Re: [Ghetto] So let's say you have three rigs... :)
Okay... Sorry for the long post. I remember learning(?) this way back in a class, so this weird problem is really eating away at me. After a nap, and spending way too much time on this, I have (likely erroneously) arrived at the following.
There's this thing called
Bayes Theorem which basically tells us how to find probabilities of things, given that other related things have happened. The form I am looking at is the theorem combined with the law of total probability, which is weirdly retyped here as:
P(A given B) = [P(B given A) * P(A)] / [ P(B given A) * P(A) + P(B given NOT A) * P(NOT A) ]
Gah. It really looks prettier on the wiki page.
So, let's let
A = "your initial choice was a slider up rig"
B = "your friend opens a rig that is slider down"
We want to solve for P(A given B), or P(your initial choice was slider up, given that the rig your friend opened was slider down). Then we can subtract the result from 1 to see whether switching would have improved your chances. Seems logical enough, right?
So we try to assign numbers to whatever we can:
P(A) = P(your initial choice was slider up) = 0.33
P(NOT A) = P(your initial choice was slider DOWN) = 0.66
P(B given A) = P(friend picks slider down, given your guess is slider up) = 1
because only one rig is slider-up and if you picked it, your friend is only left with slider down.
(this is where I took a nap so bear with me)
The last term we need to find is P(B given NOT A), which is P(friend picks slider down given that your initial choice is also slider down).
Given that your initial choice happens to be slider down, your friend has two rigs to choose from, slider up and down. This is where the question of who packed the rigs comes into play.
Case 1: If the friend packed the rigs, he knows which one is slider up, so he will always be able to select the slider down rig from the remaining two, so in this case P(B given NOT A) = 1.
Case 2: However, if the friend did not pack, he must choose at random which container to open. Thus, he only has a 50/50 chance of picking the slider down, so P(B given NOT A) = 0.5 in this case. Yes, we know that the event B has already occurred, but this expression simply calls for its probability in the general case, before anything else happened.
Anyway, if you put all the above numbers into the expression and solve, you get that:
P(A given B) = 0.33 in the first case and 0.5 in the second case, so already there is a difference. Recall that A = "your initial choice was slider up" so if you switched to the other remaining rig, it would basically flip the outcome, so then P(slider up for you if you switch) = 1 - P(A given B) = 0.66 if your friend packed, and 0.5 if he didn't.
Phew.
I checked that a few times, yet something tells me that I've made a mistake somewhere, as Mr Ghetto's solution seems more intuitively correct.
Does anyone care to chime in with a correction? What am I missing?
Steve