What is the probability of A winning in a pistol duel against B?

[KFC]

I read a funny question about probability in a magazine. Here is the problem. A and B are to fight pistol duel. They are to fire at their choice of target in succession in the order A and B cyclically until only man is left unhit. A's chance of hitting B is 0.3, B's is 0.5. What's the probability for A to win finally? Here is the solution

$$0.5*0.3 + 0.5^2*0.7*0.3 + 0.5^3*0.7^2*0.3 + \cdots$$

I am trying to understand this expression. A goes first. For the first term, A have 0.3 to shot B and 0.5 of not being killed. So if A can't kill B and not being killed in the first round. Then A is going to shot B again, so there exists the second term. But how do you understand this term? Why there is only one 0.3 and two 0.5? What does the 0.7 represent?

 

[Office_Shredder]

Each term in the summation is the probability of A killing B in the $$k$$th shot. For A to kill B in the first shot, A has to hit, probability .3. B has to miss, probability .5.

For A to kill B in the second shot, A has to miss and B has to miss on the first shot. Probabilities .7 and .5. Then A has to hit and B has to miss on the second shot, probabilities .3 and .5 respectively. These events are all independent so you multiply the probabilities.

Repeat ad nauseum... each additional shot means one more miss for each of A and B, giving an extra .7 and .5

As to why it makes sense to add all these numbers up, I'll let you think about it

 

[KFC]

make sense, thanks

Each term in the summation is the probability of A killing B in the $$k$$th shot. For A to kill B in the first shot, A has to hit, probability .3. B has to miss, probability .5.

For A to kill B in the second shot, A has to miss and B has to miss on the first shot. Probabilities .7 and .5. Then A has to hit and B has to miss on the second shot, probabilities .3 and .5 respectively. These events are all independent so you multiply the probabilities.

Repeat ad nauseum... each additional shot means one more miss for each of A and B, giving an extra .7 and .5

As to why it makes sense to add all these numbers up, I'll let you think about it

 

[Gerenuk]

Hmm, that's weird. I'd rather expect that after one shoots and hits, the other is dead and can't shoot back. Your solutions seems to indicate that A is going second!

You can also solve the question another way. Let "p" be the probability for the first player to win when it's his turn. His chance to hit is p_1. If he misses, the other one gets the chance p_2 to win. So what should "p" be? It is the chance that you hit with your first shot OR
- you miss
- the other one misses
- "you win the remaining game being in the position to shoot first" (which is exactly "p" again)

So we can write down
$$p=p_1+(1-p_1)(1-p_2)p$$
Now can solve this for p.

 

[Office_Shredder]

I assumed that they were shooting at the same time... if both die then nobody wins