# Two series of independent Bernoulli trials. Find distributions of being simultaneously successful and of first success being simultaneous.

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 $\frac{1-p}{2-p}$

• I think there may be a slip at the end. Dec 20, 2014 at 16:10
• Hi! In the third question, how did you go from the summation to the second line equation? More specifically, I'm wondering how you got the $(1-p)^2$ on the numerator for the right-hand side? My understanding of the geometric series is that it's $\frac{1}{(1-p)^2}$.
– Sean
Oct 20, 2018 at 9:11
• @Seam Hi, set $q = (1 - p)^2$ and apply the formula for the geometric series, then you arrive at the result after back substitution and solving the binomial formula. I don't know why I extended the fraction with the $(1 - p)^2$ in the first place. Oct 21, 2018 at 13:59

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: $$1-\frac{p}{2-p}=\frac{2-p}{2-p}-\frac{p}{2-p}=\frac{2-2p}{2-p}$$