Nick and Penny are independently performing independent Bernoulli trials. For concreteness, assume that Nick is flipping a nickel with probability p1 of Heads and Penny is flipping a penny with probability p2 of Heads. Let $X_1$ , $X_2$, . . . be Nick’s results and $Y_1$ , $Y_2$, . . . be Penny’s results, with $X_i \sim \operatorname{Bern}(p_1)$ and $Y_j \sim \operatorname{Bern}(p_2)$.
(a) Find the distribution and expected value of the first time at which they are simultaneously successful, i.e., the smallest n such that $X_n$ = $Y_n$ = 1.
Hint: Define a new sequence of Bernoulli trials and use the story of the Geometric.(b) Find the expected time until at least one has a success (including the success).
Hint: Define a new sequence of Bernoulli trials and use the story of the Geometric.(c) For $p_1$ = $p_2$ , find the probability that their first successes are simultaneous, and use this to find the probability that Nick’s first success precedes Penny’s.
Here is my solution. I'd like to know if I did this right.
a) Let $Z$ be the number of trials until (and including) the first time the are simultaneously successful. Then $Z \sim \operatorname{FS}(p_1p_2)$, where $\operatorname{FS}$ denotes the First Success distribution (equivalently, $Z-1 \sim \operatorname{Geom}(p_1p_2)$). Then $E(Z) = \frac{1}{p_1p_2}$.
b) Analogous to a), the results shoult be $\frac{1}{p_1+p_2 – p_1p_2}$
c) Let $p=p_1=p_2$, $S$ be the event that their first success occurs simulateously, and $S_i$ the event that their first success occurs simultaneously at the $i^{th}$ trial. Then
\begin{align}
P(S) = P(\cup_{i=1}^{\infty} S_i) &= \sum_{i=1}^\infty \left((1-p)^2 \right)^{k-1} p^2 \\
&=\frac{p^2}{(1-p)^2} \frac{(1-p)^2}{1-(1-p)^2} \\
&= \frac{p^2}{1-(1-p)^2} \\
&= \frac{p}{2-p}
\end{align}
The probability that Nick's success precedes Penny's should be -by symmetry- one half of the probability that their first successes do NOT occur simultaneously, and thus $\frac{1}{2-p}$.
EDIT
I think the answer to c) should be $\frac{1-p}{2-p}$
Best Answer
The probability that their first successes are simultaneously is $\frac{p}{2-p}. \quad \checkmark$
The probability that their first successes are not simultaneously is the converse probability: $$1-\frac{p}{2-p}=\frac{2-p}{2-p}-\frac{p}{2-p}=\frac{2-2p}{2-p}$$
Then the probability that Nick’s first success precedes Penny’s is the half of it.