Take three spoons--any three spoons. Declare one of them valuable and ask friend to split them into two groups. Ask him which group likely contains the valuable spoon, the singleton or the pair.

When he says the pair, point out to him that one of the pair is worthless. [-X

When he points out that there will always be a worthless spoon in the pair and that the other one is likely to be the valuable one, actually point to the worthless spoon. [-(

When he still insists that the other spoon of the pair is the most likely to be the valuable one, congratulate him =D> on proving the answer to Monty Hall's Puzzle. 8)

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Keeper of the Jabberwock

What does moving two of the spoons over to the other side of the table have to do with it?

Those represent the two doors that the player didn't pick the first time.

what if, the spoon by itself is the valuable spoon?

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Those that lack education have a hard time understanding its value. - Cross

1/3 of the time, it is.

The other 2/3 of the time, the valuable spoon is in the duo. That's the whole point.

I was thinking about this last night and came up with almost the same thing.

It should be obvious that if you're allowed to choose two doors, and you win if the big prize is behind either of the doors you choose, then you've got a 2/3 chance of winning - right?

That's essentially what happens when you switch. Rather than winning if the big prize is behind the single door, you win if it's behind either of the other two.

I guess you could also look at it as if you're picking a door that you think the big prize is not behind, and you win if it's not behind it.

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SeanF

"Ask to understand, but don't challenge unless you have the knowledge."--NEOWatcher

The contents of this post are ©2010 by SeanF and may not be copied or retransmitted in any form without the express written consent of SeanF

I remember one of my professors in grad school brought up this problem, and proceeded to try to prove it using Baysian statistics. While his conclusion was correct, his proof was flawed although it looked reasonable. He was unable to complete the proof in class (and wasn't too happy about me pointing out the flaw in his proof). I then pointed out that his proof was a perfect example of why I find statistics to be so frustrating a subject. You can take something that seems intuitively correct, provide an incorrect proof that looks reasonable , and not know that you are wrong. If you are dealing with more advanced statistics, involving questions for which you don't know the answer, then it is very hard to be sure that your proof is correct.

Probability theory.

The problem is that most people have flawed probabilistic/statistical intuition. However, some exceptional people would immediately give the correct answer. The rest of us need to rely on some kind of method.

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Once again, I'm here to demonstrate just how dumb I am!!! :roll:

Let's take this from the top.

You are presented with three doors. Behind one of the doors is a great prize. Behind the other two are zonkers. You don't know which door hides the great prize.

You are asked to pick one of the doors. You pick door number two. At this point, you have a 1 in 3 chance of picking the great prize.

Monty, being the showman that he is, opens door number 3 and reveals a zonker. The chance that your door hides the great prize raises to 1 in 2. The chance that you picked the great prize remains 1 in 3.

Monty then asks you if you would like to keep door number two or change to door number one.

NO MATTER WHAT YOU DO, you now have a 1 in 2 chance of making the correct choice. Why??? Because you now have to choose between ONLY TWO things; stay with your original choice, or change to door two. The effect of the option of changing doors is equivelant to starting the game over with only two doors instead of three.

IF you were not given the option of changing doors, which is not equivelant to keeping your original choice, then your chance of choosing the correct door would remain 1 in 3.

Does this make sense???

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Yes, it makes perfect sense, and that’s what I thought yesterday. But today I’m having second thoughts. I think my “method” of flipping the coin might reduce my chances from 2 in 3 to 1 in 2.

See what you think of this chart:

Three possible arrangements of the goats and prize:

g......g.......p Pick 1, switch you win, pick 2, switch you win, pick 3, switch you lose



g.......p........g Pick 1, switch you win, pick 2, switch you lose, pick 3, switch you win



p........g........g Pick 1, switch you lose, pick 2 switch you win, pick 3 switch you win.


Always switch = 6 chances to win, 3 chances to lose.


g......g.......p Pick 1, don’t switch you lose, pick 2, don’t switch you lose,
...........................pick 3, don’t switch you win


g.......p........g Pick 1, don’t switch you lose, pick 2, don’t switch you win,
...........................pick 3, don’t switch you lose

p........g........g Pick 1, don’t switch you win, pick 2 don’t switch you lose,
..........................pick 3 don’t switch you lose


Never switch - 6 chances to lose, 3 chances to win


Always Switch = 6:3 win

Never Switch = 3:6 win

Picture this

You have a thousand doors (spoons). One has a prize, the others don't. You pick one door. Then 998 doors are opened to reveal that there wasn't a prize behind those doors, which leaves two closed doors. The one you picked and one other. Do you stay with your original guess or change?

The probability is not 50:50 don't forget that you've been given LOTS of additional information about where the prize is not hidden.

What was the probability that your orignal choice was correct? You should switch doors.

This problem is very counterintuitive, but that's the nature of conditional probability

Does this help?

For more info and other math problems, check out www.askdrmath.com and do an archive search, you should find this problem and a better explanation than I gave

Later

Pete

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PJE

There's so much I don't know about astrophysics. I wish I had read that book by that wheelchair guy.

Why is not being given the option different than being given the option but not changing?

As I said in the other thread, can you describe a situation in which you'd win if you were given the choice and didn't switch, but would lose if you weren't given the choice (or vice-versa)? If not, then the odds don't change.

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SeanF

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The contents of this post are ©2010 by SeanF and may not be copied or retransmitted in any form without the express written consent of SeanF

Stop.

Just because there are now 2 doors to choose from does not mean that they are equally likely to have the prize.

In another game, I give you the choice between 2 doors, with the information that 1/3 of the time I will put the prize behind door A and 2/3 of the time I will put the prize behind door B. You would pick door B. Right? Wouldn't you?

That's what Monty's game effectively does. The door you first pick is door A. It has a 1/3 chance of hiding the prize. Door B is the other. It has a 2/3 chance of hiding the prize. If you're smart you'll switch to door B.

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Hang on. Instead of trying to do numbers and statistics, go back to my OP and really think about it. The situations between the spoons and the doors are identical.

Let me extend the idea slightly. I have two pieces of wood that have slots for a single spoon. One piece of wood has one slot and the other has two slots.

When I randomly put the spoons in the slots, I'm more likely to put the valuable spoon in the two-slotted block than in the one-slotted block.

But the two slotted block will ALWAYS contain a non-valuable spoon. It might be in the right-hand slot or it might be in the left-hand slot, but it will always be in one of them. In fact, sometimes BOTH slots will contain non-valuable spoons.

But the majority of the time, one of the slots in the larger block will contain a valuable spoon and thus one should always select the larger block.

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Keeper of the Jabberwock

My chart above shows what happens if you always switch or always don't switch.

SeanF asked:
Because in the first case, not being given the option to change, you are stuck with your first choice. You cross your fingers and hope that you chose right the first time. In the second, being given the option, you now must choose to stay with your first choice. You get to choose again.

And that is the whole point of this.

In the first round, you have to choose between three paths:

Door One or Door Two or Door Three.

Then, after door three is opened, you now are faced with a choice between two paths:

Do I stay with door number two or Do I change to door one?


01101001
Yes, it does because there are now only two doors and only one of which has the prize.

You change the situation entirely with your next statement:

I used to watch "Let's make a Deal" religiously. I don't remember Monty ever giving any information like you present. If he had, or if the original scenario that I presented contained that information, then you may be correct.

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Experience is a wonderful thing. It enables you to recognise a mistake when you make it again.

and so the wheel makes another turn. . . .

I think the best way to explain the "Monty Hall" problem is to approach it from Monty's perspective. Don't try to explain it with math -- probabilities are almost always confusing to people.

Besides, the psychology of "Monty" is the key to the problem. If you make different assumptions of how he'll act, you can come up with different answers to the puzzles.

Let's look at Sam5's example
Sam's logic works because he assumes (correctly from my recollection of the show) that Monty will always open a door to reveal a goat.

But suppose Monty decides he will always open the door to the right of the one you guess, regardless of which is in it. So then the chart becomes:

g......g.......p
Pick 1, Monty opens door 2, switch you win;
pick 2, Monty opens door three and now you now where the prize is. Chances are Monty won't let you choose it.
pick 3, Monty opens door, switch you lose

So in the three possible cases, there are two cases where you'd get a second choice. And for those two cases the odds are 50/50. So it doesn't help you to switch or stay.

He's telling you exactly that.

When you choose door A, it has a 1/3 chance of being correct. That means doors B and C together have a 2/3 chance of one of them hiding the prize. Then, Monty tells you, shows you, that door C does not hide the prize.

Yet, still, doors B and C together have a 2/3 chance of hiding the prize. What has chnged is that you know it's not door C. Therefore, you know door B has a 2/3 chance of hiding the prize.

Don't stick with A. Switch to B. You now know door B is twice as likely to hide the prize.

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One thing should be made clear.

The first choice doesn't matter. If I remember right, sometimes Monty OPENED THE FIRST DOOR FOR YOU. . .and then asked whether you wanted to open another door. BTW, the door that was opened contained a nice prize but one of the other doors contained a better one.

This makes the choice obvious, but it really isn't that much different than the original puzzle. . .think about it.

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Keeper of the Jabberwock

What's the obvious choice?

01101001 wrote:
No, he isn't telling you that.

Each door has a 1 in 3 chance of hiding the prize. I think the flaw in your reasoning is this; first, combining the chances of doors B and C and then, second, assuming that after door 3 is opened that door B has still a 2 in 3 chance of hiding the prize.

IF
THEN
That means doors A and B together have a 2 in 3 chance of hiding the prize.

THEN IF
THEN
The only thing I know is that it's not behind Door 3. I still don't know if the prize is behind Door A or Door B. Using your reasoning Door A also has a 2 in 3 chance of hiding the prize.

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Remember. Just because I'm sure doesn't mean I'm right.
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Nope. Before Monty acts, A and B together have a 2/3 chance and C a 1/3 chance. Once Monty acts and shows C, you know A and B together have a 1/1 chance of hiding the prize. That is not useful information. It does not mean that A has a 1/2 chance and B has a 1/2 chance. You've already agreed that having two doors does not mandate that they are equally likely.

A has a 1/3 chance and B has a 2/3 chance, again, by virtue of the fact that B and C together have a 2/3 chance and you know it's not behind C.

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GO BACK TO MY ORIGINAL POST AND DISPROVE IT!

Any other statistical argument (for and against) is math. Mine is a front and center real-life equivalent of Monty's puzzle. Disprove my example.

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Keeper of the Jabberwock

I agree. It was quite illustrative.

Astro, would you pick the singleton spoon, or the pair? The pair, I'm sure. And once you know which of the pair has no value, I certainly hope you'd persist in going with the remaining half of the pair. You'd be right 2 out of 3 times.

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I think the problem here is that approximations do not hold their accuracy before and after the addition of information.

That "2/3" for "B and C" is only an approximation*. It really means (before you get the extra clue) that B has a 1/3 chance and C has a 1/3 chance. (It's not like B is 1/6 and C is 3/6, or something)

So when you know that C is worthless, you are left with both A and B having their 1/3 chance (which are now of course 1/2 of the remaining choices).

---

As was noted, you could also see it as A and B having 2/3 chance, and C 1/3. That fact that you picked A is irrelevant. When C is shown to be worthless, A and B are still 2/3 of the original chance, and are now both 1/2 of the remain chocies.

---

In any case, statistics (here) is about predicting the likely outcome.

If you study a comple population you will get the "real" chances. And it's easy to show (I just made a table in Excel) that whether you change your choice or not, you have a 50% success rate.

Reality beats statistics every time.


[edit, forgot to add my footnote:]

* I'm not at all sure this is the right word to use here.

AstroRockHunter and pzkpfw, I suggest you go to the web and try one of the many simulations of this. It's easy to try to argue the math, but there's nothing quite so convincing as seeing it in action.

No, it isn't, because what you picked has influence on Monty's decision over which door to eliminate. That's why it's relevant.

If you started with 2 doors then you are correct in saying that it doesnt matter which door you choose.

However, you started with 3 doors, and you probably picked the wrong door. One of the other two doors probably contains the prize. By opening one of those two doors, Monty, nice guy that he is, just showed you which of the other two doors you should have picked.

It might be more clear if instead of 3 doors you have 100, but just one prize. You pick a door (which has a 99% chance of being wrong), and then monty opens 98 others, leaving one unopened and you are asked if you want to switch.

I'm starting to regret that I brought this topic up. I think this is will be my last post on this topic. I know there are some simulations on the web where you can try it yourself and see the results. Maybe you can try that and at least see that switching is the correct strategy, even if you don't fully understand why.

I meant it's irrelevant to whether A and B have a 2/3 chance, before door C is opened.

I agree it influences the choice of which door is opened, since if you picked an incorrect door, Monty only has one choice of which door to open and show it has no prize. But YOU still don't know whether you picked the wrong door and whether Monty had only one or two choices.

Copy and paste the following into Excel. The use menu item "Data | Text to columns" to split it:
[
Door A,Door B,Door C,Pick,Open,Change?,Final Choice,Wins?
P,Z,Z,A,B,Y,C,N
P,Z,Z,A,B,N,A,Y
P,Z,Z,A,C,Y,B,N
P,Z,Z,A,C,N,A,Y
P,Z,Z,B,C,Y,A,Y
P,Z,Z,B,C,N,B,N
P,Z,Z,C,B,Y,A,Y
P,Z,Z,C,B,N,C,N
,,,,,,,
Outcomes:,,,,,,,
,Changed,Did Not,,,,,
Win,2,2,,,,,
Lose,2,2,,,,,
]
Assumptions:
Monty will only open a non-prize door.
(This is why there are only two cases each for when B or C is picked)

Where did I go wrong in this?

Note that I'm only looking at one of the three cases for where the prize is. That does not matter.