A hustler presents you with three boxes into one of which he places a valuable prize, but you don't know which one. You get to choose one box and keep the prize if it's there. You choose, but he doesn't tell you if you won. Instead, he says to you "I'm going to open one of the boxes and show you what's inside." He does (and you see that there's nothing inside), and then says, "Now, would you like to stay with your first choice, or would you like to switch."
Should you switch or stay?


"It is well, when judging a friend, to remember that he is judging you with the same godlike and superior impartiality."~Arnold Bennett

Well now, let me see - my first instincts I have to say would be to go with my first choice - but that's just how I am!

~~~~Littera scripta manet~~~ the written word remains. (the saying continues; The weak word perishes)

I was in a hurry when I posted this and neglected to mention that my post was "copy and paste."
It is from
http://explorepdx.com/mathcube.html#dvd
And this one
http://explorepdx.com/montyhallpuz.html

My boss brought this up for us to discuss, and the discussion was rather lively for a bit.
The short answer is: switch!
This question is a little bit like an example of conditional probability.


"It is well, when judging a friend, to remember that he is judging you with the same godlike and superior impartiality."~Arnold Bennett

UGH - I could wrestle with that for hours! (indecisive with things like that) but more than likely - I would switch.

"Flatter me, and I may not believe you. Criticize me, and I may not like you. Ignore me, and I may not forgive you. Encourage me, and I will not forget you."
~William Arthur Ward

Very interesting, TN.
I'm sure I would've stayed with my first choice.. now I'll think twice!


--------Sanya--------
Stella Splendens
December 22, 1985-March 27, 2003
Rest In Peace
..lost time is gone forever

When I was in high school, my math teacher presented me with the same problem. The class was about a 50-50 split on staying and switching, and then our teacher informed us that statistically there were higher odds if you switched. That idea always bugged me, but yet the math seemed solid, so I couldn't realistically challenge it. But, now I have studied more and broadened my understanding of math and probability and have come up with my own theory on it, which would state that its equal probability with either choice.

If you have ever decided something by flipping a coin, then you will know that the probability of you winning is 1-2. So you flip a coin, and choose tails. The coin lands heads. The next day you flip the coin again, but this time you pick heads, and the coin lands tails. This defies standard laws of probability, because even though had one-in-two odds, and you played twice, you still lost, which according to the fundamentals of probability is impossible. This is because there is a common problem with people's perception of probability. The odds of winning a coin toss are 1-2, but probability has no memory, hence you could lose hundreds of times in a row in a game where the odds are 1-2.

The same principle applies in this game: you have to pick one of three boxes, which translates to 1-3 odds. You pick a box, and one is removed, leaving you with your box and one other, one of which containing a prize. Now this is where most people tend to think that the odds are 2-3 if you switch boxes, and 1-3 if you stay with the same, but if we go back to the coin toss, and apply the rule that probability has no memory, then the odds become 1-2 for either of the remaining boxes, which leaves them at even odds, with neither keeping or switching boxes giving you an advantage.

For those math hounds out there, that will fight me on the "memory theory", there is also another possible solution. If probability did have a memory, then the odds of the other box containing the prize are 2-3, but most people overlook the fact that either box would then have 2-3 odds on it containing the prize. For, if you did switch boxes, then your original box just jumped odds from its original odds of 1-3, and is now the "other box" and would now have an odd of 2-3 itself. Hence, no matter how you look at it, neither switching or keeping the original box will make a difference for the probability.

Unless I'm misunderstanding the premise, it makes no difference whether you "switch" or "stay":

Three boxes: A, B and C... The prize is in one of them. The probability is 1/3, evenly spread between A, B and C.

You choose randomly, say Box A.

You are then shown that Box B has nothing in it. Box B plays no further part in the matter.

So, the prize is either in Box A or in Box C.

Beyond that, nothing is known about the contents of Box A or Box C.

So the probability is now evenly spread (ie 1/2) between the two "remaining" boxes A and C.

Therefore, switching or staying makes no difference - I think this puzzle is a "hoax" of some sort.

Davdoodles
XXX

"I'm going to open one of the boxes and show you what's inside."
This implies that the chosen box could have been the one opened.

"I disapprove of what you say, but I will defend to the death your right to say it." --Voltaire

quote:

---

"I'm going to open one of the boxes and show you what's inside."
This implies that the chosen box could have been the one opened.

---

Read the description here. It makes it clear that the "chooser" is aware that the box he/she chose is not the box shown to be empty. But, good thinking anyway Twister.


The more I think about this the more I'm confident it IS a "Hoax". I'm certain in fact:

Thenostromo's links (above) show "wins" and "loses" represented by plastic columns full of nuts. Each "game outcome" results in one nut being thrown in one of four columns. the columns are divided into two "groups" - Stay" and "Switch".

Forget about comparing the numbers of "wins" to "losses" in each (Stay/Switch) group - it's a red herring.

Instead, add together and then compare the total number of wins (ie in both the "stay" and the "switch" groups) with the total number of losses in both group. Figures from an experiment can be found under the photos on this page.

In that example, it looked like this:

Switch:
win 107
lose 48

Stay:
win 94
lose 170

Lets total the "wins": 107+94= 201

Lets total the "loses": 48+170= 218

After four hundred or so throws, almost precisely half (about 201) are "wins" and half (218) are "loses". The slight discrepancy (201-218) is statistically insignificant).

"So far so good, Smarty-Doodles" you say, "but 'wins' and 'loses' are clearly different in proportion to each other in the two (Stay/ switch) groups. Look at those plastic column/tubes full of nuts. Isn't it true that by switching you increase your chance of winning, and by staying you decrease your chance of winning?

Short answer: Nope, it just looks that way.

Here goes: Ignore the "lose" columns for the moment. Look only instead at the the "wins" columns in the example I linked above and several examples on this page.

Notice that in each example, the "wins" columns are roughly equal to each other (in "height"). In the example above it is 94-107. Mildly but not statistically significantly in favor of switching. This shows that, whether you "stay" or "switch", your nut is really just as likely to end up in a "win" column. In other words, the chance of winning is the same whether you "stay" or "switch".

Almost there. I agree that the "loses" columns are remarkably different to each other (and very distracting ), but this merely suggests that more people CHOSE to "stay" rather than "switch" - why that is is interesting in itself I guess, but it has nothing to do with whether the prize is more liklely to be in Box A or Box C.

You may go now .

Davdoodles
XXX

GamzPlayer: "The odds of winning a coin toss are 1-2, but probability has no memory, hence you could lose hundreds of times in a row in a game where the odds are 1-2."

Three Box Puzzle is a simple problem of conditional probability. It has nothing to do with that Memory Theory you came up with. Memory Theory you mentioned is about *series* of games, and here, we are talking just about this ONE game and this ONE game involves simple conditional probability, ie given a particular situation, what is the chance of this or that happening.

I gave my boss some of this feedback and this was his reply

quote:

---

Too bad his explanation doesn't account for all the experiments conducted that PROVE the 2/3 probability if you switch! The problem in his thinking lies in his improper analogy of the "coin flip" and the mis-application of the "probability has no memory" statement. His analogies are only valid if he randomly decides to switch his choice (= 50/50).

By extension of his reasoning, I could hand him a deck of cards (52), and ask him to randomly try to select the "Ace of Spades" out of the deck. Not good odds (1/52), right? Then I remove one card at a time from the deck (that is not the Ace of Spades) and show the card to him (just like the box experiment), until there are only 2 cards left: the card he selected, plus one other card, one of which by definition is the Ace of Spades.

At this point it should be easy to see that if one never switches from his original card selection throughout the process, one has an extremely slim chance of being correct (1/52), not 50/50!

Please note that if he flips a coin to (randomly) decide what to do at this last decision point (2 cards left), his odds then instantaneously become 50/50 (just like the coin tossing analogy). But why do that when he has a ~98% chance of being correct if he switches cards only at the last decision point!

It took the substitution of a "deck of cards" analogy in place of the "three boxes" experiment to help get me mentally over the hump and accept that the probabilities are not as they first seem (50/50). Such is the power of cognitive illusions. This is possibly yet another reason why Las Vegas makes so much money!

---

To which I must reply that expanding the size of this puzzle changes the parameters being discussed. A choice of one card that I must stay with out of 52 and a process begins where 49 cards are eliminated and I am allowed to decide if I want to stay with my original choice or switch to the one remaining card on the table that could possibly be the Ace of Spades, of course one should switch. But I think this example is non-sequiter and does not fit.
[This message was edited by thenostromo on 04-20-04 at 01:04 AM.]
Back for another edit. Darn. I think I'm seeing it. Even in the larger 52 card comparison, your first choice has got 1/52 written all over it, the third remaining card that is going to be shown to be not the Ace of Spades suddenly gets removed from contention, and the last remaining unknown card had 1/2 written all over it. I think I'm seeing it.

[This message was edited by thenostromo on 04-20-04 at 01:10 AM.]

Another explanation of the puzzle can be found here: http://www.encyclopedia4u.com/m/monty-hall-problem.html

I still disagree. There are only four possibilities:

1. Switch - Win
2. Switch - Lose

3. Stay - Win
4. Stay - Lose

Each has an equal chance of being "chosen". There are two "wins" and two "loses". Simple maths says the chance of a "win" is one in two.

This stuff about 2/3 v 1/3 is rubbish. It's a red herring. There are only two boxes left.

The prize is either in one box or it is in the other box.

So again, the only relevant statistic is 1/2

It makes no difference whether you switch or stay - look closely at the columns of pebbles (links in my earlier posts) - the "stay-win" column and the "switch-win" column are basically the same height as each other - this shows that you chance of winning is the same whether you change or switch.

TN, your card example supports this outcome - you can start out with a million cards if you like, but when only two are left, you said it yourself: it's 1/2...

Davdoodles
XXX

The card example is illustrative. Davdoodles, I think you're overlooking one thing:

The card selected by the imaginary protagonist is presumably selected at random, hence the 1/52 odds. So, as the antagonist removes cards he knows are not the ace of spades, two cards are exempt from this selection process - the card the protagonist selected, and the ace of spades, which is chosen with foreknowledge, not at random.

Imagine a similar deck of cards in which the ace of spades' logo is printed on both sides, and then let's take a blind protagonist who cannot see this and an antagonist with clear vision who can. If both must select different cards and the protagonist selects first, who has the better chance of selecting the ace of spades? Clearly, the antagonist, unless the protagonist happened to select it first - a 1/52 probability. Barring this possibility, the antagonist is certain to select the ace of spades as the remaining card next to the protagonist's selection.

This example illustrates the principle you're overlooking - the protagonist chooses at random, whereas the antagonist, because the outcome of his selection is predetermined, does not. Hence normal rules of probability no longer apply and the advantage falls in favor of the 'other' choice in proportion to the size of the initial pool of choices.

I should clarify one thing in the previous post - when I refer to the antagonist 'selecting' a card, I mean that, since the outcome is predetermined that the ace of spades will be among the remaining two final cards, then if the protagonist does not select this card then the antagonist is sure to hold it back (select it) to be among the final two.

I've been away for a while, but see this is still running, so...

Aren't we looking at two things here? The puzzle only relies on the fact that my first choice was wrong! Nobody has factored in the number of times (1 in 3) that I chose the right box in the first place. If that choice is wrong and I make another choice, my maths tells me that I've a 2 in 3 chance of picking the right box: I've just chosen two boxes out of three.

I'm with Davdoodles on this one. If I've eliminated one box because it is empty then I start another choice process: the choice between two boxes is 1 in 2.

Does that make sense?

Get Curious!

No, no - the first choice isn't necessarily wrong in the box puzzle. If I'm reading it correctly, the idea is that the hustler opens one of the boxes that you didn't choose, and it was empty. Hence, your original choice could still be correct - a 1/3 probability.

"The prize is either in one box or it is in the other box.
So again, the only relevant statistic is 1/2"

If you toss a coin, there can be two outcome, heads or tails. You probably notice that I have said nothing about the nature of the coin that is being toss, that is, whether it is a fair coin or a biased one.

If the chance for tossing a coin is 50/50, the coin has to be fair one. However, it is clear that the converse is not necessarily true, that a coin with two sides does not make a fair coin.

Consider, the rule of the game is that you choose only one box from the start (with 1/3 chance of winning), and the game host has the remaining two boxes (with 2/3 chance of winning). This rule is saying: if there is one prize contained in one of three boxes, and that the three boxes are divided into a group of one box and a group of two boxes, in which group do you think the prize would more likely be found.

This situation is the same as tossing a biased coin. Opening of an empty box of the host's is only a nominal act to cause confusion, because there MUST be an empty box in the host's lot. But the host still has a 2/3 chance of winning, because he was the group with the two boxes to start with. Just like the host has been given the 2/3 side of a biased coin! True, there are only two boxes left, but as shown before, having two outcomes do not warrant a 50/50 outcome. The probability at the start remains the same for both the play and the host.

"It makes no difference whether you switch or stay - look closely at the columns of pebbles (links in my earlier posts) - the "stay-win" column and the "switch-win" column are basically the same height as each other - this shows that you chance of winning is the same whether you change or switch."

In your examples, there are many more Stays than Switches. It would help if you simulate this experiment with an equal amount of Stays and Switch. Then you can really see how your results stand with your theory.

I encountered this before in a logic course. It's one of those things that I accept grudgingly because I know that I have an elementary understanding of the issue. But I always figured it would not matter.

Using the deck of cards example: Picking out any card, knowing that 50 are to be discarded as junk, ultimately yields the decision to switch with the final remaining card or keep the card in hand. This indicates 50/50 odds to me since it is essentially a choice between these final two contenders.

I feel like wiping the dust off Visual Studio and writing a simulation for this, because I think that's the only way I'll accept it. Hmmm...

-----------------------------
Pobody's Nerfect.

i would have a little fun with the husler and say, what if i told you that thats the box i chose? then after he replies, i would say, maybe it was and maybe it wasn't.
then i would shake his hand then walk away.

that way we are equal

quote:

---

Originally posted by thenostromo:
A hustler presents you with three boxes into one of which he places a valuable prize, but you don't know which one. You get to choose one box and keep the prize if it's there. You choose, but he doesn't tell you if you won. Instead, he says to you "I'm going to open one of the boxes and show you what's inside." He does (and you see that there's nothing inside), and then says, "Now, would you like to stay with your first choice, or would you like to switch."
_Should you switch or stay?_

http://www.earthlife.net/insects/images/lines/owl01.gif
_"It is well, when judging a friend, to remember that he is judging you with the same godlike and superior impartiality."~Arnold Bennett_

---