1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Probability involving coin flips

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data
    We are flipping a coin with probability p of getting heads n times.
    A "change" occurs when an outcome is different than the one before it. For example, the sequence HTHH has 2 changes.
    If p=1/2 what is the probability that there are k changes?

    2. Relevant equations
    I've been working with the probability mass function of a binomial random variable:
    (n C k) pk(1-p)n-k

    3. The attempt at a solution
    For the n flips there are n-1 possible "gaps" between flips when change could occur.
    I then reasoned that at the end of every flip since you a flipping a fair coin, there is a 1/2 chance of getting a change and a 1/2 chance of not getting a change. My resulting formulation for probability of k changes in n flips was:
    (n-1 C k)((1/2)k)((1/2)n-k)
    but I worked out explicitly the probabilities of k changes for n=2, 3, and 4 and this function did not give me at all correct answers. I'm not sure how I should approach it differently.
     
  2. jcsd
  3. Sep 21, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You've the right idea with (n-1 C k) counting the numbers of ways to get k flips in n throws in terms of where it flips. So how many total ways are there of throwing the coin n times and getting k flips? Now how many ways of throwing the coin n times without that restriction? Isn't the ratio going to be the probability?
     
  4. Sep 21, 2009 #3
    "So how many total ways are there of throwing the coin n times and getting k flips? "
    I say it's (n-1 C k) again.

    "Now how many ways of throwing the coin n times without that restriction?"
    2n

    So this would give (n-1 C k)/2n, but when I run it against my calculated probabilities this gives half the value of the original answer. So it seems I should multiply by 2 in the formula. Not sure how to justify that though. Is it because you can have changes from H to T and T to H?
     
  5. Sep 21, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, if you have k flips, you can either start with H or T. 2*(n-1 C k) total ways, right?
     
  6. Sep 21, 2009 #5
    Okay that was my line of reasoning. I just wanted to be sure I wasn't making a major miscalculation.
    Thanks very much for your clear explanation.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Probability involving coin flips
  1. Coin flip Probability (Replies: 1)

Loading...