P(H heads in a row) given N coin flips

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October 3rd 2009, 03:04 PM
lambda

P(H heads in a row) given N coin flips

Hi,

I recently thought of the following problem and my basic grasp of probability and combinatorics is not enough to handle it:

You flip a coin N times. What is the probability P(N,H) that you get at least one sequence of at least H heads in a row?

Since this question deals with the number of times a result will happen, I tried to use a Poisson Distribution, but I'm having trouble separating the problem into mutually exclusive events.

I wrote a program to do a simulation and found that P(20,7) ≈ 0.058, so please make sure your answer fits with this result.

Thank you for your help!
October 3rd 2009, 04:42 PM
HallsofIvy

Quote:

Originally Posted by lambda

Hi,

I recently thought of the following problem and my basic grasp of probability and combinatorics is not enough to handle it:

You flip a coin N times. What is the probability P(N,H) that you get at least one sequence of at least H heads in a row?

Since this question deals with the number of times a result will happen, I tried to use a Poisson Distribution, but I'm having trouble separating the problem into mutually exclusive events.

I wrote a program to do a simulation and found that P(20,7) ≈ 0.058, so please make sure your answer fits with this result.

Thank you for your help!

I would do it this way: treat the "H heads in a row" as a single object and call it "R" for "row". With N flips and "H heads in a row", there are N- H other flips. How many different ways are there to order 1 object and N- H other objects?
October 3rd 2009, 09:31 PM
lambda

Problem - counts certain cases twice

Quote:

Originally Posted by HallsofIvy

I would do it this way: treat the "H heads in a row" as a single object and call it "R" for "row". With N flips and "H heads in a row", there are N- H other flips. How many different ways are there to order 1 object and N- H other objects?

Thank you for your reply.

I tried this approach and got $P(n,h) = \frac{acceptable\_series}{total\_possible\_series} = \frac{2^{n-h}(n-h + 1)}{2^n} = \frac{n-h + 1}{2^h}$ but kept getting answers that were roughly 2x too large.

I've realized that this approach counts certain answers twice. For example, for n=7 and h=3, the series HHHHHHH is counted 5 times, as RHHHH, HRHHH, HHRHH, HHHRH, and HHHHR. Do you have any tips on how to fix this?