We get the event E = "someone brings a bomb on the plane you are on", with a p(E) proability. Now, you are afraid of having a bomb on your plane, so you bring along your own. The probability of having TWO bombs on your plane is p(E)² (because the two events are independent), therefore you have p(E) less chances to get a bomb on your plane! What''s wrong? ToohrVyk

The presumption that it takes both bombs to kill you. One will do

Also, the probability of you bringing a bomb on the plane should be p(1), because you say, "so you bring along your own [bomb]." It's p(E) for someone else on the plane, but in the context of the question if you always bring a bomb onboard, then that's p(1). It's essentially the probability of an event that always happens.

I'm sure the essence of the issue is whether or not a bomb is detonated, and I'm pretty sure someone who is figuring this out in their head and mistakenly believing that bringing a bomb along with reduce their changes of getting killed (because that's where the real fear is), will not detonate that bomb. So what's it gonna be? The change you will be killed, or the chance there will be a bomb? If it's the chance of there being a bomb, well then you're pretty screwed, because you always bring your own. Yet if it's the probability of being killed, then you need to decide the probability of the bomb being detonated. I personally wouldn't take any chances and assume that if the guy brought the bomb on board, the chance of detonation is p(1). At the same time, I certainly wouldn't detonate my bomb, so that's p(0). Yet common sense alone says that bringing along your own bomb won't help the situation, because as you said it's an independent event from the other guy. If anything, it would probably increase the chances of disaster, since there's certainly a random factor where the chance of you accidentally detonating your bomb is p(x > 0).

To make an already long story shorter, ride the train.

[edited by - Zipster on May 23, 2003 3:22:24 AM]

We are not talking about life or death, simply about the probability of having another bomb on the plane, which is said to be p(E)².

On your second try you got it right, but explained it wrong. There is indeed a p(E)² probability of having TWO bombs on a certain plane at any given time - this part is correct.

The idea is we want the probability of having ANOTHER bomb on the plane, which is equal to the probability of having TWO bombs on a given plane, KNOWING one is there.

This is written as p(E & E | E) and resolves to p(E), which sounds natural.

So what was wrong is : I was asking for p(E & E | E) and answering with p(E & E)... Computing the probability for the wrong problem!

ToohrVyk

There is something terminally wrong with you people...

--SuperRoy
[ Author:: Linux GameDev Articles ] [ Programmer:: WhitespaceUnlimited ] [ Webdesigner:: CTH3.com Webdesign ]

Well of course you were computing the probability for the wrong problem (which I know was part of the trick), because you overlooked that you always bring a bomb. p(E & 1).

But if you think about it, would someone really be afraid of there being a bomb, or just the bomb going off? If the probability of there being a bomb was p(1), yet the probability of a bomb being detonated was p(0), I really wouldn''t be scared, would you?

Yeah, I think too much about this stuff

quote:

---

This is written as p(E & E | E) and resolves to p(E), which sounds natural.

---

Ehh, isn''t it:
p(E & E | E)=1

Coz, probability that ''someone brings..'' and ''someone brings..'' given ''someone brings..'' seems to be 1.

To be really correct:
A="someone else than you bring a bomb"
B="you bring a bomb"
P(A&B|B)=P(A&B)/P(B) => P(A&B|B)=P(A&B)=P(A)P(B)= P(A)

/regards


/Mankind gave birth to God.

Oops... Yeah, indeed

ToohrVyk