0
$\begingroup$

An extension of flipping a fair coin and evaluating the number of changeovers (i.e. a head to a tail or a tail to a head) in $n$ trials, we can work out the distribution to be binomial: $${{n-1} \choose k} \cdot \left(\frac{1}{2}\right)^{n-1}$$ for $k$ changeovers in $n$ flips of the coin.

How would we go about modifying this for use with a biased coin, i.e. $P(\text{head})=\frac{3}{5}$?

$\endgroup$
  • $\begingroup$ Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Please consider rewriting your post. $\endgroup$ – mzp Apr 3 '18 at 1:20
  • 1
    $\begingroup$ P.S. Sorry people! I understand that the above holds as you are equally likely to choose a head or a tail if the coin is unbiased. When the coin is biased, however, naturally this fails. Is it necessary to take two separate distributions, one representing when you go from a tail to a head and vice-versa, or is that over-complicating the issue? $\endgroup$ – user548298 Apr 3 '18 at 1:41
0
$\begingroup$

Yes, I would consider two cases, depending on whether you start with heads or tails.

Suppose the probability of heads is $p$, but different flips are still independent. Let $A_n(k|f)$ be the conditional probability of $k$ changeovers in $n$ trials given that the first flip is $f$ (either H or T). Of course in $1$ flip we have $0$ changeovers, so $A_1(k|H) = A_1(k|T)$.

By conditioning on the second flip, we get

$$\eqalign{A_n(k|H) &= p A_{n-1}(k|H) + (1-p) A_{n-1}(k-1|T)\cr A_n(k|T) &= p A_{n-1}(k-1|H) + (1-p) A_{n-1}(k|T)\cr}$$

In terms of the probability generating functions $G_n(t|f) = \sum_{k=0}^{n-1} A_{n}(k|f) t^k$, we have

$$ \eqalign{G_n(t|H) &= p G_{n-1}(t|H) + (1-p) t G_{n-1}(t|T)\cr G_n(t|T) &= p t G_{n-1}(t|H) + (1-p) G_{n-1}(t|T)\cr}$$ or $V_n(t) = M V_{n-1}(t)$, where $V_n(t)$ is the vector $\pmatrix{G_n(t|H)\cr G_n(t|T)}$ and $M$ is the matrix $ \pmatrix{p & (1-p) t\cr p t & 1-p\cr}$. You can diagonalize $M$ and get explicit formulas for $G_n(t|f)$ and the combined probability generating function $G_n(t) = p G_n(t|H) + (1-p) G_n(t|T)$, but it will be complicated. Certainly not a binomial.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy