Nick and Penny are independently performing independent Bernoulli trials. For con- creteness ... Nick and Penny are independently performing independent Bernoulli trials. For con- creteness, assume that Nick is flipping a nickel with probability p1 of Heads and Penny is flipping a penny with probability p2 of Heads. Let X1, X2, · · · be Nick’s results and Y1, Y2, · · · be Penny’s results, with Xi ∼ Bern(p1), and Yj ∼ Bern(p2).
a) Find the distribution and expected value of the first time at which they are simul- taneously successful, i.e., the smallest n such that Xn = Yn = 1.
b) For p1 = p2, find the probability that their first successes are simultaneous, and use this to find the probability that Nick’s first success precedes Penny’s.

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*For part A, I made X = # of successful trials, got the geometric distribution to be X~Geom(p1p2), and got the expected value to be E(x) = 1/(p1p2).

*For part B, I did X = # of trials for Nick's 1st success, Y = # of trials for Penny's 1st success, X,Y~Geom(p1) and using LOTP,

P(X=Y) = ($p1^2$)/($1-(1-p)^2$)
Then finally P(X < Y) = (1-P(X = Y))/(2) = [ 1 - (($p1^2$)/($1-(1-p1)^2$) ] / [2]

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*I feel like both of my answers are off but I do not know where they are off. Any help would be very much appreciated.


Your answers are both right, but the results can be further simplified. Set $q=1-p$
$$\mathbb{P}[X=Y]=\sum_{k=1}^{\infty}p^2(q^2)^{k-1}=p^2\sum_{k=0}^{\infty}(q^2)^{k}=\frac{p^2}{1-q^2}=\dots=\frac{p}{2-p}$$
$$\mathbb{P}[X>Y]=\frac{1-\frac{p}{2-p}}{2}=\dots=\frac{1-p}{2-p}$$
