We have a rigged coin, the probability of obtaining heads is triple that of obtaining tails.
Consider the variable $X$ defined as follows: We flip the coin twice in a row.
- If Heads are obtained on the first roll and Tails on the second, we take $X = 1$.
- If you get Tails on the first roll and Heads on the second, we take $X = 2$.
- Otherwise, the coin is tossed twice (If you get any outcome other than or you repeat the process from the start.)
I want to find the density function of the variable but I'm confusing with that "Otherwise" means, the probability of getting head first and then tail is $\frac{3}{4}*\frac{1}{4} = \frac{3}{16}$, same for the first tail and then head. So $P[X=1] = 3/16$ and $P[X=2] = 3/16$, is this enough or I'm missing something? Any help is appreciated!
First, if the probability of heads is three times the probability of tails, let $\Pr[H] = 3\Pr[T] = p$. Then since $\Pr[H] + \Pr[T] = 1$, we obtain $$3(1-p) = p,$$ or $p = 3/4$. Therefore, for a sequence of two independent coin flips, $$\Pr[HH] = (3/4)^2 = 9/16, \\ \Pr[HT] = (3/4)(1/4) = 3/16, \\ \Pr[TH] = (1/4)(3/4) = 3/16, \\ \Pr[TT] = (1/4)^2 = 1/16.$$ If, after two coin flips, the outcome is neither $HT$ nor $TH$, then the coin is flipped two more times--i.e., the process is repeated--then we can see that because $\Pr[HT] = \Pr[TH]$, we must have $$\Pr[X = 1] = \Pr[X = 2] = 1/2.$$ This can also be computed as $$\Pr[X = 1] = \frac{\Pr[HT]}{\Pr[HT] + \Pr[TH]} = \frac{3/16}{6/16} = \frac{1}{2},$$ and similarly for $\Pr[X = 2]$.
The question becomes more interesting (and difficult) if one simply flips until the last two flips is either $HT$ or $TH$; e.g., if the sequence of flips is $HHT$, then $X = 1$ because the last two flips was $HT$. Similarly, $TTTTH$ will yield $X = 2$. These are not allowed in the original phrasing of the question. I leave the exploration of this situation as an exercise for the reader.