Cool Probability Question

The probability of Tom missing and Brad hitting the target in the first try should be:

$(0.2)(0.8)$

If they both miss the first shots then there is again the chance of Tom missing and Brad hitting. If this goes on for an infinite number of times then the probability can be given by:

$(0.2)(0.8)\quad +\quad (0.2)(0.8){ (0.8) }^{ 2 }\quad +\quad (0.2)(0.8){ (0.8) }^{ 4 }\quad +\quad (0.2)(0.8){ (0.8) }^{ 6 }\quad ....$

Observably, this is a geometric series. The sum to infinity of a geometric sequence is;

${ S }_{ \infty }=\frac { a }{ 1-r }$

Now compute;

$=\frac { (0.8)(0.2) }{ 1-{ 0.8 }^{ 2 } } \\ =\frac { 4 }{ 9 }$

The probability that they both take the same number of shots to hit the bull's-eye is $(0.2)^2 + (0.8)^2(0.2)^2 + (0.8)^4(0.2)^2 + \cdots$ , where the $n$ th term is the probability that they both hit the target for the first time after $n$ shots.

The sum of this series is $\frac{(0.2)^2}{1-(0.8)^2} = \frac19$ . So the probability that one of them takes more shots than the other is $\frac89$ , and the probabilities are symmetric so the answer is $\frac49$ .

Redoing the calculations with $0.2$ replaced by $p$ , I get a general answer of $\frac{1-p}{2-p}$ .

Given that the probability to hit the target $p=0.2$ and hence that of failing to hit the target $q=0.8$ . The probability that Brad hits the target on his $i^{th}$ shot (after failing $i-1$ times) before Tom hits the target (that is failing all $i$ times) is given by: $P(i)=q^{i-1}pq^{i}=pq^{2i-1}$ And for all cases, $i =1, 2, 3... \infty$ is: $P =\sum _{ i=1 }^{ \infty }{ p{ q }^{ 2i-1 } } =pq\sum _{ j=0 }^{ \infty }{ ({ q }^{ 2})^{j} }$ $=pq\frac { 1 }{ 1-{ q }^{ 2 } } =\frac { 0.2\times 0.8 }{ 1-0.64 } =\frac { 0.16 }{ 0.36 } =\boxed{\frac { 4 }{ 9 }}$

The probability, $p$ , that they hit the target at the same time satisfies: $p=\frac{1}{25}+\frac{16p}{25}$ , so $p=\frac{1}{9}$ .

In the case when they do not draw either player is as likely as the other to take more shots. So the probability Andy takes more shots is $\frac{1}{2}(1-\frac{1}{9})=\frac{4}{9}$

The equation in the first paragraph is obtained by noticing that they hit the bullseye at the same time either by both hitting on the first shot (with probability $\frac{1}{25}$ ), or both missing the first shot (with probability $\frac{16}{25}$ ) and hitting together at some later time. If they both miss the first shot we can think of the game as having started over again and the probability they hit together at some later time is just $p$ .

the probability both of them take the same number of shot is $(0.2)^2+(0.8)^2(0.2)^2+.......=1/9$

so the probability that tom took more shots is $1-(1/9)/2=4/9$