three articles instead of one :) The second article is the one with the correct calculation.  Mark 



#14
Nov1512, 10:07 PM

P: 18

A good way to visualize all the possibilities is to think of the possibilities in sets, where the run of at least five consecutive heads starts at some toss, n, and a set includes all possibilities for some n. The sets of permutations can be specified by placing the first head of the run of at least five consecutive heads at some place, n, right after a tail (excluding n=1), where the sets are ordered for n=1,2,3,4,5, and 6 in the following way:
H, H, H, H, H, ..., ..., ..., ..., ... respresents the set of all combinations whose runs of at least five consecutive heads start with the first toss. The following five tosses can occur in any way; consequently, there are 2x2x2x2x2 = 32 ways these can occur. T, H, H, H, H, H, ..., ..., ..., ... represents the set of all combinations whose runs of at least five consecutive heads start with the second toss. The following four tosses can occur in any way; consequently, there are 2x2x2x2 = 16 ways these can occur. ..., T, H, H, H, H, H, ..., ..., ... represents the set of all combinations whose runs of at least five consecutive heads start with the third toss. The first toss can occur in two ways, and the last three can occur in any way; consequently, there are 2x2x2x2 = 16 ways these can occur. ..., ..., T, H, H, H, H, H, ..., ... represents the set of all combinations whose runs of at least five consecutive heads start with the fourth toss. The first two tosses can occur in any way, and the last two can occur in any way; consequently, there are 2x2x2x2 = 16 ways these can occur. ..., ..., ..., T, H, H, H, H, H, ... represents the set of all combinations whose runs of at least five consecutive heads start with the fifth toss. The first three tosses can occur in any way, and the last one can occur in two ways; consequently, there are 2x2x2x2 = 16 ways these can occur. ..., ..., ..., ..., T, H, H, H, H, H represents the set of all combinations whose runs of at least five consecutive heads start with the sixth toss. The first four tosses can occur in any way; consequently, there are 2x2x2x2 = 16 ways these can occur. By adding 32 + 16 + 16 + 16 + 16 + 16, it then follows that there are 112 distinct combinations (permutations; order matters) of ten tosses, each with a run of at least five consecutive heads. Also, there are 2x2x2x2x2x2x2x2x2x2 = 1024 ways a coin can be tossed ten times. Therefore, the probability of getting a run of at least five consecutive heads in ten tosses of a coin is 112/1024 = .109375 or 10.9375 %. Keep in mind that probability is a fancy term for the long term relative frequency of an event of a random phenomenon and is what one would tend to observe in a very long series of trials. If anyone is so inclined to verify this, try tossing a coin one thousand times and recording the number of sets of ten tosses out of one hundred where there were runs of at least five consecutive heads. If possible, schedule a coin toss event and have a group of friends, classmates, etc. carry out sets of ten tosses simultaneously to expedite this otherwise slow process. Then report back with your numbers. 


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