# Geometric Distribution - Biased Coin Flip

Whilst studying I stumbled on this problem, which I wish to check if my understanding is correct.

Imagine we have a biased coin with probability 'k' of getting a head when flipped. Now A defines the number of times a biased coin is thrown until getting the 1st Head. On the other hand, we have B which represents the number of times a biased coin is thrown until getting the 1st tail.

Question: Show that $p(1,b) = k^{b-1}(1-k)$, where $p(a,b)$ is the joint distribution of a & b.

My Understanding:

From the problem I believe that we can say that A ~ Geo(1) & B ~ Geo(b). Moreover, we can say that A & B are two dependent random variables as the number of flips in A affects the number of flips of B and vice versa.

Now, to find $p(1,b)$

$p(1,b) = P(A=1, B=b)$

$= P(B=b|A=1)P(A=1)$

{Using the Geometric distribution: $P(X=n)=(1-p)^{n-1}p$} $= [(1-(1-k))^{(b-1)-1}(1-k)] [(1-k)^{1-1} (k)]$

$= [(1-(1-k))^{b-2}(1-k)][(1-k)^{0}(k)]$

$= k^{b-2}(1-k)(k)$

$= k^{b-1}(1-k)$

Can someone verify if my understanding is correct? Thanks :)

EDIT:

Just to make sure that I have understood the answer given correctly: In case that this time we have $p(a,1)$

A~Geo(k)and B~Geo(1-k)

$p(a,1) = P(A=a, B=1)$

$P(B=1 | A=a)P(A=a)$

$=[(1-(1-k))^{(1-a)-1}(1-k)][(1-k)^{a-1}k]$

$=[k^{-a}(1-k)][(1-k)^{a-1}(k)]$

$k^{1-a}(1-k)^{a}$

\begin{align} p(1,b) &= P(A=1, B=b) \\ &= P(B=b|A=1)P(A=1) \\ &=P(B=b|A=1)k \\ &=\begin{cases}0 & , b=1 \\ k^{b-2}(1-k)k &, b > 1\end{cases} \\ &=\begin{cases}0 & , b=1 \\ k^{b-1}(1-k) &, b > 1\end{cases} \\ \end{align}

• Try to compute $p(a,1)$ similarly as an exercise.

Comment on your current answer to compute $p(a,1)$, well, the two problems are actually quite symmetrical, so the lack of symmetric in the answer that you obtained should warns us.

Your mistake was actually when you deal with $P(B=1|A=a)$. If $a=1$, well, it is equal to zero since we can't get two outcome simultaneously. If $a>1$, well, the first toss is not head, hence the first toss must be tail. Hence in summary:

$$P(B=1|A=a)=\begin{cases}0 &, a=1 \\ 1 &, a > 1\end{cases}$$

• As another exercise, try to compute $p(1,b)=P(A=1|B=b)P(B=b)$ to get the answer that I provided you earlier.

• Remark: A sanity check, you want to make sure the joint pmf sums up to $1$.

• Thanks a lot! Now I have understood what i was doing wrong! I will re-do the exercise with you suggestions. Again, thanks – vic12 Mar 3 '18 at 11:03

A variable X~Geo(p) where p is the probability of success for each trial. Since the probability of getting heads is 'k', A ~ Geo(k) And since B is the complement of A, their probabilities must sum to one. So, B~Geo((1-k)). The rest of your work seems correct.

• Thanks for the correction. So to make sure that I have understood correctly, I had added another section in the question under EDIT. Would you be so kind to correct my reasoning if need be. Thanks :) – vic12 Mar 3 '18 at 9:06