Understanding Bayes' Theory: Examples and Confusion

[WCMU101]

I'm struggling with Bayes' theory. Please consider the following:

Example 1:
Submarine sinks if one missile hits it. Two ships aim at submarine and fire one missile each. Ship 1 shoots missile X1, ship 2 shoots missile X2.
P(X1 hitting = 0.8)
P(X2 hitting = 0.5)
P(X1 and X2 hitting | submarine seen sinking) = P(submarine seen sinking | X1 and X2 hitting)*P(X1 and X2 hitting)/P(submarine seen sinking)
P(X1 and X2 hitting | submarine seen sinking) = 1*(.4)/(1-.5*.2) = .4/.9 = .444444...

Example 2:
Same as example one except:
P(X1 hitting = 0.5)
P(X2 hitting = 0.5)
Thus,
P(X1 and X2 hitting | submarine seen sinking) = P(submarine seen sinking | X1 and X2 hitting)*P(X1 and X2 hitting)/P(submarine seen sinking)
P(X1 and X2 hitting | submarine seen sinking) = 1*(.25)/(.75) = .3333...

However, some would argue that in example two, P(X1 and X2 hitting | submarine seen sinking) = 0.5 - since you could say we know for sure that 1 hit, thus the probability that the other one hit is still 0.5 (independent events). Who is correct?

The reason I showed the first example, is because I don't see how you could come up with an answer without Bayes' in the first example - you don't know which one hit?

Let me show you another example which I posted in an earlier thread (https://www.physicsforums.com/showthread.php?t=607946):

Example 3:
2 fair dice. What is the probability of both showing six if I have observed
at least one six.
Thus, the way I did it was:
P(2 6's | observing at least 1 6) = P(observing at least 1 6 | observing 2 6's)*P(2 6's)/P(observing at least 1 6)
P(2 6's | observing at least 1 6) = 1*(1/36)/(1-25/36) = 1/11

However, as the person who replied to my post pointed out, an answer of 1/6 here would appear to be the logical answer - knowing that the first dice came up 6, the probability that the second dice came up 6 is still 1/6 - independent events.

So, as you can see, I really am struggling with Bayes' theory here. Could somebody please help me out of the darkness!

Thanks.

Nick.

 

[haruspex]

you could say we know for sure that 1 hit, thus the probability that the other one hit is still 0.5 (independent events).

No, that doesn't work. Knowing that at least one hit is not the same as knowing a particular one hit. This is a classic stumbling block in probability. It probably has a name.
Let's make it as simple as possible. I toss a coin twice. It came up heads at least once. What's the probability it was heads both times?
A priori, we have HH, HT, TH, TT, equally likely.
The given information rules out only TT, leaving two head & tail combinations to one HH.
But if instead you'd been told the first was a head that would rule out TT and TH, leaving HH and HT.

 

[viraltux]

However, some would argue that in example two, P(X1 and X2 hitting | submarine seen sinking) = 0.5 - since you could say we know for sure that 1 hit, thus the probability that the other one hit is still 0.5 (independent events). Who is correct?

The reason I showed the first example, is because I don't see how you could come up with an answer without Bayes' in the first example - you don't know which one hit?

precisely because you don't know which one hit you have these results, you can calculate both "without" Bayes dividing all favorable outcomes between all possible outcomes, for instance:

0.8*0.5 / ( 0.8*0.5 + 0.8*0.5 + 0.2*0.5) = 4/9

0.5*0.5 / (0.5*0.5 + 0.5*(1-0.5) + (1-0.5)*0.5) = 1/3

Your problem with the theorem of Bayes is in how you interpret it, when you wonder who is correct, well, both, because both measure different things, 1/3 is the probability that both missiles hit if the submarine is sinking and you don't know if it was hit by both missiles or just one, if you know one missile hit, let's say X1, then 1/2 is the probability X2 hit the submarine too.

 

[WCMU101]

Thank you both very much - those answers were excellent! My headache has disappeared. Also, thank you haruspex for replying to my other posts - I should get you to do my phone interview for me haha

Thanks again!

Nick.

 

[haruspex]

1/3 is the probability that both missiles hit if the submarine is sinking and you don't know if it was hit by both missiles or just one, if you know one missile hit, let's say X1, then 1/2 is the probability X2 hit the submarine too.

That doesn't seem right. If you know exactly one hit then the probability the other hit as well is zero.
The key issue is whether you know that a particular missile hit (the first one, the red one, the one that had further to go, whatever) or merely that at least one did. This seems paradoxical because the puzzle is usually worded as for volunteered information. If Joe tells me the red missile hit, why did he choose to word it that way? The interpretation, for the purposes of the puzzle, is that Joe decided to tell me the fate of the red missile, whatever that was. But I could instead assume that he was really just telling me that at least one had hit, and if both hit then he would have mentally tossed a coin to decide which one to mention. In that view, specifying the red missile changes nothing.
The puzzle is more robust if structured as question and answer:
Scenario 1: Joe, did the red missile hit? A: Yes.
Scenario 2: Joe, did either missile hit? A: Yes.

 

[viraltux]

That doesn't seem right. If you know exactly one hit then the probability the other hit as well is zero...

I think haruspex wants to go philosophy on my ***

Be careful WCMU101, don't do that in your interview!