Expected value and probability

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SUMMARY

The discussion revolves around calculating the expected value E[X] in a shooting game scenario involving two players, A and B. Player A has a probability P1 of hitting player B, while player B has a probability P2 of hitting player A. The random variable Y is defined to represent the outcomes of the first shots: Y=0 if A hits B, Y=1 if B hits A, and Y=2 if both miss. The expected values for these outcomes are calculated as E[X|Y=0]=1, E[X|Y=1]=2, and E[X|Y=2]=2 + E[X], leading to a recursive relationship for E[X].

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yevi
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I have the following question:
2 persons shoot it each other. Person A shoots at Person B, if A misses, B shoots at A and so on.
The Game continues until one of them hits the other one.
Probability that A hits B is P1 ,and probability that B hits A is P2.

I need to find the E[X].

They give some guidance:
Declare another Random variable Y as follows:
Y=0 if person A hits person B on fist shot.
Y=1 if person B hits person A on fist shot.
Y=2 if none of them hits on first shot.

I also have a solution, that I don't understand:

The idea is: E(X)=E[E[X|Y]]=\sumE[X|Y=y]P(y) (this I understand)
What I don't understand is how they've calculated the following:
g(0)=E[X|Y=0]=1
g(0)=E[X|Y=1]=2
g(0)=E[X|Y=2]=2+E[X]
 
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yevi said:
g(0)=E[X|Y=0]=1
g(0)=E[X|Y=1]=2
g(0)=E[X|Y=2]=2+E[X]

Hi yevi! :smile:

I don't undestand what "g(0)" means, but the rest is fairly clear:

If Y = 0, then that means that the game stops at 1, so the expected value is 1.

If Y = 1, then that means that the game stops at 2, so the expected value is 2.

If Y = 2, then that means that the game starts all over again, as if there had been no shots, and so the expected value is 2 more than it was at the start. :smile:

(But this is a really cumbersome way of doing it. :frown:

It would be more straightforward to say:
E[X|X=1]=1
E[X|X=2]=2
E[X|X>2]=2 + E[X])
 
tiny-tim said:
If Y = 0, then that means that the game stops at 1, so the expected value is 1.

If Y = 1, then that means that the game stops at 2, so the expected value is 2.

What do you mean stops at 1 or at 2?
 
yevi said:
What do you mean stops at 1 or at 2?

I mean the game stops at 1 shot, or at 2 shots.

eg, if Y=0, then A hits B on first shot, so game stops with only 1 shot. :smile:
 
yevi said:
What do you mean stops at 1 or at 2?

Person A shoots first. Suppose he hits person B. Then it stops at 1 (one turn)

Suppose not (i.e. Person A misses), then now it's person B's turn. He shoots. Suppose he hits person A. Then it stops at 2(two turns)

Now, suppose not again (i.e. person B misses), then now it's back to person A's turn. Hence, it starts over and therefore you have to add 2(two turns) to the expected value. Hope that helps.
 
Thank you both!
 
Interesting! Apparently Tiny Tim and Konthelian helped you solve this problem: find E(X), yet it was never stated what "X" means!
 
HallsofIvy said:
Interesting! Apparently Tiny Tim and Konthelian helped you solve this problem: find E(X), yet it was never stated what "X" means!

Hi HallsofIvy! :smile:

Konthelian and I knew because … we were playing the game yesterday! :wink:

:eek: … nurse … !
 

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