Expected value and probability

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Homework Help Overview

The discussion revolves around a probability problem involving two individuals shooting at each other, where the goal is to determine the expected number of shots taken until one person hits the other. The probabilities of each person hitting the other are denoted as P1 and P2.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the use of a random variable Y to represent different outcomes of the shooting game. They discuss the expected values associated with each outcome and question the meaning of certain terms like "g(0)". There is also a discussion about the implications of the game stopping after a certain number of shots.

Discussion Status

The conversation is ongoing, with participants clarifying their understanding of the expected values and the conditions under which the game stops. Some guidance has been provided regarding the interpretation of the random variable Y and its outcomes, but there is still some confusion about specific terms and the overall setup.

Contextual Notes

There is a lack of clarity regarding the definition of the variable X, which is central to the problem. Participants are also navigating the complexity of the expected value calculations and the implications of the game's rules on those calculations.

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]]=[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]
 
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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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