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09-19-2002, 07:36 PM
2 cards red and 1 card black, are face down on a table. Let's say this nice man tells you that if you choose black, you win \$100.
So you point one card. Let's say the card in the middle. The man gives you a favor and turn up one red card from the other two cards that you didn't choose. Now do you want to stick with the cards that you chose in the first place (the middle one,) or it doesn't matter?

Soh

p.s. no trick question, rather logic.

Ed Miller
09-19-2002, 07:40 PM
I love this question.. so much so that I ask a variant of it as an interview question.

lorinda
09-19-2002, 08:34 PM
Do we know that the guy always points to one of the cards for this question, or is this the only time we have ever seen it done?
If it was a one off, and I knew he knew i was a mathmatician then I would stay with the same card, because I would assume he was misleading me to think that I should switch with a 67% chance.
However, if I have seen the guy doing this many times then I would switch....see my post "Game Show" on Internet gambling a few weeks back /forums/images/icons/laugh.gif

BruceZ
09-19-2002, 08:35 PM
Yes of course switch. Now, when can I start, how much will I make, and what will I do?

09-20-2002, 05:50 PM
Chances of picking the correct card start out a P=.33

So chances are that .66 that the correct card is 1 of the 2 that you haven't picked. By showing you a dead card, he is increasing your chances by switching.

irchans
09-21-2002, 09:22 AM

2 cards red and 1 card black are face down on a table. Let's say this nice man tells you that if you choose black, you win \$100. ASSUME HE KNOWS NOTHING ABOUT LOCATION OF THE BLACK CARD. So you point to one card. Let's say the card in the middle. The man gives you a favor and RANDOMLY CHOOSES A CARD FROM THE TWO REMAINING. HE TURNS IT UP AND IT IS RED. Now, do you want to stick with the card that you chose in the first place (the middle one,) or it doesn't matter?

09-21-2002, 11:15 AM
change

09-22-2002, 04:39 AM
if he doesn't know if the card is red or black, it doesn't matter if you switch or not.

if he does know, then you switch.

irchans
09-22-2002, 10:34 AM
sdplayerb said:
&gt;if he doesn't know if the card is red or black, it doesn't
&gt;matter if you switch or not.
&gt;
&gt;if he does know, then you switch.

I believe that sdplayerb's conclusion is correct. What reasoning could support this conclusion?

baggins
09-23-2002, 05:16 AM
if he doesn't know, then he could just as easily have shown you a black card. showing you a red card means that now you both know that there is one red and one black left, but neither knows where. whats the point in switching then?

if he knows, however, then of course he's going to show you the red one. every time. and then give you the option to take the worst of it, instead of giving you even money.

DJA
09-23-2002, 12:30 PM
This has been posted before... as you can see by the posts you should switch...

The best explanation I have read that is very intuitive is this... Say there are 100 cards 99 red and 1 black and you have to pick the black card. You pick 1 and he does you a favor and turns over 98 red cards... would you switch your pick?

karlson
09-24-2002, 02:10 AM
I think you guys are full of it.

You should switch whether he flipped a random card or not.
Think of it this way: you are given the option of sticking with your card or picking both of the other ones as a set. By flipping a red card, he is giving you the second option.

I am anxious to hear why I am wrong.

lorinda
09-24-2002, 09:51 AM
Three people have a bet...
Player A picks a card a random
Player B picks a card a random (and flips it over if you like)
Player C picks the other card
You are telling me player C is favourite now to win the game?

heihojin
09-24-2002, 12:06 PM
Whether or not the propositioner could have turned over a black card is irrelevant, so long as the probability of that event occurring is less than 1.

Let the following events be defined:

A: You select the black card on your first attempt.
B: The propositioner turns over a red card.
C: You hold the black card at the end of the game.

The question is, given that the propositioner turns over a red card, do you increase the probability of event C occurring by switching or not switching?

The probability of event C occurring given that event B occurs is expressed as P(C|B).

If you do not switch:

P(A) = 1/3
P(C|A*B) = 1

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

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

If you do switch:

P(A) = 1/3
P(C|A*B) = 0

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

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

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.

heihojin

irchans
09-24-2002, 02:29 PM
I loved your very clear argument. I am worried that your formula

P(C|B)=P(A)*P(C|A*B)+P(!A)*P(C|!A*B)

might not always be correct.

I think the formula is correct if A and B are independent events because of the following reasoning (using ! for not and * for "and"):

Spitting the set B*C into two parts gives:

P(B*C)=P(A*B*C)+P(!A*B*C)

From the definition of | :
P(C|B)*P(B)= P(B*C),
P(A*B*C) = P(A*B)*P(C|A*B), and
P(!A*B*C) = P(!A*B)*P(C | !A*B).

Substituting gives

P(C|B)*P(B) = P(A*B)*P(C|A*B) + P(!A*B)*P(C | !A*B).

IF A AND B ARE INDEPENDENT, THEN THIS REDUCES TO

P(C|B)=P(A)*P(C|A*B)+P(!A)*P(C|!A*B).

A and B are independent if P(B)=1, but would they are not be independent if the "propositioner" has no knowledge of the location of the black card. Does the formula hold if A and B are not independent?

irchans
09-24-2002, 02:31 PM

lorinda
09-24-2002, 03:07 PM
Okay, Im gonna lose this argument even though I understand the concepts /forums/images/icons/frown.gif

My notation is non-existent Im afraid having been self-taught only in probability so even though I can understand arguments, I cant follow notation /forums/images/icons/frown.gif

In the original question , we have cards ABC where we always pick card A,
The propositioner is considered to know where the black card is and so removes a non-black card.

ABC A is black, propositioner removes B switch loses
ABC B is black, propositioner removes C switch wins
ABC C is black, propositioner removes B switch wins

This is elementary

Now to the random case

ABC A is black, prop removes B, and it is red switch loses
ABC A is black, prop removes C, and it is red switch loses
ABC B is black, prop removes B, game cancelled
ABC B is black, prop removes C, switch wins
ABC C is black, prop removes B, switch wins
ABC C is black, prop removes C, game cancelled

six equally likely events, two draws, two wins, two losses, what am I missing?

lorinda
09-24-2002, 03:13 PM
The longhand version of the non-random case reads

ABC A is black, prop removes B, switch loses
ABC A is black, prop removes C, switch loses
ABC B is black, prop removes B, changes mind, removes C,win
ABC B is black, prop removes C, switch wins
ABC C is black, prop removes B, switch wins
ABC C is black, prop removes C, changes mind, removes B, win

4 wins, 2 losses, as we know

karlson
09-24-2002, 03:43 PM
Of course you are right.
I was somehow incorporating the events where a black card is turned over into wins.
By the way, this is the same thing that heihojin's formula does, as far as I can tell.

irchans
09-24-2002, 09:17 PM
I think lorinda's reasoning correct .

In the case where the propositioner always picks a red card, it is best to switch. (We can use lorinda's enumeration or heihojin's formula to show that this works.)

In the "randomly choosing propositioner" case, it does not matter whether you switch or not. (We can use case lorinda's enumeration or her three gambler post on 9/24 to understand this. Heihojin's formula seems to fail in this case due to lack of independence. Baggins explanation for this case also works.)

We can modify Heihojin's formula to make it work for the "random" case. If we don't assume independence, Heiholjin's formula becomes:

P(B) * P(C|B) = P(A*B) * P(C|A*B) + P(!A*B) * P(C|!A*B)

As Heiholjin points out,

P(C|A*B) = 1 and P(C| !A *B) =0 if you don't switch, so

P(B) * P(C|B) = P(A*B)
P(C|B) = P(A*B)/P(B) = P(A|B).

Recall that in Heiholjin's notation C = winning. In the "random" case, P(B) = 2/3 and P(A*B) = 1/3, so P(C|B) = 1/2. So you win 1/2 the time when you don't switch. You win the other half the time if you switch. So it does not matter which strategy you choose in the random case.

heihojin
09-24-2002, 10:08 PM
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
09-24-2002, 11:09 PM
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
09-25-2002, 12:02 AM
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 &lt;= n &lt;= 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 &gt; 1, or n &gt; 1/2.
If n &lt; 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
09-25-2002, 12:17 AM
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
09-25-2002, 07:45 AM
heihojin writes:

"If n &lt; 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.

09-25-2002, 11:56 AM
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
09-25-2002, 01:11 PM
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....... /forums/images/icons/frown.gif

irchans
09-25-2002, 10:36 PM
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
10-02-2002, 01:37 AM