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3 Door Logic Problem

From the Wikipedia article, here is the problem with the constraints made explicit:

A thoroughly honest game-show host has placed a car behind one of three doors. There is a goat behind each of the other doors. You have no prior knowledge that allows you to distinguish among the doors. "First you point toward a door," he says. "Then I'll open one of the other doors to reveal a goat. After I've shown you the goat, you make your final choice whether to stick with your initial choice of doors, or to switch to the remaining door. You win whatever is behind the door."

You begin by pointing to door number 1. The host shows you that door number 3 has a goat.

~~ Paul
 
This is my flow of thought about the coin puzzle

agreed. Marilyn is right on that one.

There are 4 possible results for two tossed coins:

HH HT TH TT

Person A has ruled out TH and TT


So far so good.

Person B has only ruled out TT

I can't see it like that

For me, Person B has also ruled out either HT or TH, we just don't know which one.

So Persons can have following scenarios

Person A:

HH or HT

Person B

HH or (TH or HT)

The coins don't know whether they were the first or second.

Person A provided information to make us assured the first coin was heads, Person B provided information to make us assured either the first or the second coin was heads.

For both persons A and B , there's still 50% chance of the remaining coin being heads.

However, if we tossed coins and counted the cases where at least one of the coins was heads, we would find that 33% of those cases had HH. Also if we counted only the cases where first coin was H we would end up with 50% of those cases being HH.

Doubts arise in my mind...


Additional mystery for person B's scenario is only about which of the coins was heads. Or is it? I'm not so sure anymore...

Try it with cards version.

Each of Persons are given a deck of four cards. two black cards, two red ones in each deck.

Person A says his first card was black.

Person B says one of his cards was black

lets examine what there is in the remaining deck in every possible scenario:

Person A

hand deck chance
BB RR 50%
BR BR 50%



Person B

hand deck chance
BB RR 50%
BR BR 50%
RB RB 50%

wait... something's fishy.




Person B

hand deck chance
BB RR 50%
BR BR 25%
RB RB 25%


I guess my point is, both have certainly removed one black card from the deck, it doesnt matter which red card they remove if they're gonna do it.

But eh... so if there's two red cards remaining then he would pick the black only 1 out of 3 times so the correct chart would be


Person B

hand deck chance
BB RR 33%
BR BR 33%
RB RB 33%

But how do the coins/cards know which one of them is first? What if the coins were tossed to two boxes ("box for first coin" and "box for second coin") and I were to "cheat" and place the head coin of person B to the "box for first coin" even if it was the second coin. Would that move person B actually to same situation as where person A is?

I'm confused.



but wait, to make it more like the coin puzzle, I would also have to take one of the other color off of the deck after they made the initial selection of cards.

So if they both take one black, I will remove one red (because throwing a heads will remove the possibility of there being tails for that coin)

So after whichever the card they drew there is only one red and one black card remaining in deck for both, right?
Equally, after whichever coin was heads, there's only one possibility for tails left, be it from either coin. There can't be two more tails left in the "deck" after one coin is known to be heads.

So Marilyn isn't removing one of the "cards" from the "deck" or the "tails that can't be"


I think I'm on to something but still confused.
 
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Nope. Doesn't matter.

If the only reason he offers a switch is because you picked the car, then why would you switch?

Because some people who think they are smart will change automatically, without realizing their logical mistake?

Read Monty Hall's comments quoted here

http://www.wiskit.com/marilyn.gameshow.html

...the only person who can answer this version of the Monty Hall Problem is Monty Hall himself. Here is what should be the last word on the subject:
"If the host is required to open a door all the time and offer you a switch, then you should take the switch," he said. "But if he has the choice whether to allow a switch or not, beware. Caveat emptor. It all depends on his mood."

For example:
On the first, the contestant picked Door 1. "That's too bad," Mr. Hall said, opening Door 1. "You've won a goat." "But you didn't open another door yet or give me a chance to switch." "Where does it say I have to let you switch every time? I'm the master of the show. Here, try it again."
 
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The short answer is that the situation _as given in the original post_ does not provide enough information to solve the problem.

Hmmm. That reminds me of the "plane sitting on a conveyor belt runway problem".
 
This is my flow of thought about the coin puzzle
I can't see it like that

For me, Person B has also ruled out either HT or TH, we just don't know which one.

If we don't know which one, then we can't rule it out. He is able to answer "yes" to "is at least one of the coins heads up" with HH, HT or TH. The reason that it is important to count both HT and TH is because one head/one tail is twice as likely as two heads.

The coins don't know whether they were the first or second.

They don't have to. Numbering the coins is a convention to make analyzing the puzzle easier. There are two ways to get one head/one tail and one way to get two heads, with each way being equally likely. That makes B's chances of having two heads half as likely as having one tail.

A, on the other hand, has removed one coin from the equation by telling us what it was, so the only probability left is what the other coin is, which is 1/2 each way. He therefore has a 50% chance of having two heads.

Although, what bothers me about this puzzle is why would A bother to specify the first coin was heads up if they both were?
 
Although, what bothers me about this puzzle is why would A bother to specify the first coin was heads up if they both were?

Maybe A only looked at the first coin.

If I was to toss two coins independently and hidden from the observer and then had to look for one tails, I just might look at them individually - and not look any further if I saw tails after checking the first coin.

Perhaps A is covering one coin in each hand, whilst B tosses in a way that he cannot tell which coin was first at all.
 
Let's extend the coin puzzle with a couple more persons.

Person C says: Yes, actually the second coin was a heads.

Now person C has 50% chance of having two heads, same as A, right?

Person D says: In my situation, either Person A's or person C's comment applies.



Now, what are the odds of person D having two heads? To me, it seems he essentially said "yes" and has the same probability as person B

But then again D has only two options, he is either exactly like A or exactly like C and should have the same odds.
 
For me, Person B has also ruled out either HT or TH, we just don't know which one.

So Persons can have following scenarios

Person A:
HH or HT

Person B
HH or (TH or HT)

The coins don't know whether they were the first or second.

Person A provided information to make us assured the first coin was heads, Person B provided information to make us assured either the first or the second coin was heads.

As Marylin's explanation goes, based on the answers to the question "do you have at least one head"

A - First coin was H
B - Yes

From this we know that
- niether A or B has "TT"
- A does not have "TH" (first coin was H)

Which leaves the following possible combinations for each persons
A - HT HH
B - HT TH HH

Within each persons combinations, the combinations are equally likely
A - HT (1/2) HH (1/2)
B - HT (1/3) TH (1/3) HH (1/3)

So the probability of A having HH is 1/2, the probability of B having HH is 1/3, based on the information supplied.
 
Although, what bothers me about this puzzle is why would A bother to specify the first coin was heads up if they both were?

Perversity on the part of the puzzle creator, to make the problem a puzzle for those intuitive thinkers to stumble over.
 
Maybe A only looked at the first coin.

If I was to toss two coins independently and hidden from the observer and then had to look for one tails, I just might look at them individually - and not look any further if I saw tails after checking the first coin.

Perhaps A is covering one coin in each hand, whilst B tosses in a way that he cannot tell which coin was first at all.

Except that the way the question is worded, it sounds like they were only asked after they had flipped both coins and observed the result. But I could be reading too much into it.
 
Let's extend the coin puzzle with a couple more persons.

Person C says: Yes, actually the second coin was a heads.

Now person C has 50% chance of having two heads, same as A, right?

Assuming he's not choosing the second coin only because the first was tails, yes.

Person D says: In my situation, either Person A's or person C's comment applies.

Now, what are the odds of person D having two heads? To me, it seems he essentially said "yes" and has the same probability as person B

But then again D has only two options, he is either exactly like A or exactly like C and should have the same odds.

If we can rule out the possibilty that D's statement is an exclusive or (in which case he can't have two heads) then the odds should be the same as B because he hasn't given us enough information to rule out another outcome besides two tails.
 
The Woo View of Monty Hall

There's a woo aspect to the Monty Hall Problem we haven't touched on yet:

The average person believes he or she has some kind of faint ESP and would make a gut decision on which of the three doors to pick initially. Therefore, when Hall opens a goat door and offers the contestant a chance to change his choice, he's more likely to hold than to change, because changing shows he doesn't trust his gut.

So you have three possible attitudes for that contestant after one door is opened:

1) Stick with the first choice, because I trust my intuition.
2) It's now 50/50 so it doesn't matter.
3) Change my choice, because my intuition is probably wrong.

The woo choice would always be #1, easily proven to be the worst choice.
 
The fact that the door he opens isn't the one I chose makes me think of it as irrelevant to the outcome.
 
How does "The Angel's Gift" go?
You decide, on a warm clear day, to go to the beach. You resolve to go to an island that has three beaches. You recall that one of the beaches is nice, and two are badly polluted, but can't remember which is the nice one. You decide randomly on one of the beaches, but halfway there, you recall clearly one particular beach you didn't choose that was polluted. Should you change your choice of beaches, or continue to the one you've already decided upon? Assume you have no hint of any unrecovered memory.

Now the correct answer to THIS one is that both remaining choices are equally likely, right? The two problems are not analogous?
 
Okay, I heard it explained 500 times, and I STILL CAN'T WRAP MY HEAD AROUND IT. All I can understand is simply a chart laying out all the options. The conclusion is clear, I just can't wrap my head around WHY it is clear, at all.

And all of you giving examples like "well imagine there are a billion boxes, then it's obvious isn't it?". Nope, that didn't help at all, because all billion of them are no longer a part of the puzzle once they've been opened. No matter how many extra boxes are added and opened, I'm still left with box I picked and box I didn't pick.

Somehow though, the math works out saying I need to switch.

I guess all I can really do is accept it.
 
Okay, I heard it explained 500 times, and I STILL CAN'T WRAP MY HEAD AROUND IT. All I can understand is simply a chart laying out all the options. The conclusion is clear, I just can't wrap my head around WHY it is clear, at all.

[snip]

Somehow though, the math works out saying I need to switch.

I guess all I can really do is accept it.

That's the whole point. If you follow your intuition (wrapping your head around it), you get the wrong answer. You have to go by hard mathematical logic. The neat thing is that even if you know how to get the right answer by logic, you can never "feel" the right answer in your gut. Logic=Randi. Gut=woo.
 
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Now the correct answer to THIS one is that both remaining choices are equally likely, right? The two problems are not analogous?

Good call! I actually thought they were analogous when I made up the "Three Beaches Puzzle" and wrote it into the forum. An hour later while driving my car I realized the flaw in my analogy. It's remarkably subtle, and is actually another manifestation of the principle of the Monty Hall Puzzle.

So you guys can figure it out for yourselves, I've put the answer in a spoiler window:


In the original game, after the contestent picks a door, the MC has to use the information of which door was chosen to decide which door to open. This gives the contestant information which indicates which door to switch to. The Three Beaches Puzzle is different because the recollection of one of the beaches should be independent of which beach was randomly chosen. You could have recalled the beach that was nice, or the beach you randomly selected. So, after a polluted beach was recalled, the probability of having been right on the first guess changes from 1/3 to 50/50.
 
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In the French version of "Deal or No Deal" (called A Prendre ou a Lesse) the banker knows the contents of the boxes.

This makes the show more diabolical because the player know has to guess if the banker is bluffing, double bluffing or not.

The French version predates UK version by many years. In fact the reason there are 22 boxes is to represent the 22 regions of France.
The UK format differs quite a lot then.

There is only one person who knows the content of the boxes - an independant adjudicator. The host, the banker and the contestants all have no idea.

Over here we have 26 (or maybe 25) boxes as well.
Our version is based on the Australian format.

I'd be interested to see the French format in action actually as it would be likely to remove all the mumbo-jumbo that people try to invoke in our version if it's a straight battle of wits with the banker, rather than one person trying to beat the odds and (almost invariably) losing.
 
If, instead of opening a goat door, the host opens one of the two remaining doors at random, how does this affect the benefit of switching?

If, instead of opening a goat door, the host opens one of the two remaining doors at random and we only consider the cases where a goat is revealed, how does this affect the benefit of switching?

If we're playing 'Deal or No Deal' what's the benefit of switching the last two boxes?
 

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