Probabilities of archers A and B hitting the bull's eye

1. Jan 10, 2013

jinhuit95

1. The problem statement, all variables and given/known data
Two archers A and B take turns to shoot, with archer A taking the first shot. The probabilities of archers A and B hitting the bull's eye is 1/6 and 1/5 respectively. Show that the probability of archer A hitting the bull's eye first is 1/2.

2. Relevant equations

3. The attempt at a solution
Well, i thought about drawing tree diagram but the problem is I have no idea when to stop because I don't know when he will hit.

2. Jan 10, 2013

dx

Re: Probability

First, what are the possible ways for this to happen?

A could hit the bulls eye on his first attempt. Or, A misses, B misses and then A hits the bulls eye. Or, A misses, B misses, A misses, B misses and then A hits the bulls eye... and so on.

Find the probability for each of these ways (for A to hit the bull's eye first), and add them up.

You will need to use the following:

1 + a2 + a3 + ... = 1/(1-a) for |a| < 1

3. Jan 10, 2013

jinhuit95

Re: Probability

the problem is, if he hits on the first attempt that's 1/6. so If i go on, it will be 1/6 + ( 5/6 * 4/5 * 5/6 * 4/5 * 5/6 * 4/5....) I have no idea when to stop. Moreover, the product of probability he misses will be come smaller and smaller and that number + 1/6 will not give me half i think.

4. Jan 10, 2013

dx

Re: Probability

Let's take the case when A gets it on his second try: A misses, B misses, and then A hits. What is the probability for this?

5. Jan 10, 2013

jinhuit95

Re: Probability

that will be 5/6 * 4/5 * 1/6 right?

6. Jan 10, 2013

jinhuit95

Re: Probability

I'm sorry but what's this?
'1 + a2 + a3 + ... = 1/(1-a) for |a| < 1'

7. Jan 10, 2013

dx

Re: Probability

Yes, correct.

Now what is the probability that A gets it on his 3rd try? (A misses, B misses, A misses, B misses, A hits bull's eye)

Ignore that for the moment. We'll get to it soon.

8. Jan 10, 2013

jinhuit95

Re: Probability

5/6* 4/5 * 5/6 * 4/5 * 1/6

9. Jan 10, 2013

dx

Re: Probability

Correct.

So do you see the pattern?

First try: 1/6

Second try: (1/6)(4/6)

Third try: (1/6)(4/6)2

Fourth try: (1/6)(4/6)3

...

What is the probability that he gets in on the n'th try? Each time you are simply multiplying the previous one by (5/6)(4/5) = 4/6

10. Jan 10, 2013

jinhuit95

Re: Probability

1/6 * 4/6 ^ n-1!

11. Jan 10, 2013

dx

Re: Probability

Exactly.

Now you simply add them up. Let a = 4/6

(1/6) + (1/6)a + (1/6)a2 + (1/6)a3...

which is (1/6)(1 + a + a2 + a3 + a4...)

This is where you use the formula I mentioned before.

12. Jan 10, 2013

jinhuit95

Re: Probability

Got it! Thanks alot!!

13. Jan 10, 2013

dx

Re: Probability

No problem.

14. Jan 10, 2013

Ray Vickson

Re: Probability

Here is another approach: let x = the probability that A will win, given that A is just about to shoot.

If A hits the target on his first shot, he wins; the probability of this is 1/6. If A misses his first shot (prob. = 5/6), then for A to win it must be the case that B misses his first shot (prob = 4/5); at that point, A is just about to shoot and we are back to the starting point (win prob. = x at that point). So, altogether we have
$$x = \frac{1}{6} + \frac{5}{6} \frac{4}{5}\, x = \frac{1}{6} + \frac{2}{3} x.$$
Solving, we get x = 1/2.

15. Jan 10, 2013

CAF123

Re: Probability

@Ray
Is there a reason why doing: $$x = \frac{1}{6} + \frac{2}{3}x + \frac{5}{6} \frac{4}{5} \frac{5}{6} \frac{4}{5}x,$$ and solving is incorrect? (I am just considering the possibility of both A and B losing until A hits on his third go).

16. Jan 10, 2013

Joffan

Re: Probability

My friend Kim and I had a similar competition, but we're much worse. I am slightly worse than Kim, so I went first. On any one shot, my chance of hitting the bull is 1/39, and Kim's chance is 1/38.

Show that, once again, this is an even competition - my chance of getting the bull's eye first is 1/2.

17. Jan 10, 2013

CAF123

Re: Probability

Are you stuck? If so, just use exactly the same method as you used to solve the first one. Alternatively, say you win on your $i$th shot. This means you (or Kim) cannot have hit the bull on the previous $i-1$ attempts.
What is the prob of you not hitting the bull on the $i-1$ attempts?
What is the prob of Kim not hitting the bull on her $i-1$ attempts?
What is the prob of you hitting on the $i$th trial given that you and Kim failed on the $i-1$ trials?
By the independence of trials you can then multiply these together and sum.
(This is really the same method by dx just reworded a bit differently)

18. Jan 10, 2013

haruspex

Re: Probability

Just use Ray's method again. It's much the simplest way to solve these problems. Did you understand Ray's method?

19. Jan 10, 2013

Joffan

Re: Probability

Sorry for the misunderstanding guys. I was giving an extension question to illustrate the reasoning; I'm not stuck.

20. Jan 10, 2013

Ray Vickson

Re: Probability

We can see it is incorrect by solving it and getting the wrong answer (x = -3/2 instead of x = 1/2). To understand WHY it is wrong, go back and look at the definition of x and the argument I gave for the equation satisfied by x. The point is that x already includes all the possibilities of winning on the first, third, fifth, .... go.

However, it would be correct to write
$$x = \frac{1}{6} + \frac{2}{3}\frac{1}{6} + \frac{2}{3}\frac{2}{3}\,x$$
for similar reasons. This equation does have the correct solution x = 1/2.

Last edited: Jan 10, 2013