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

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In a pistol duel between A and B, A has a 30% chance of hitting B, while B has a 50% chance of hitting A. The probability of A winning can be expressed as a summation of independent events where A hits B in successive shots. Each term in the summation accounts for the probabilities of misses and hits in a cyclic manner, reflecting the sequence of shots taken by both players. An alternative approach to solving the problem involves defining a probability variable for the first player to win based on their chances of hitting and the subsequent outcomes of misses. The discussion highlights the complexities of calculating probabilities in sequential events, emphasizing the need to consider independent shot outcomes.
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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?
 
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Each term in the summation is the probability of A killing B in the kth 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
 
make sense, thanks
Office_Shredder said:
Each term in the summation is the probability of A killing B in the kth 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
 
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.
 
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I assumed that they were shooting at the same time... if both die then nobody wins
 
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