Expected value and probability

AI Thread Summary
The discussion revolves around calculating the expected value E[X] in a shooting game between two players, A and B, where each has a specific probability of hitting the other. The random variable Y is defined to represent the outcomes of the first shot, with values indicating whether A hits B, B hits A, or neither hits. The expected values for each scenario are clarified: if A hits B, the game ends after one shot; if B hits A, it ends after two shots; and if both miss, the game restarts, adding two to the expected value. The participants express confusion over the notation and the implications of the game stopping after certain shots, but they ultimately clarify the mechanics of the game. The conversation highlights the complexity of calculating expected values in probabilistic scenarios.
yevi
Messages
65
Reaction score
0
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]
 
Physics news on Phys.org
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 … !
 
Back
Top