Is this really valid?

Suppose you set down three cards face down: two hearts, and a spade. You ask someone to pick one of the cards; if they pick the spade, they win. They pick card 1. You then reveal card 3: a heart. You offer to switch their card with 2. Is it favorable to switch?

Apparently, it's supposed to be, but I'm a little confused: there are two lines of reasoning that both seem to make sense to me, but they contradict each other:

I. There was a 2 in 3 chance you were wrong, and since you know 3 is wrong, you are basically asking yourself a question: were you right or wrong on your first pick? If you pick right you keep your card, which is a 1 in 3 chance; if you pick 2 you are betting on wrong, 2 in 3.

Seems to make sense, but...

II. We know the third card is a heart. That means the face-up setup is either spade-heart-heart, or heart-spade-heart; what is it really that favors one over the other? When we're deciding what completes the threefold heart-spade complex (i.e., the first two cards, which we know contains a spade and heart), can we not, for all intents and purposes, instead focus on the first two cards, the ones we don't know? We know that one has to be a spade, and one has to be a heart. Either combo seems plausible, doesn't it? Like in algebra, it seems you could cancel out heart, since we know that one; it really boiles down to heart-spade versus spade-heart; the fact that there is an equivalent third card added to both sides doesn't seem to mean much. It's like saying x = y, but x+1 does not equal y+1.

Couldn't we just treat this as a new game, one where you put two cards face down, one a spade, other a heart, and ask what they choose? When the heart is revealed, isn't that essentially what you're doing? Why, really, should it matter what you picked before? You're still choosing the completion of the heart-spade complex, aren't you? Isn't that neutral?

"I" seems very logical, yet I can't quite refute "II."

You know what I think it is? I think it's that when 3 is revealed, your #1 is not eliminated right off the bat, but #3 is. So for example, if you were told that #1, your first pick, was wrong, then the chances of being correct would indeed be 1 in 2 since you know no distinct differences between 2 and 3. But since you know 3 is wrong, you can compare 1 in 2, which, with 3 out, is the choice between right and wrong, reverting back to I. Because 3 is wrong, the value of 1 and 2 I guess do have different values, unlike if 1 was wrong when there is no conflict between 2 and 3.

The thing about I is, it is taking into account slightly more information (what you picked) than the more objective postulating in II; that extra information I guess is surprisingly useful!

II. doesn't apply, they are not equally plausible. The other person showing you a heart which isn't your original pick gives no additional information about your card, because you already knew there was at least one heart out of those two cards. So 1/3 of the time, they are revealing just one of the other hearts in those two cards, and 2/3 of the time they are revealing the only heart which isn't yours.

With a caveat: the other person has to be forced to reveal a non-yours-heart. If the revealed card is random (and therefore has the chance of being the spade), then it's back to 50-50, because then it actually is new information about your card.

Consider this: let's say you actually do this 3 million times. On average, you will pick correctly 1 million times, and incorrectly 2 million times (initially). Every time you pick correctly initially, swapping loses. Everytime you initially picked incorrectly, swapping wins. Swapping gets you the spade 2/3 of the time.

"With a caveat: the other person has to be forced to reveal a non-yours-heart. If the revealed card is random (and therefore has the chance of being the spade), then it's back to 50-50, because then it actually is new information about your card."

Yes, this makes sense. If nothing is being said about the chosen card, rather, only a non-chosen card, then this indeed can be useful information. With no new information on the chosen card itself, I suppose there is no reason to change the thought that there is a 1 in 3 chance that you were right, thus maintaining from the very beginning that there is 2 in 3 chance of your choice being wrong? That would in turn make 2, i.e., wrong, indeed the better choice.

"Consider this: let's say you actually do this 3 million times. On average, you will pick correctly 1 million times, and incorrectly 2 million times (initially). Every time you pick correctly initially, swapping loses. Everytime you initially picked incorrectly, swapping wins. Swapping gets you the spade 2/3 of the time."

Yeah, this is an even simpler way to view it.

I knew there had to be something wrong with II, it was just a little confusing at first.

There is an essential piece of information that is missing from the statement of this problem. Is the host able to see both of the other cards (and therefore always show you a heart), or did he take a look at one of the other cards and then show the competitor only if it is a heart? If he is able to see both of the other cards and pick one of them that is a heart, the standard answer is correct. If he merely looked at one of the other cards and, finding it was a heart, told you, the "wrong" answer is correct.

This sort of problem is basically about Bayesian probability, where you calculate a probability based on some relevant information which restricts the range of possibilities.

I thought of another interesting way to help the intuition beat the 50-50 assumption.

When you are offered the switch, essentially you are being offered two cards at once, as opposed to your initial one. Or, at least, the probabilistic fruits of two cards. You already knew there would be at least one heart out of the two cards you don't hold, so that's nothing new when it is flipped up, at least about the combined two cards. You're not being given a choice between your card and another card, you're being given a choice between your card or the combined chances of the other two.

If, after you pick a card, asked to switch, but with no cards revealed, it doesn't help your chances then, right? Because then when you bet on wrong, you have to both be right that your first choice was wrong, yet still have to worry about picking the right card anyway (say you're right that your original choice, 1 is wrong, you still have to pick between 2 and 3). So yeah, when that card is revealed, it's helping you a bit.

Arguement II is indeed valid. Whether you pick card 3 randomly, or you knew where the hearts were, there is one absolute truth: When the player you asked to pick a card reaches the switching stage, card 3 is no longer part of the game. Hence he has to pick one of two cards faced down, and he has even odds to pick the right one. In other words, the probability that the other card is the right one is 0.5, and switching is not favorable, but doesn't change your odds.

That being said, I will now explain why arguement I is incorrect:

Case #1 - You know where the spade is and chose card 3 to be heart for sure:

If you think about it carefully, in this case, the player could actually choose only 2 cards, and not 3. Elaboration:

We have 3 cards: 1 spade and 2 hearts. Since the player chooses only 1 card, then there is always a heart that he doesn't choose. Hence let heart1 be a heart that the player didn't choose and heart2 be the other heart. By definition, the player can't choose heart1 out of the 3 cards in the beginning. In addition, card 3 is always heart1, because you always reveal a heart that the player didn't choose, which is the definition of heart1. Thus, after revealing card 3, heart1 cannot be chosen, for it is revealed and is out of the game. The conclusion is that heart1 cannot be chosen by the player at any stage, namely, heart1 is never part of the game, and the player can only choose the spade or heart2 at any point, meaning he has a 1:2 chance to begin with, and not 1:3 as the arguement suggests.

Case #2 - You don't know where the spade is and choose card 3 randomly:

Let's call the spade card "spade", one of the heart cards "heart1" and the other heart card "heart2". Since you choose card 3 randomly, it's like letting the player choose 1 card (card 1) that he thinks is a spade, and 1 card (card 3) that he thinks is a heart. Then you reveal card 3 (which the player chose) and let him pick one of the two remaining cards. Since there are 3 cards, there are 6 combinations the player can pick. Assuming he got to the switching stage, the player didn't pick (heart1, spade) or (heart2, spade), since otherwise, when card 3 is flipped, he has already lost. Hence, in order to get to the switching stage, the player may choose only (spade, heart1), (spade, heart2), (heart1, heart2) or (heart2, heart1). Since 2 of these 4 options lead to victory, and all options are chosen with equal probability, there is a 1:2 chance for him to have chosen the correct card (again, assuming that card 3 is indeed a heart).

I hope you managed to understand what I'm trying to say, 'cause it's hard to explain logic. xD

I used to have your arguments, Mikoro.

By not switching, you fail to exploit the "betting I'm wrong" idea. If you were simply asked to switch, with no cards revealed, then indeed it won't help you, but there is actually a specific reason why. It's because, if you switched, you both have to be correct that your first choice was wrong (2/3), and, be correct as to which one you switched to was correct (if it's one of the two other cards, you also have to survive a 1 in 2 chance).Thus you must survive a 1/2 chance, and a 2/3 chance; that multiplies to 2/6 or 1/3; just as good as keeping.

However, when one of the wrong cards is revealed, this makes betting on wrong better: Now you only have to be right that you were wrong (2/3); if you're right about that, you are correct. That 1/2 that was multiplied to 2/3 has been eliminated.

See, if you don't switch, you don't exploit the extra information given to you... think about it: You are told something new, but you end up making the same decision as you did in the beginning. In this case, it didn't matter how much information you had, because you ended up picking the same one anyway. Switching seeks to exploit this information.

If you pick card one the first time, and you pick card one after new information is given, have you improved on your decision? No: you made the same one. You had more information, but your decision had the same value as before, seeing that it's the exact same one.

It is really counter-intuitive though, I know. For some reason though, that wrong card being revealed gives away more information than it seems.

Sorry mate, still haven't convinced me. ;)

I believe your argument is flawed for two reasons:

1. No new information was given to the player about the card he picked or about the card he can switch to, but rather about the card that was revealed. By revealing one of the hearts, you simply removed it from play and turned the game into "pick a card out of two".

2. Let "card" be the card the player picked, "revealed" the card revealed by you, and "other" the card neither revealed nor originally picked by the player. Suppose it is like you say, and at the switching stage, there is a 2:3 chance that "other" is the spade. Then, if he chose "other" in the beginning and you revealed "revealed", "card" would also, according to your arguement, have a 2:3 chance to be the spade, however that is not possible. Or in simple words, as I wrote in 1., you reveal new information only about the card that has been removed from play, thus anything the player can deduce from it about any of the two other cards has to be the same (because they were the same to him before the revelation), and their odds have to remain equal.

EDIT: Okay, apparently you are right. I still don't understand why, I'll try thinking about it a bit more.

EDIT #2: I think I understand why you are right. Now for the tricky part: understand why I was wrong.

It is tricky. It doesn't seem like the new information would give you something useful, but... it does. :)

Let's put it this way:

When you pick card one, at this moment whether your original choice is right or wrong has been determined, even though we don't know what that determination is yet (face down cards).

Would you agree that you are screwed if it was 2 or 3? Would you agree that your chances of winning if a card is not revealed is 1 in 3 then? Hopefully.

When one other card flips over, you're still screwed if it was 2 or 3 -- that actually does not change. Just because we know that 3 is wrong -- and this is an important thing to understand -- does not change the fact that there is a 2 in 3 chance that the correct card is either 2 or 3. Since we know it's not 3, we can simplify this and cross out 3: we say there is a 2 in 3 chance that the correct answer is 2. We essentially made the same statement, but crossed out three since there was no point in including something we knew was wrong (though the  "either 2 or 3 statement" is still technically correct!).

I think the instinctive obstacle for most of us when first coming across this is that we want to assume that if you take one thing away from the game it's just 1 in 2; but in fact, the fact that we know something new about a card we did not pick makes it a bit more special than that, for reasons explained above.

Wait a minute... why? Why would that change the fact that there was a 2 in 3 chance that your first pick was wrong? Schubomb said that the random reveal being a heart (rather than the non-random reveal in my original problem) gives new information about the picked card, but how exactly?

I'm not sure if you guys know this show, but does this mean that on "Deal or no Deal," at the end of the game, switching the case and keeping the case are equally good? Because that would entail the cases being revealed, but not knowing if they contained the desired $1,000,000, seemingly equivalent to, in the other game, revealing one of the mystery cards at random (that could be heart or spade).

It turns out that in Deal or no Deal specifically, switching does not improve or worsen your chances.

This is a great article to explain the distinction: http://mathfactor.uark.edu/2008/01/dealornodeal/

That article is easier to get if it's looked at mathematically. In (A), the combinations with an a at the end of course mean it's favorable to switch, whereas any combination with b or c at the end means it's favorable to keep the original, since the prize is in a. The four combinations given are the possibilities that happen assuming the game doesn't end prematurely (what we want to know), which is why the other two possibilities don't apply to the question. Within a finished game, there is a 1 in 4 chance of each combination occurring. Two 1/4's ending in a; two 1/4's not. Thus, 50% odds for switching; 50% odds for keeping.

In (B), however, in a finished game, things are different. There is a 1 in 6 chance of acb occuring, because the contestant both has to pick the right one, and the host has to pick one of the specific choices out of two. Basically what it does is cut the 1/3 chance (a being chosen) into two slices, depending on which incorrect answer the host chooses to reveal. There is a 1 in 6 chance of abc occuring for the same reason. Added together, that would be a 1 in 3 chance of a not being at the end, meaning that there is a 1 in 3 chance that it's favorable to keep.

From there, we can look at what happens with b being picked originally. If that is the case, bca is the only result; every time b is picked first, bca follows. Since there is a 1 in 3 chance the contestant will pick b, and a 1 in 1 chance the host will then reveal c, then there is also a 1 in 3 chance of bca occurring. The same goes for cba, which also has a 1 in 3 chance of occurring, based on the same logic as before. Since there are two 1/3's ending in a, which indicates "switch," there is a 2 in 3 chance that it's favorable to switch.

What makes this confusing is that in (A), there is of course the possibility of the game not finishing. But, breaking it down logically/mathematically, we determine that if the game does finish, there are only four possible outcomes to look at, and we work from there.

I guess one more way to put it is that there is no reason to freak out when you are showed one of the wrong answers -- well duh, no matter which card you pick, at least one of the two cards remaining has to be wrong (they can't both be right ). If a wrong answer is being cherry-picked by the host from the two remaining cards, basically he's telling you that at least one of the cards is wrong, something that we kind of already knew . Thus it shouldn't change your attitude in any way that there is a 2 in 3 chance that the correct answer is one card that you didn't pick.

Let's say there were 10 cards instead. You pick one. The host is essentially, by eliminating all of the wrong answers, picking 9 out of the 10 cards -- he chooses from these 9 the correct answer remaining. The only way the correct answer won't be remaining is if you picked the correct answer -- a 1 in 10 chance. His eliminating is biased towards the correct answer -- if the correct answer is in his pile of 9 cards, he'll leave that one.

But if it was a deal or no deal type situation where you picked each card one by one (with the risk that you will eliminate the correct card) there is no discrimination. From your vantage point, any of the 9 cards you could eliminate are equally likely to contain the prize. You eliminate one of those 9 cards (and it doesn't contain the prize). Now there are 8 cards left. Again there is no reason to believe why any of those 8 cards should be favored. Then you get down to 7, same thing. And the same applies when you get down to just the final card and your card. Sure, it was unlikely that the card you picked happened to be the right one, but it was just as unlikely that the card remaining was the right one too -- there is no reason to distinguish it from the card you have. So they are equally likely as well: 50-50.

This fact is not true when the host eliminates the cards instead -- in that case, the card he leaves remaining is distinguished in probability from the card you originally picked because, while your card was a blind 1/10 pick, the card left by the host was the result of picking the best out of the 9 cards (and there is of course a 90% chance that the "best" is the one that contains the prize). Thus while there are only two cards left -- yours and the one left remaining, the latter is much more reliable.

The host is picking 9 out of the 10 cards in the first situation, which will favor the correct card 90% of the time. In the deal or no deal scenario you are first picking 1 card out of the remaining 9, then 1 card out of the remaining 8, etc, and there is never a rational reason for you to favor any card over the rest, thus at every point all of the cards in play have the same chance of being correct.

I mean no pretension by any of this; I just like to get all of this down because it fascinates me. Even though the question is easy to answer mathematically it's much harder to understand why the math would work out that way. In fact reading over Schubomb's much more concise comments they seem to make perfect sense now, but it has been a long process for me to intuitively understand all of this reasoning :)

Hmmm it's been almost two years.

I just wanted to say that what we generally call "understanding" is, in fact, having the ability to guess the correct answer without proving it (the difference from "knowing" being the guess part). While this ability is highly useful, it is simply an illusion of understanding, as real understanding comes down to knowing the answer, and knowing how to prove it.

A close friend of mine once had his brother tell him he cannot understand why De Morgan's laws are correct. My friend simply replied "I understand they are right, because I can prove they are right."

Basically what I'm trying to say is our emotions and thoughts are not real. They do not dictate reality, nor reflect it. Math is what reflects reality.

On a side note: I found it interesting that I a quote I encountered reading about clinic depression ("our emotions are not real") seems to fit here so well. I suppose relying on our emotions is a global handicap (or, at times, advantage) of our kind.

I know what you mean. On the other hand, I could argue a lot of math is supposed to represent what we understand intuitively. I guess this could lead to another debate, but I think what we have as math is our ideas just put into variables -- it's the ideas in our minds that reflected the meaning we assigned to variables and operations. For example in the case of the number zero, it can only be understood if we understand what "nothing" is, and the grasping of "nothing" seems to come from outside of the mathematical system. I would argue math is simply another way of expressing our ideas, just in a more quantifiable way. Of course we don't have math for "happy" or "upset," and that's probably only because it seems a system of numbers isn't effective at expressing those things. So instead we simply say "I'm happy" instead of spitting out a formula to represent that feeling.

The mathematical proof to the question might very well be based on human ideas. What I am calling "the intuition" of it may simply just be some other human ideas. The mathematical proof contains human ideas that we can represent in numbers. In fact the only key distinction between this and the intuitive proof is that the ideas that are contained in the intuitive proof are difficult or impossible to express in numbers. Sometimes math may represent something better; other times not. I'm not sure it's set in stone. In the case of the questions on this thread, the math does a good job of representing reality; but the intuition may allow us to deepen our understanding of that reality. Granted, it's also possible it may instead cloud our understanding of that reality, but again, I don't think it's set in stone one way or the other -- some human ideas, however they are expressed, are right, others are wrong :) I'm not willing to assume that the ideas expressed in the mathematical answer leads to a full understanding of the situation (it could, but I'm not willing to assume it).

So in summary, if our understanding is an illusion, then our knowledge from math should be too, because math, as being created by us, reflects us as well.

That is an interesting point of view. The way I view math is somewhat different, though. I used to think of mathematics as a sort of religion - a belief that the world functions according to a set of certain basic assumptions (axioms) and rules of deductive reasoning. But when I thought about the assumptions themselves, my perspective changed a bit.

The axioms in mathematics are what I would call "philosophically natural" assumptions. It is a bit hard to define, since the foundation of math is, as you said, in human ideas, but if I had to, I would say philosophically natural assumptions are assumptions that you cannot possibly imagine being wrong under any hypothetical circumstances. Considering that, math is nothing but an attempt at defining all that is true in the universe.

Our current math, however, is not enough to determine everything. There are certain claims that cannot be proven or refuted, and a natural axiom that can change that has yet to be found. Furthermore, there are other assumptions we have to make in order to talk about our physical world, which I would call "physically natural" assumptions and are certainly not "philosophically natural". One could definitely imagine a world without gravity, for example, but predicting what would happen if you throw a rock into the air without assuming it exists would be very difficult. The need for discussing our own physical world (and not any imagineable world) and, the necessity of making assumptions on how it works to do so, bred the science of Physics.

This is how I perceive math and science in a nutshell.

yeah you're trading a 33 percent chance with a 50/50 if you switch. 

 Game shows know people tend to stay with there original pick so they offer this option. If everyone did the smart thing they may not.

 It was on an episode of mythbusters.