Karin has an unfair coin; the probability of heads is $p$ $(0 < p < 1)$. She tosses the coin until she obtains heads. She then tosses a fair coin as many times as she tossed the unfair one. For every head she has obtained with the fair coin she finally throws a symmetric die. Determine the expected number and variance of the total number of dots Karin obtains by this procedure.
Let Biased Head be $X$, Fair Head be $Y$ and $one$ in dice throws be $Z$.
$X$~$Bern(p)$; Number of times she tosses biased coin $N$ ~ $Geom(p)$;
Total times she gets a head in $N$ trials will be: $S_N=X_1+X_2+...+X_N$,
thus expectation of getting head will be: $E[S_N]=E[X]E[N]=p\frac1p=1$
Now the number of trials $N$ is the average number of times she tossed a biased coin to receieve her first head $N=\frac 1 p$ is this right?
To get the number of time she obtained head in fiar coin game we note that the total number of heads is $F=Y_1+Y_2+...+Y_{\frac 1 p}$, so the expected number of heads is $E[F]=E[Y]E[1/p]$, $Y_i$~$Bern(1/2)$.
Let $U=\frac 1 p$, $F(U)=P(U\le u)=P(\frac 1 p \le u)=1-P(p\le \frac 1 u)=1-\frac 1 u$, if $u>1$, $0$ if $0\le u \le1$ and $1$ if $u<0$.
So, $f(u)=1/u^2$ if $u>1$; $0$ if $0\le u \le1$ and u if $u<0$
And I can't compute expectation because the integrals are convergent!
Can you give me a hint?
I think you have a mistake. She is tossing the die based on how many $\textit{fair}$ heads she gets, not $\textit{biased}$ ones. The expected number of fair heads is $\frac{1}{2}$ the expected number of fair flips, or $\frac{1}{2p}$. Therefore she expects to roll the die $\frac{1}{2p}$ times. Each roll has an expected value of $3.5$ dots, so the total expected value of dots is $$\frac{3.5}{2p}$$
I'll leave the variance for you to compute. I'll also remark that this concept of multiplying expectations where the number of terms is also a random variable, while intuitively logical, requires some justification. This is precisely what Wald's Theorem says.