Expected value and probability

  • Thread starter yevi
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  • #1
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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]]=[tex]\sum[/tex]E[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]
 

Answers and Replies

  • #2
tiny-tim
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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])
 
  • #3
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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?
 
  • #4
tiny-tim
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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:
 
  • #5
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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.
 
  • #6
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Thank you both!
 
  • #7
HallsofIvy
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Interesting! Apparently Tiny Tim and Konthelian helped you solve this problem: find E(X), yet it was never stated what "X" means!
 
  • #8
tiny-tim
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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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