Does It Matter?

[heihojin]

irchans states: "Heihojin's formula seems to fail in this case due to lack of independence." irchans then assumes that events A and B are independent.

This is not correct, because event B is only independent of event A in the case where P(B) = 1, i.e. the propositioner will always turn over a red card. In all other cases, event B is not independent of event A.

To demonstrate, let P(B|!A) = n, where n is not equal to 1 (I will use != to denote "not equal to"). This means that if you do not choose the black card on your first attempt, then there is a non-zero probability that the propositioner will choose the black card. This excludes all cases where the propositioner "must" turn over a red card.

By definition, event B is independent of event A if and only if P(B|A) = P(B).

P(B|A) = 1, because the propositioner has only red cards from which to choose.

P(B) = P(A) * P(B|A) + P(!A) * P(B|!A) = 1/3 * 1 + 2/3 * n = 1/3 + 2n/3.

Thus, P(B|A) = P(B) if and only if n = 1. But because n != 1, this is a contradiction and thus event B is not independent of event A.


heihojin

[heihojin]

As a correction to my last post, I wrote "Because P(C|B) is greater for switching than not, you should switch. Notice that this is regardless of the probability of event B occurring, so long as it is non-zero."

What I should have said is:

"Because P(C|B) is greater for switching than not, you should switch. Notice that this is regardless of the probability of event B occurring given event A does not occur, so long as it is non-zero, i.e. P(B|!A) != 0."

[heihojin]

This is a recant of my previous post. This was not immediately obvious to me when I posted it this morning; rather, a counterexample to my previous argument occurred to me earlier today after I woke up, and I just spent two hours working it out on a whiteboard to prove it to myself.

The correct answer seems to be that whether or not you should switch depends on the probability of event B occurring.

Let P(B|!A) = n, where 0 <= n <= 1.

P(B) = P(A) * P(B|A) + P(!A) * P(B|A!) = 1/3 + 2n/3

If you don't switch, the probability model is:

P(A*B*C) = 1/3
P(A*B*!C) = 0
P(A*!B*C) = 0
P(A*!B*!C) = 0
P(!A*B*C) = 0
P(!A*B*!C) = 2n/3
P(!A*!B*C) = 0
P(!A*!B*!C) = -2n/3 + 2/3

And so P(C|B) = P(B*C) / P(B) = (1 / 3) / ((2n + 1) / 3) = 1 / (2n + 1).

If you switch, however, the probability model is:

P(A*B*C) = 0
P(A*B*!C) = 1/3
P(A*!B*C) = 0
P(A*!B*!C) = 0
P(!A*B*C) = 2n/3
P(!A*B*!C) = 0
P(!A*!B*C) = -2n/3 + 2/3
P(!A*!B*!C) = 0

And so P(C|B) = P(B*C) / P(B) = (2n / 3) / ((2n + 1) / 3) = 2n / (2n + 1).

Thus, you should switch so long as 2n > 1, or n > 1/2.
If n < 1/2, then you should not switch.
If n = 1/2, then it doesn't matter, and this is the answer to the original question.

Sorry for the error. :-(


heihojin

[heihojin]

I'm just full of errors today. I misunderstood irchans; in fact, in my original post, I had assumed independence of events A and B by assuming P(B|!A) = 1.

Again, sorry for the error.


heihojin

[irchans]

heihojin writes:

"If n < 1/2, then you should not switch.
If n = 1/2, then it doesn't matter, and this is the answer to the original question."

where

n = probability that the propositioner picks red after the player has picked a red.

heihojin's reasoning is great because it requires no assumptions about the motivations of the propositioner! In the typical monty hall question n = 1, so switch. In the random case, n = 1/2 so it does not matter.

Even if he doesn't know, its still kind of like hitting a runner runner if you don't change (weird comparison but it works in my messed up head). By switching doesn't one increase thier odds by switching (i think for some reason they increase by 17%)? Eh, I'm probably wrong.

[lorinda]

Thanks to heihojin and irchans for their posts, I have taken the trouble to wade through the notation, and not only have I learned something, but am able to appreciate why this is such a good answer.
Keep up the good work guys, I really like the honesty on this forum, it is unlike any other forum (as I have stated before).
Now....to that calculus book that's been sat on my shelf for twelve years....... [img]/forums/images/icons/frown.gif[/img]

[irchans]

If the propositioner does know the colors of the cards and always turn over a red card, then switching wins (2/3)= 67%.

It is difficult to explain why the knowledge and motivation of the propositioner affects the odds. I suggest you read all of the posts. That might help, but it is a difficult concept.

[baggins]

the knowledge of the propositioner is crucial. if he knows which of the remaining cards is red (and one must be), then he can flip over a red (therefore meaningless, since black is the goal) card every time. your odds of having chosen the black card do not change because the propositioner knows which cards are which. your odds improve because you KNOW that he will HAVE to flip over a red card if he knows which are which. if he doesn't know and flips black, he can't offer you to switch, obviously. if he doesn't know and flips red, he still doesn't know where the black card is. the odds of one of the remaining cards is 1/1. the odds of him flipping over a red card from the 2 (unknown) remaining cards is EITHER 1/2 or 2/2. out of the three possible choices you had in the beginnning, 1/2 is twice as likely as 2/2. since you don't know which one for sure (since you haven't seen the cards face up yet) and your prop. doesn't know either (our main assumption here), if he flips a red card, neither of you will know if the other one could be black or not. so it does not matter whether you switch or not, as your opponent has no clue (and, therefore, no psychological leverage in the game) where the black card is. if he flips over black, the game must be over (which the prop doesn't want, otherwise he wouldn't have offered in the first place). so, knowing that the prop (who is giving you this offer without any risk to your 'bankroll', so you're really on a freeroll here...) is interested in making you choose whether or not to switch (which can be THE only real entertainment in making such an offer, if you ask me...) you must assume that he knows that it is in his best interest to KNOW which card is black, and follow his endgame strategy from there (which can only be maximized in one way, obviously. if you pick the black card, he can only show you a red, and then make the offer to switch. if you pick a red card, he can show you the black card, at which point the game is over <which we already determined the prop is vehemently against>, or show us the remaining red card and offer a switch. either way, it is in the best interest of the prop <and his sick mind games!!> to know which of the 2 remaining cards is red, so that he can appropriately flip over a red card. so the scenario only gets played out one way if the prop intends to derive any enjoyment at all from the game. and the only way he can ensure this proper endgame scenario is to know which cards are which. which is why he MUST know, which is why your odds are better switching. i'm not going back to see if i need to close a paranthetical phrase here or not. im sure you all followed. or at least eLROY did if he still reads this board.