# Probability of at least 5 consecutive heads in 10 coin flips

by Biotic
Tags: coin, consecutive, flips, heads, probability
 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 1-6 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.
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PF Gold
P: 26,122
Hi Biotic!

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 Quote by Biotic I thought it was like this: Those 5 heads can start at spots 1-6 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)
No, you're double-counting …
for example, you've counted 10 heads 6 times!
Try again!
 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 1-6 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.
P: 635

## Probability of at least 5 consecutive heads in 10 coin flips

 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 24 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 23 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?
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P: 4,504
 Quote by Biotic 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 1-6 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.
Do you know about Markov chains? If so, you can model this using a Markov chain with states 0,1,2,...,10, where at the end of toss n we are in state k if we have just finished a run of k heads in a row. The one-step transition probabilities are P(0 --> 0) = 1/2, P(0 --> 1) = 1/2, and for i >= 1, P(i --> 0) = 1/2 (we get tails), P(i --> i+1) = 1/2 (we get another head to add to the run). You want to know the probability of reaching states 5,6,7,8, 9 or 10 at least once by toss n = 10. We can model this by amalgamating states 5,6,7,8,9,10 into a single absorbing state (end) and compute the probability that we are in state 'end' at toss n = 10. In the modified chain the transition probabilities are as before for i = 0,1,2,3, but P(4 --> 0) = 1/2, P(4 --> end) = 1/2 and P(end --> end) = 1. You get the answer forming the 6x6 transition matrix P = (P(i --> j)) and then taking the 10th matrix power P10 = P^10 (= P*P* ... *P, which is matrix multiplication 9 times); the element P10(0,end) is the required probability.

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

RGV
 Sci Advisor HW Helper Thanks PF Gold P: 26,122 i don't think so most probability questions don't have a neat solution
 P: 325 The answer is related to the "Fibonacci n-step numbers", where n=5 in your case. See http://mathworld.wolfram.com/Fibonaccin-StepNumber.html
 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
Mentor
P: 14,243
 Quote by Biotic 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?
A bit simpler is to go back to the idea you had in the original post.
 Quote by Biotic I thought it was like this: Those 5 heads can start at spots 1-6 in 10 flips, so there are 6 possibilities. That means 6*2^5 in total (favorable outcomes)
2^5=32 possibilities is the correct number for a sequence of 5 or more heads starting with the first flip. This is incorrect for a sequence starting at flip #2 to flip #6. Rhetorical question: What does it mean for a sequence to start at (for example) flip #2? The answer is that the first flip must have been tails. You have already accounted for the first six flips being heads in that 2^5 possibilities for a sequence of five or more heads starting at flip #1. This the total number of possibilities is 2^5 + 5*2^4 = 112.
 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
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