[Guest]

This is a two part question:

1. You toss a fair coin four times in a row and it lands on tails all four times. What is the probability that your next coin toss will be tails?

2. A carnie at the fair has hidden a gold ring under one of three cups, marked A, B, and C. He lets you name one cup as your choice. You pick cup A. He then lifts cup C, showing you that it's empty, and allows you to choose again. What are the odds that the ring is under cup B?

[Guest]

Yes definitely, but it is accurate more times than people give it credit for. It comes under a lot of stick because it can (and quite often is) wrong.

It's not that it's often wrong. That's why I used to tell my students not to use it.

I tell them not to use it now because I have discovered that the people who run it have a proven history of distorting the information presented to suit their own agendas, and at times falsifying their own credentials in order to cow people into accepting their interpretation of 'the facts'. (Famously, a senior editor claimed to have a PhD in Theology, which he used to brow-beat people into accepting his views for the presented information. Turns out, he was a 24 year old community college dropout. Another editor uses her position to control the information surrounding the Lockerbie plane crash, in order to make sure the accused man stays in jail. Pretty nasty stuff.)

[Guest]

This is what I was getting at in my previous post. The question talks about an isolated incident and doesn't give any rule about what the carnie will or won't show, therefore we can't actually deduce anything from what we see. If we observed his trick enough times to be able to identify what rule he was using, or if we otherwise knew (e.g. he told us) then we would be able to deduce something.

The reason why it's worth stressing this is that people often just take the 2/3 answer as the truth no matter what the circumstances and this is simply wrong. I'll give the following answers depending on different reasons why we were shown the empty cup:

1. The carnie will always show us an empty cup - answer is 2/3 (several people have already explained why).

2. The carnie will pick one of the remaining cups at random. Sometimes this will result in us being shown the ring (and knowing we've lost) and sometimes this will result in us being shown an empty cup - the answer is 1/2. (I can explain maths behind this if necessary)

3. The carnie will only show us an empty cup if we have already picked the cup with the ring, otherwise he shows us nothing - answer is 0.

4. The carnie will only show us an empty cup if we have picked an empty cup, otherwise he shows us nothing - answer is 1.

5. The carnie will secretly roll a dice and if it comes up 6 he will show us an empty cup, otherwise he shows us nothing - answer gets more complicated (can't be bothered to work it out)

6. The carnie will show us an empty cup if he likes your hairstyle - answer gets more complicated

7. The carnie will try to guess whether you are good at probabilities and, if he thinks you are, will only show you an empty cup if you have already picked the correct cup in the hope you will fall into the common trap of always thinking you are better off switching - answer may be close to 0, but depends on how good the carnie is at guessing your skills.

and so on...

I hope this all makes sense.

LOL! Okay, neida, I see where you're coming from. My intent was not to illustrate the difficulties of second-guessing the illusive motives of an experienced trickster, but to show that knowing more information sometimes plays into the probability of a choice, and sometimes it doesn't. Clearly, you're showing me the same thing, only with more detail and thought to the matter.

[Guest]

That may be, but in order to know when to trust Wikipedia, you either have to be looking at a priori info (math, for example), or you have to already know what it is that you're looking at, right? To know that their article on probability is accurate, you have to know something about probability already. So the only way you can trust the info there is if you already know the info.

An interesting notion. Couldn't you say that about nearly anything regarding trust? It doesn't have to be right all the time in order to be deemed trustworthy. I have high confidence in Wikipedia to be right more often than not because topics of interest to a wide audience have a high probability of being corrected by smart people in the community who know better and have an interest in its upkeep. The more relevant the topic is to a wide audience, the more opportunity there is for misinformation to be policed and corrected. I would even assert that information in Wikipedia is more trustworthy than information published in my local newspaper.

[Guest]

So you all don't think that, given the statement of the original puzzle, this was intended to be the MH Paradox, but was rather indended to be a question in which we are required to know something about the carnie's method of choosing which cup to hide the prize under, and choosing which cup to reveal?

In that case, I think the odds are 1/14, because the carnie is receiving his instructions through a hidden earpiece, manned on alternating days by two brothers, one of whom always tells the carnie to hide the ring in his palm and not under the cup, and the other of whom always tells the carnie to hide the ring under cup A. One day out of every two weeks, both brothers get drunk and the carnie is left to his own devices. Because he's rather unimaginative, on those days he always hides the ring under cup B.

Thus, the chances that the ring is under cup B are 1/14.

Wow, that was easier than trying to do a logic puzzle.

Brotherbock, you are a card. And someday somebody's gonna have to deal with you.

[Guest]

Brotherbock, you are a card. And someday somebody's gonna have to deal with you.

Thank you, thank you <bow>

Regarding Wiki, though, the difference between Wiki and my local paper is that far fewer people take my local paper to be 'an authority' on what they print. All too many people view Wiki as an authority, which makes it far more dangerous. I have seen any number of people, in academic papers (undergrads, at least, not grad students) cite Wiki 'stub' articles as evidence for their claims, even though that stub might have been written five minutes ago, with no editing.

In fact, I'd claim that the subjects that need to be researched are more likely to be the ones that get less editing, not more. A thousand people will correct you if you post the wrong birthdate for Brittany Spears, but how many people will catch a mistake you make in an article about probability? There are far fewer people reading it, to begin with, and far fewer people who will know that the mistake has been made. Those most likely to know aren't likely to be reading it...because A) they already know, and B) Wiki has a very bad reputation among academics around the world, making them likely to not even be reading it in the first place.

And even if you do catch a problem and post a correction...if one of the senior Wiki staff has a claim that they really want to push, they can just delete your edits, ban your account, and continue to print their version of the truth. They have done exactly this in the past.

That, to me, makes Wiki far less reliable than my local paper.

[Guest]

Thank you, thank you <bow>

Regarding Wiki, though, the difference between Wiki and my local paper is that far fewer people take my local paper to be 'an authority' on what they print. All too many people view Wiki as an authority, which makes it far more dangerous. I have seen any number of people, in academic papers (undergrads, at least, not grad students) cite Wiki 'stub' articles as evidence for their claims, even though that stub might have been written five minutes ago, with no editing.

In fact, I'd claim that the subjects that need to be researched are more likely to be the ones that get less editing, not more. A thousand people will correct you if you post the wrong birthdate for Brittany Spears, but how many people will catch a mistake you make in an article about probability? There are far fewer people reading it, to begin with, and far fewer people who will know that the mistake has been made. Those most likely to know aren't likely to be reading it...because A) they already know, and B) Wiki has a very bad reputation among academics around the world, making them likely to not even be reading it in the first place.

And even if you do catch a problem and post a correction...if one of the senior Wiki staff has a claim that they really want to push, they can just delete your edits, ban your account, and continue to print their version of the truth. They have done exactly this in the past.

That, to me, makes Wiki far less reliable than my local paper.

Perhaps we could move the discussion on the validity of Wikipedia to the "Others" forum? I have really enjoyed this topic and would hate to see it locked for off-topic content.

[Guest]

1/1 because there only 1 coin needed to toss but on rare cases 1/2 if the coin can land on its side. as for odds under the cup, there is a 1/2 chance it is under cup B as theres 2 cups left to choose from.

[Guest]

Sorry, this bit isn't correct. Being able to choose but knowing you won't doesn't make any difference to the probabilities than not being able to choose. Think about it like this:

Label the 3 cups A, B and C and say you pick cup A (if you pick another cup it's just a rotation of this). There are 6 possibilities, each with equal probability:

1: Ring is under cup A, carnie shows empty cup B - don't switch to win

2: Ring is under cup A, carnie shows empty cup C - don't switch to win

3: Ring is under cup B, carnie shows cup B - game over

4: Ring is under cup B, carnie shows empty cup C - switch to win

5: Ring is under cup C, carnie shows empty cup B - switch to win

6: Ring is under cup C, carnie shows sup C - game over

You see here that, as you put it, in cases 3 and 6 you are shown the ring and you "merely lose" or, as I have put it "game over". So, on the occasion that he shows you an empty cup there are only 4 possible ways this could have happened (cases 1,2,4 and 5 above, each with equal probability) and 2 cases where you should switch and 2 where you shouldn't (again each with equal probability). So the probability of cup B being the winning cup is 1/2.

I agree, but on this occasion the Wikipedia page is actually fairly good. It even has a diagram showing what I've just explained above for anyone needing to think it through some more.

Happy to oblige!

The question was :

"2. A carnie at the fair has hidden a gold ring under one of three cups, marked A, B, and C. He lets you name one cup as your choice. You pick cup A. He then lifts cup C, showing you that it's empty, and allows you to choose again. What are the odds that the ring is under cup B?"

Here is what i beleive is the correct analysis for this question. Just use common sense people! comon!

-Common sense states that the carnie does not want to win it for you. so naturally he will never open a winning cup.

-Also, the carnie knows which cup has the ring, he "has hidden" it.

-If like under options 3 and 6 the silly carnie wanted you to win (u bribed her?), then you dont need to switch do you? and this question will be not necessary

-The goals of the player is to win the ring

Most importantly neida, ur analysis is flawed. here's why:

1: Ring is under cup A, carnie shows empty cup B - don't switch to win

2: Ring is under cup A, carnie shows empty cup C - don't switch to win

These 2 are BOTH the same possibility in the most "common sense" not 2. you chose cup A and won regardless of switching option or carnie opening anything. its 1 out of 3 chances if you dont switch.

3: Ring is under cup B, carnie shows cup B - game over (You dont even need to switch to win do you? please remember that winning is the goal and its not possible the carnie will win it for you) - (So you need to Eliminate this possibility)

4: Ring is under cup B, carnie shows empty cup C - switch to win

5: Ring is under cup C, carnie shows empty cup B - switch to win

6: Ring is under cup C, carnie shows sup C - game over (Eliminate whis one too...)

So only option 1&2 (both 1 option really) and options 4 and 5 are relevant in this example.

So again, you win 2/3 if u switch.

I do however understand where you come from, but on this particular example, you are acting too much like lawyers, creating and looking for loop-holes that really dont need to exist unless u really want them too. Good judges always solve this problem with common sense and juris prudence

Also for other replies, adding too many ifs, such as maybe the ring didnt exist, or maybe the carnie always cheats, or maybe the carnie doesnt know etc. etc. all dont make any sense. This does not look like a trick question, its a staight foreward problem. Why waste any more time with pointless analysis

[Guest]

Why waste any more time with pointless analysis

Heh... Welcome to the Den.

[Guest]

Here is what i beleive is the correct analysis for this question. Just use common sense people! comon!

-Common sense states that the carnie does not want to win it for you. so naturally he will never open a winning cup.

-Also, the carnie knows which cup has the ring, he "has hidden" it.

If you were to read my response properly you would see that I was responding to the case raised by Grayven where the carnie did not know where the ring was hidden, not the original question. To me common sense says that if I am told the carnie doesn't know where it is, then it is safe to assume he doesn't know where it is.

1: Ring is under cup A, carnie shows empty cup B - don't switch to win

2: Ring is under cup A, carnie shows empty cup C - don't switch to win

These 2 are BOTH the same possibility in the most "common sense" not 2. you chose cup A and won regardless of switching option or carnie opening anything.

Interesting that you think that because they have the same outcome they are the same thing. If the cups really were labelled A, B and C do you think that lifting up B is the same as lifting up C (you don't further down your analysis). Just because two different events result in the same outcome, this does not make them the same and automatically give them the same probability as everything else.

If you really do think that, then let me ask you this. Roll a dice and tell me if the outcome is a 6 or not a 6. I don't care what number it is, I just want to know if it is a 6 or not. According to your reasoning, rolling a 1, 2, 3, 4 or 5 are all the same thing, as they all result in not a 6. Therefore, the chances of rolling a 6 are 1/2. What does common sense tell you here?

3: Ring is under cup B, carnie shows cup B - game over (You dont even need to switch to win do you? please remember that winning is the goal and its not possible the carnie will win it for you) - (So you need to Eliminate this possibility)

Mmmm yes, that is what I was implying by saying game over and referring to when I later worked out the probabilities. It is precisely because you have to eliminate these possibilities that you get to the correct answer of 1/2. Maybe I should have been more explicit on this point.

Also for other replies, adding too many ifs, such as maybe the ring didnt exist, or maybe the carnie always cheats, or maybe the carnie doesnt know etc. etc. all dont make any sense. This does not look like a trick question, its a staight foreward problem.

You made an assumption yourself at the start of the post that the carnie does not want you to win. If that is the case, then if you have picked an empty cup, why is he bothering to show you a different empty cup? Surely he just takes your money and says "bad luck".

I don't think anybody is contesting that, properly worded, the intended question should have an answer of 2/3. The reason for the ongoing debate is that a lot of people think that would always be the case, no matter whether the carnie knew or not or any other rule that was to be applied and I have simply tried to point out the error in that way of thinking.

Why waste any more time with pointless analysis

Good point well made.

[Guest]

If you really do think that, then let me ask you this. Roll a dice and tell me if the outcome is a 6 or not a 6. I don't care what number it is, I just want to know if it is a 6 or not. According to your reasoning, rolling a 1, 2, 3, 4 or 5 are all the same thing, as they all result in not a 6. Therefore, the chances of rolling a 6 are 1/2. What does common sense tell you here?

+

You made an assumption yourself at the start of the post that the carnie does not want you to win. If that is the case, then if you have picked an empty cup, why is he bothering to show you a different empty cup? Surely he just takes your money and says "bad luck".

Too many things to quote but i'll stop on 2 things:

1. Your analysis is faulted. The "outcome" is the number on the dice. While in our problem the "outcome" is winning or losing and not which cup is opened in the process. So there is huge difference. I was approaching the answer form the hypothesis i has assumed based on what i called "common sense".

This is clearly explained by the fact that you win 1/3 of the times if you dont switch in the classic monty hall problem, regardless of which door is opened, and not 1/4 (adding the 2 if you chose the right one and the chance that 1 of the other 2 doors is opened, with the other 2 known possibilities)

(edited: added "and the chance...opened)

Other than that, if you change the hypothesis as you assumed, the rest of the problem is in my opinion correct. (e.g. you dont eliminate the "game over" options, since the "game over" is an outcome added to the win or lose)

2. I did not make an asumption that the carnie does not want you to win, i assumed does not want to "win it for you". Also makes for a huge difference.

I didnt say that you didnt understand the problem. I was just pointing out using your dense explanation my point, plus the 1/4 error.

Edited January 19, 2009 by Ronin

[Guest]

as far as the probability of the classic monty hall paradox, the answer on the surface certainly is that cup b has a 2/3 chance of having the ring. HOWEVER: if the carnie knows where the ring is, then the only reason he would give you the option of switching choices is to give you a chance to move your hand to where the ring is NOT, he would never give you the option of choosing the right cup if he knew you were wrong to start. In this case you should not switch, (you have to remember this guy is a carnie). If he does not know where the ring is, then he revealed cup C knowing that there was a 1/3 chance that cup C had the ring, that leaves you with a 1/3 chance with cup A and the rest of the probabilities with cup B, so if you have the option to switch, by all means you should.

the monty hall paradox only applies when the host, or "carnie" in this case, has a purely unbiased stance, i.e. he does not know where the prize is. that is why you should never play 3 card monte in the subway. you will lose nearly every time.

Furthermore, since this guy is a carnie, and we know he is a carnie, the odds that the ring is in cup B is 0, because the carnie is trying to rip you off.

[Guest]

the monty hall paradox only applies when the host, or "carnie" in this case, has a purely unbiased stance, i.e. he does not know where the prize is.

Whilst I agree that the carnie could well be fiddling the odds (in which case its probably best to assume that he'll only offer you the chance to switch if you've picked the winning cup), the above statement isn't quite right...

It monty didn't know where the prize was, then some (1/3) of the time he would pick the winning door, and your odds would return to 1/3 chance of winning if you switched.

The important thing is that he always gives you the opportunity to swap and always picks a goat (losing door).

Edited January 19, 2009 by armcie

[Guest]

as far as the probability of the classic monty hall paradox, the answer on the surface certainly is that cup b has a 2/3 chance of having the ring. HOWEVER: if the carnie knows where the ring is, then the only reason he would give you the option of switching choices is to give you a chance to move your hand to where the ring is NOT, he would never give you the option of choosing the right cup if he knew you were wrong to start. In this case you should not switch, (you have to remember this guy is a carnie). If he does not know where the ring is, then he revealed cup C knowing that there was a 1/3 chance that cup C had the ring, that leaves you with a 1/3 chance with cup A and the rest of the probabilities with cup B, so if you have the option to switch, by all means you should.

the monty hall paradox only applies when the host, or "carnie" in this case, has a purely unbiased stance, i.e. he does not know where the prize is. that is why you should never play 3 card monte in the subway. you will lose nearly every time.

Furthermore, since this guy is a carnie, and we know he is a carnie, the odds that the ring is in cup B is 0, because the carnie is trying to rip you off.

I disagree. If other people, potential marks, are watching, it's a dead giveaway to only offer the chance to switch to some of the people.

In the end, this is just a bad game for the carnie to offer, as he won't make a lot without cheating. BUT, given that he is offering the game, and if we say that others will be watching (it's a circus/carnival, after all), it only makes sense for him to offer the switch, if not to everyone, then at least in some of the cases where the switch would be bad for him. And now if he's offering the switch, but you know that the switch can either be bad or good for you (e.g. you've seen people both win and lose from switching), then you need to switch.

Edited January 19, 2009 by brotherbock

[Guest]

1. Your analysis is faulted. The "outcome" is the number on the dice. While in our problem the "outcome" is winning or losing and not which cup is opened in the process. So there is huge difference.

No, I stated that the outcome of my question was it is either a six or not. Let me rephrase that as "If it is a 6 you win, otherwise you lose". Interesting that you determine the outcome in one case as an intermediate step (the number on the dice) but not in the other (which cup is shown and what is observed). Also in one you say the outcome is "winning or losing" (with the cups) but not in the other (the rephrased version where now the outcome is also "winning or losing"). Anyway...

I was approaching the answer form the hypothesis i has assumed based on what i called "common sense".

This is clearly explained by the fact that you win 1/3 of the times if you dont switch in the classic monty hall problem, regardless of which door is opened, and not 1/4 (adding the 2 if you chose the right one and the chance that 1 of the other 2 doors is opened, with the other 2 known possibilities)

Where do you get 1/4 from? I never mentioned 1/4? If you are trying to apply my reasoning to the original Monty problem then you can't, as the two scenarios where there is "game over" are not possible outcomes and the entire table of possible outcomes changes. You can't do a like for like comparison.

2. I did not make an asumption that the carnie does not want you to win, i assumed does not want to "win it for you". Also makes for a huge difference.

Whoever said that the carnie lifting the cup with the ring would "win it for you"? In my opinion if he did that you would automatically lose, as you didn't select the winning cup.

[Guest]

If he does not know where the ring is, then he revealed cup C knowing that there was a 1/3 chance that cup C had the ring, that leaves you with a 1/3 chance with cup A and the rest of the probabilities with cup B, so if you have the option to switch, by all means you should.

the monty hall paradox only applies when the host, or "carnie" in this case, has a purely unbiased stance, i.e. he does not know where the prize is. that is why you should never play 3 card monte in the subway. you will lose nearly every time.

No, both of these statements are incorrect. This is the point I have been trying to make that a lot of people think that this is true when in fact it isn't.

I will try to explain in a different way, hopefully appealing to the "common sense" approach favoured by Ronin...

Let's consider the two cases separately. Case 1 where Monty/carnie knows where prize is and will always show you an empty door/cup. Case 2 where Monty/carnie does not know where prize and randomly chooses from two remaining doors/cups. (I will just use carnie and cup for the rest of this explanation)

Now in both cases I don't think there is any argument that the probability of cup A having the prize is 1/3 and the probability of cup A not having the prize is 2/3. To work out the ultimate probability of cup B having the prize we therefore need to look at the following:

Case 1:

For cup B to have the prize, cup A must not. Probability of this is 2/3. We know in this case that, if cup A does not have the prize then carnie MUST show the other empty cup. Therefore, in the case that cup A does not have the prize, the probability that the one cup remaining (after you have picked one and carnie removes one) contains the prize is 1 (it must be in the remaining cup). Therefore the overall probability of the remaining cup containing the prize is 2/3 x 1 = 2/3.

Case 2:

For cup B to have the prize, cup A must not. Probability of this is 2/3. In this case, if cup A does not have the prize then carnie will show us one of the other cups at random. Therefore, in the case that cup A does not have the prize, 1/2 the time it will be game over (carnie uncovers prize) and 1/2 the time the prize will be under the remaining cup. Therefore, the overall probability of the remaining cup containing the prize in this case is 2/3 x 1/2 = 1/3. However, in this case, once we are shown the empty cup, we know we can rule out the possibilities that result in game over (which also = 1/3 through the same reasoning), so the only alternatives now are prize is under cup A (with overall probability 1/3) and prize is under the remaining cup (with overall probability 1/3), so as we are left with two equally likely events, the amended probability of the prize being under the remaining cup is 1/2.

Note that the key difference comes in being able to distinguish between two separate events in Case 2 and then eliminate some possibilities based on what we see. In Case 1 we can't distinguish between events, but we know that the reason why is that the carnie is having his options limited by the rules.

I hope this helps, as I don't know how many different ways I can try to explain this before I start repeating myself.

Edited January 19, 2009 by neida

[Guest]

neida, i understand why the paradox works the way it does. it is pure mathematics. ideally, and mathematically, you should always switch your choice if given the option.

In a real-world situation, there is more going on than what meets the eye. the host may or may not know where the prize is. logically, he will not offer you the prize if he can at all avoid it. again, he is a carnie, carnies are notorious for being swindlers and cheats, if he lets you win, he is hussling you.

the most correct answer to the question "should you switch your choice" should not be based off of pure mathematics, but off the style of prior games (does tha carnie always give the player the opportunity to switch?) and off of the knowledge of the carnie (does he know where the prize is/is not?)

Many other factors affect whether you should switch your choice in a real life situation. theoretically, you should always switch based on the pure mathematic workings of the problem.

to summarize: if this were a math problem: you should switch choices, if it is a logic problem, you should not trust the carnie.

great article on the monty hall paradox is on wikipedia. it gives all the in and outs of both why it works mathematically, but why it rarely works practically, and also the psycology behind it.

[Guest]

No, both of these statements are incorrect. This is the point I have been trying to make that a lot of people think that this is true when in fact it isn't.

I will try to explain in a different way, hopefully appealing to the "common sense" approach favoured by Ronin...

Let's consider the two cases separately. Case 1 where Monty/carnie knows where prize is and will always show you an empty door/cup. Case 2 where Monty/carnie does not know where prize and randomly chooses from two remaining doors/cups. (I will just use carnie and cup for the rest of this explanation)

Now in both cases I don't think there is any argument that the probability of cup A having the prize is 1/3 and the probability of cup A not having the prize is 2/3. To work out the ultimate probability of cup B having the prize we therefore need to look at the following:

Case 1:

For cup B to have the prize, cup A must not. Probability of this is 2/3. We know in this case that, if cup A does not have the prize then carnie MUST show the other empty cup. Therefore, in the case that cup A does not have the prize, the probability that the one cup remaining (after you have picked one and carnie removes one) contains the prize is 1 (it must be in the remaining cup). Therefore the overall probability of the remaining cup containing the prize is 2/3 x 1 = 2/3.

Case 2:

For cup B to have the prize, cup A must not. Probability of this is 2/3. In this case, if cup A does not have the prize then carnie will show us one of the other cups at random. Therefore, in the case that cup A does not have the prize, 1/2 the time it will be game over (carnie uncovers prize) and 1/2 the time the prize will be under the remaining cup. Therefore, the overall probability of the remaining cup containing the prize in this case is 2/3 x 1/2 = 1/3. However, in this case, once we are shown the empty cup, we know we can rule out the possibilities that result in game over (which also = 1/3 through the same reasoning), so the only alternatives now are prize is under cup A (with overall probability 1/3) and prize is under the remaining cup (with overall probability 1/3), so as we are left with two equally likely events, the amended probability of the prize being under the remaining cup is 1/2.

Note that the key difference comes in being able to distinguish between two separate events in Case 2 and then eliminate some possibilities based on what we see. In Case 1 we can't distinguish between events, but we know that the reason why is that the carnie is having his options limited by the rules.

I hope this helps, as I don't know how many different ways I can try to explain this before I start repeating myself.

Neida,

I just wanted to say that I do appreciate your explanations and have learned to think differently about problems as a result. While I can also appreciate anyone's revulsion against legalistic requirements to meticulously spell out whether cheating can occur, I don't at all believe that to be your point. What you are teaching is that the rules by which changes are made to the landscape of a problem can affect probabilities in profound ways. If those rules are stated, then they are useful, if they are not stated, then assuming them to be known leads to error (or surprise answers, when they are left out deliberately).

One of the most obvious illustrations I've seen in support of your same point was in bonanova's probability problem where a magician rolls two dice for a blindfolded subject, and tells the subject that least one of the dice shows a 6 and asks what is the probability that both dice add up to exactly 7. The assertion was that the chances were 2/11 (which led to lengthy debate). Most simply, as bonanova explained later, if the magician always chooses the lowest-numbered die, then the probability of both dice adding to 7 would be zero in that case.

Phatfingers

[Guest]

This is a two part question:

1. You toss a fair coin four times in a row and it lands on tails all four times. What is the probability that your next coin toss will be tails?

2. A carnie at the fair has hidden a gold ring under one of three cups, marked A, B, and C. He lets you name one cup as your choice. You pick cup A. He then lifts cup C, showing you that it's empty, and allows you to choose again. What are the odds that the ring is under cup B?

1. 50%. The probability is 50/50 for each flip. The gambler's falacy leads one to believe that they are "due" for a heads, but that is not the case. If you asked the probability that it will land on heads on five straight flips the answer would be different.

2. 33.33%. Just because one of the choice was eliminated does not discount that it was an original option. There is still just a 1/3 chance

[Guest]

neida, i understand why the paradox works the way it does. it is pure mathematics. ideally, and mathematically, you should always switch your choice if given the option.

to summarize: if this were a math problem: you should switch choices, if it is a logic problem, you should not trust the carnie.

No, this is not the point at all - it is not a paradox. Both the cases I presented above are pure math problems. In one you should switch, in the other it makes no difference. I could easily present you with slightly different variations (as I did earlier in the thread) where you definitely shouldn't switch. So saying that, whenever presented with a mathematical version of this question you should always switch is incorrect and that is the point I have been trying to make.

In the real world, the carnie would no doubt be playing all sorts of mind games too, so there's no way of knowing what's going on (unless you start getting into game theory) and you simply can't treat it as a math problem at all.

great article on the monty hall paradox is on wikipedia. it gives all the in and outs of both why it works mathematically, but why it rarely works practically, and also the psycology behind it.

I agree - I pointed it out a few pages ago in this thread, but I suggest you read it again. It doesn't point out why it works mathematically but rarely practically, it points out why subtle changes to the wording (or the situation) actually make the answer to the mathematical problem very different.

[Guest]

Neida,

I just wanted to say that I do appreciate your explanations and have learned to think differently about problems as a result. While I can also appreciate anyone's revulsion against legalistic requirements to meticulously spell out whether cheating can occur, I don't at all believe that to be your point.

Thanks Phatfingers! I must admit that I was starting to feel like I was just being awkward, but you are right. I don't have a problem with the answer to the properly worded question being 2/3, or even us making assumptions as to how the question was intended to be worded. But I do have a problem with the fact that many people think the wording doesn't matter and that, even if the carnie chose completely randomly the answer would still be the same. That is simply incorrect and that's what I've been trying to counter!

I agree with your comment on the 2 dice question posed by Bonanova. It was actually that question that opened my eyes on this (and the two children) question, so I would recommend others to read it.

[Guest]

Neida, To see what im saying, please refer to my original analysis of neida' your post, where i showed the reasoning of going from the 6 posibilities you presented to combining the 2 into 1, then to eliminating the 2 game overs, and that reasoning correctly lead to the classing monty hall chances. It was not a coincidence. I think we're just disagreeing on the way to look at it. the results are the same.

Back to the subject, and adding to what u're saying, also creating different new hypothesis, im gona explain 3 different cases:

Hypothesis: Carnie will always open any cup after you choose, and he doesnt know which is the winning cup. Then:

-you have 1/3 chance to win if you switch

-you have 1/3 chance to win if you dont switch

-plus a 1/3 chance for a "game over" (if game over is a win, then you win 2/3 of the time)

>>>therefore, if you chose a cup, and the canie opened any cup that was by chance empty, and you were offered to switch, there is a 50/50 chance you win by switching, since now all probablilities are equally distributed.

Most importantly, the incorrect analysis of most people under the classic monty hall problem corresponds to the hypothesis above! which is notibly 50/50 if you switch (you can see that on some posts in this thread).

Another hypothesis is, the carnie knows which is the winning cup, and will only offer you to switch if she could fool you, meaining offer you to switch to an empty cup. Then:

-Simply put, never switch if you were offered to, since you won already.

-therefore you have 100% chance to win if u dont switch

-you therefore have a 1/3 chance to win...

A third hypothesis is, carnie knows the winning cup (similar to classing monty hall), but could or could not offer you to switch (say 50% chance he offers u to switch, by opening another cup). then:

IF the wining cup is B, the options are as follows:

1stChoice [] Offer switch? [] Outcome [] Win% [] % to Switch [] Final Probability

A[] Yes[] Won[] 1/3[] 1/2[] 1/6

B[] Yes[] Lost[] 1/3[] 1/2[] 1/6

C[] Yes[] Won[] 1/3[] 1/2[] 1/6

A[] No[] Lost[] 1/3[] 1-(%sw)[] 1/6

B[] No[] Won[] 1/3[] 1-(%sw)[] 1/6

C[] No[] Lost[] 1/3[] 1-(%sw)[] 1/6

Total chance to win in this game = 1/6 x 3 = 1/2

(edit: fixed the shape of the table)

Logically, the more he/she offers u to switch, the closer you get to 2/3 chance to win, since the bigger the 1/2 and more final probability for yesses between the 3 first rows in the table.

Edited January 20, 2009 by Ronin

[Guest]

neida,

Thanks. I think we are both getting at the same idea. You are better at articulating it. Thanks for clearing that up. Have a good one.

oh and also: the only reason I called it a paradox (which it technically isnt) is because of the term "monty hall paradox." I should have been more clear.

[Guest]

Great - I think we're all finally agreed!

What I think I find most interesting about this puzzle isn't the answer, but actually how people interpret the question - and not just the different ways it can be interpreted, but how odd the "generally accepted" interpretation is. Let me explain...

The way I like to think of this problem is that it is actually about two probability events, not just one. The first is which cup the ring is under. The second is which cup we are shown (and whether it is empty or not). What I think is really interesting is that the "generally accepted" interpretation of the puzzle is that the ring is placed under a cup at random, but the cup we are shown is predetermined to be an empty one. The reason why this is accepted is simply because we are shown an empty cup, and that leads to the assumption that act was predetermined. But that would be like saying "I just rolled a dice and got a 6, so it was definite that I was going to roll a 6."

Basically, what I think is interesting is not the fact that an assumption is made, but the fact that two assumptions are made and they are both different - we just don't realise we are doing it. If we assumed everything was random or everything was predetermined then that would ordinarily make more sense. But in this case it is more commonly accepted that we should make two different assumptions and it is rarely even questioned!

Neida, To see what im saying, please refer to my original analysis of neida' your post, where i showed the reasoning of going from the 6 posibilities you presented to combining the 2 into 1, then to eliminating the 2 game overs, and that reasoning correctly lead to the classing monty hall chances.

Just to clarify, in the classic Monty scenario you can't combine those two events into one. It does lead to the correct answer, but not through correct logic. If you take the approach I've just mentioned above then the probability of each of the outcomes 1-6 I previously listed are P(ring under specified cup) x P(cup shown being shown), which in each case in the example where the cup is shown at random is 1/3 x 1/2 = 1/6. In the case where we are always shown an empty cup (or the generally accepted traditional Monty Hall problem), the probabilities of the different outcomes change to the following (using same numbering to avoid typing it all out again):

1: 1/3 x 1/2 = 1/6

2: 1/3 x 1/2 = 1/6

3: 1/3 x 0 = 0

4: 1/3 x 1 = 1/3

5: 1/3 x 1 = 1/3

6: 1/3 x 0 = 0

Cases 1 and 2 you win if you don't switch. Cases 3-6 you win if you switch. So totalling cases 3-6 your chances of winning if you switch are 2/3.