Will These Brain Teasers Stump You?

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This just reveals why probability questions are always stupid. Mathematicians may well understand math, but their ability to frame questions in the English language is always embarrassingly lacking.

Also if you want a pure math problem, stop using real world scenarios in which any non brain damaged person knows that there will be other factors (like human agency) affecting the outcome. Of course I am going to consider whether or not I can trust the warden or the game show host. I'd have to be emotionally stunted not think about that aspect of the scenario.

The prisoner problem is particularly poorly written: there is no opportunity to flip a coin in secret, so no flip can have occurred.

wulfy42:
Not true at all, in fact, it's actually the opposite in some ways.

Lets look at the 3 door problem. You have a 1/3 chance of getting the car to start. That is, in the end, your exact chance of getting the car no matter what else happens along the way. Why?

Well lets look at what is really happening here real quick.

You initially choose a door, there is a 33% chance you chose the care. The game announcer knows if you have the car or not, and knows which other door (or doors) do not have the car.

Either way, there is a door without a car for him to open.

Now, at that exact point in time, you have a 50% chance to have the right door. In no way does the actions of the annoucer opening a door make any difference in your probability of initially choosing the right door. If you do choose again, you now have a 50% chance of being right....but...you have that SAME 50% chance of being right if you don't choose again (it's still the exact same chance).

Choosing again, does not change your odds. Saying it does it just plain silly. The annoucer always is going to offer you another pick, and one of the other doors is always doing to have a goat. Your probability of choosing the correct door is not going to change, but....if you choose again at this point, your NEW probability of choosing the correct door will be 50% (instead of the original 33%). That is the same if you choose the door you currently have, or choose the other door.

In effect you are starting off with a 33% chance (your initial choice)...which for all intents and purposes is pointless (as long as the announcer always has to give you the option to choose again after getting rid of one door), and then ending up with a 50% chance to win. Your net over all chance to choose the car in this case is 50% (as one goat is always removed). Keep the same door, choose the other door, it doesn't matter, you have a 50% chance to win.

Consider a variant on the Monty Hall problem. Lets have 1001 doors. The car is behind a door, 1000 goats are behind the others. You pick a door at random - lets call this door 1. Most likely, door 1 is a goat.

The announcer, who knows where the car is, then proceeds to open 999 of the remaining doors, revealing goats behind all of them. Except he, rather suspiciously, leaves door 752 closed.

Would you switch from door 1 to door 752?
I would. Because either you got the car on your first guess - which is 1 in 1001 odds - or it's behind door 752.

(And, the more important question - who is going to look after our thousand goats?!)

Likewise for the standard Monty Hall problem. Either you got the car on your first guess - which is 1 in 3 odds - or it's behind the other door.

The "solution" to the "Three Prisoners Problem" is wrong, and why it's wrong creates its own interesting logical/literary paradox. Don't misunderstand, the logic is sound and the math is correct, but there are key details in the text and the nature of the exercise that must be considered.

Because this is a "thought exercise" and math problem, we have to concede that anything which didn't happen in the story didn't happen (damn creativity-stifling thought exercises...). In the story, the warden didn't flip a coin, nor did he walk away from A before revealing the condemned. We cannot make the assumption that the warden flipped an imaginary coin (didn't happen in the text = didn't happen), therefore the coin flip never took place, eliminating the possibility of A's pardon. Similarly, if the story did contain a coin flip, we cannot assume the flip was to obfuscate the "real" answer (because that would be an assumption) and therefore the existence of a coin flip in the story states outwardly that A is being pardoned, else it would not have occurred.

2xDouble:
The "solution" to the "Three Prisoners Problem" is wrong, and why it's wrong creates its own interesting logical/literary paradox. Don't misunderstand, the logic is sound and the math is correct, but there are key details in the text and the nature of the exercise that must be considered.

Because this is a "thought exercise" and math problem, we have to concede that anything which didn't happen in the story didn't happen (damn creativity-stifling thought exercises...). In the story, the warden didn't flip a coin, nor did he walk away from A before revealing the condemned. We cannot make the assumption that the warden flipped an imaginary coin (didn't happen in the text = didn't happen), therefore the coin flip never took place, eliminating the possibility of A's pardon. Similarly, if the story did contain a coin flip, we cannot assume the flip was to obfuscate the "real" answer (because that would be an assumption) and therefore the existence of a coin flip in the story states outwardly that A is being pardoned, else it would not have occurred.

True. The three prisoners problem would work better if the Warden stepped into a side-room for a moment before giving an answer. Then we'd have no knowledge over whether or not a coin was flipped.

2xDouble:
Because this is a "thought exercise" and math problem, we have to concede that anything which didn't happen in the story didn't happen.

No we don't.

The problem says nothing about whether the warden flipped a coin or not. It never says he didn't flip a coin; assuming he didn't is as much of an assumption as assuming he did.

It's possible to assume that anything not expressly stated in the story doesn't happen, but we don't "have" to, and in fact doing so creates a needless issue in the problem.

The problem lays out a process through which the warden tells A who is going to be executed. The process may or may not involve flipping a coin. The problem is silent on whether the warden actually flips a coin or not. I think the most reasonable explanation is that the warden flipped a coin if necessary. This is explicitly a probability problem, not a logic riddle; if the person who will be executed could be figured out exclusively from whether the warden flipped a coin or not, it defeats the entire problem. So while you "can" assume that he doesn't flip a coin because it doesn't say he did (or you can assume he did flip a coin because it doesn't say he didn't), doing so immediately destroys the issue at question, and so it seems a lot more reasonable to simply read the problem in a way that corresponds to the information provided (that is, the warden follows the agreed-upon procedure) and that properly retains the issue.

tl;dr: assuming he didn't flip the coin because it didn't explicitly say so is just as much of an assumption as anything else, and it's needlessly creating a problem where one doesn't need to exist.
I think people are trying to hard to outsmart the question rather than just solve the probability problem.

wulfy42:
As far as the prisoners one, actually I don't agree with it, as the initial random chance to determine who was pardoned was 33% for each of them. That means that each prisoner had a 1/3 chance to be pardoned. When the warden says that bart is being executed (so is not the one pardoned), that means the initial chance has changed from 1/3 to 1/2.

At that point in time, when the warden said bart was going to be executed, if the random selection was done then, it would be a 66% chance for charlie to be pardoned. The truth is though, that ship had already sailed and the initial 33% chance for each still meant that both charlie and alried had the same chance of being pardoned (but now there is only 2 of them, so it's a 50% chance).

Just because there are more situations where the warden would have said what he did, does not mean it actually changes the initial probability of each of the prisoners being pardoned. What does change the probability is removing bart from the chances to be pardoned, which changes the chance from 33% each, to 50% each for charlie and Alfried.......so I do not agree with that puzzle.

No, his initial chance hasn't changed at all because:

1. He never gave to the warden the choice of saying directly: "Alfred, you're going to be executed".
2. When who was going to be pardoned was decided, Alfred's chances were 1/3. Bart's going to be executed? Too bad for Bart. But that didn't change the initial decision of who would be pardoned. They didn't reset the odds by randomly selecting between Charlie and Alfred again after it was revealed that Bart was going to be executed.

wulfy42:
Not true at all, in fact, it's actually the opposite in some ways.

Lets look at the 3 door problem. You have a 1/3 chance of getting the car to start. That is, in the end, your exact chance of getting the car no matter what else happens along the way. Why?

Well lets look at what is really happening here real quick.

You initially choose a door, there is a 33% chance you chose the care. The game announcer knows if you have the car or not, and knows which other door (or doors) do not have the car.

Either way, there is a door without a car for him to open.

Now, at that exact point in time, you have a 50% chance to have the right door. In no way does the actions of the annoucer opening a door make any difference in your probability of initially choosing the right door. If you do choose again, you now have a 50% chance of being right....but...you have that SAME 50% chance of being right if you don't choose again (it's still the exact same chance).

Choosing again, does not change your odds. Saying it does it just plain silly. The annoucer always is going to offer you another pick, and one of the other doors is always doing to have a goat. Your probability of choosing the correct door is not going to change, but....if you choose again at this point, your NEW probability of choosing the correct door will be 50% (instead of the original 33%). That is the same if you choose the door you currently have, or choose the other door.

In effect you are starting off with a 33% chance (your initial choice)...which for all intents and purposes is pointless (as long as the announcer always has to give you the option to choose again after getting rid of one door), and then ending up with a 50% chance to win. Your net over all chance to choose the car in this case is 50% (as one goat is always removed). Keep the same door, choose the other door, it doesn't matter, you have a 50% chance to win.

Think well what you just posted: "In no way does the actions of the annoucer opening a door make any difference in your probability of initially choosing the right door."

What are your initial odds? 33% if the host opens your door right after you pick the first time.

Hm, quite a bit of debate over the Monty Hall problem here. I'll take my crack at explaining it.

1.) you pick a door. This door has a 33% chance of car and a 67% chance of goat.

2.) The announcer opens an UNPICKED door, revealing a goat. (That's what he always has to do)

This action has no bearing on the odds of the door you picked. it is still 33% chance of car and 67% chance of goat.

The best and and most accurate prediction you can make is that your initial door was a goat door (67% probability).

Working with this prediction, the announcer has revealed the other of the 2 goat doors, and the third door has a 67% chance of being the car.

Therefore, switching improves your odds from 1/3 to 2/3.

Zakarath:
Hm, quite a bit of debate over the Monty Hall problem here. I'll take my crack at explaining it.

1.) you pick a door. This door has a 33% chance of car and a 67% chance of goat.

2.) The announcer opens an UNPICKED door, revealing a goat. (That's what he always has to do)

This action has no bearing on the odds of the door you picked. it is still 33% chance of car and 67% chance of goat.

The best and and most accurate prediction you can make is that your initial door was a goat door (67% probability).

Working with this prediction, the announcer has revealed the other of the 2 goat doors, and the third door has a 67% chance of being the car.

Therefore, switching improves your odds from 1/3 to 2/3.

Correct. What's interesting about the Monty Hall problem is how easy it is to understand if you do it yourself at home.

Take three cups and three objects (two the same, one different. Dice or coins work well enough). Now place the objects under the cup in a manner you remember and do the experiment.

The crux of the problem lies on the fact that you MUST reveal one of the two failure doors. Without revealing anything the odds don't change but because the host is forced to reveal a dud they do.

It is honestly so obvious if you play the scenario out. If you're struggling to get it (not you Zakarath, general 'you') just try physically replicating the game, trying it on a buddy is even better, play the part of the host.

Hugga_Bear:

FoolKiller:
And this is how wars are started. Because the problem with these is that a lot of mathematicians don't agree with the way the math is setup for it. But that's none of my business.

Could you elaborate? To my knowledge nothing here is a mathematical trick.

The monty hall problem is famous and all 3 of these are key to understanding probability theory and why 'bits' of information matter.

I don't think it's fair to consider these problems 'tricks'. The second one was a little obfuscating (possibly because of the sheer quantity of words) but they're perfectly logical, the useful information can be separated and you can work from there...in fairness I have a hard time imagining how I thought before learning Bayesian probability...still...it seems obvious. Not all the information is useful, so remove the extraneous data and do the maths...

Well, technically I've only ever heard the first and second today, but as I said, and as the thread has shown, this shit starts serious arguments.

The second one is not up for debate as the choice is made and that is the end of it.

All I will say is that the debatable part is how the question is set up. The math is sound if you approach it from the way that the solutions are set. The debates are about whether the approach is correct to begin with. But like I said, I'm not getting involved and I'm not going to go into the math.

EDIT:
The Monty Hall problem currently exists in Deal or No Deal with a contestant picking 1 of 30 briefcases. Each case has a different monetary value ranging from $1 to $1,000,000. The contestant then gets to open a few of the remaining 29 to be opened. Then the "banker" offers an amount between the two extremes based on what values have been removed. It goes on until the player stops and accepts the offer or has no cases left to open but his/her own. In this game, the case selected initially is the one that remains. The Monty Hall solution applies to this one.

FoolKiller:

EDIT:
The Monty Hall problem currently exists in Deal or No Deal with a contestant picking 1 of 30 briefcases. Each case has a different monetary value ranging from $1 to $1,000,000. The contestant then gets to open a few of the remaining 29 to be opened. Then the "banker" offers an amount between the two extremes based on what values have been removed. It goes on until the player stops and accepts the offer or has no cases left to open but his/her own. In this game, the case selected initially is the one that remains. The Monty Hall solution applies to this one.

I don't think so, because the boxes being opened are under no obligation to hold back a specific number. The $1,000,000 can show up straight away, for example. In the traditional Monty Hall problem, the fact that the door to the car is *never* opened is part of the reason why switching is always favourable.

As a result, in Deal or No Deal the contestant has no more information about their box than any other, so the Monty Hall problem doesn't apply, even though the banker does sometimes offer to let the contestant switch the final two boxes.

The_Darkness:

FoolKiller:

EDIT:
The Monty Hall problem currently exists in Deal or No Deal with a contestant picking 1 of 30 briefcases. Each case has a different monetary value ranging from $1 to $1,000,000. The contestant then gets to open a few of the remaining 29 to be opened. Then the "banker" offers an amount between the two extremes based on what values have been removed. It goes on until the player stops and accepts the offer or has no cases left to open but his/her own. In this game, the case selected initially is the one that remains. The Monty Hall solution applies to this one.

I don't think so, because the boxes being opened are under no obligation to hold back a specific number. The $1,000,000 can show up straight away, for example. In the traditional Monty Hall problem, the fact that the door to the car is *never* opened is part of the reason why switching is always favourable.

As a result, in Deal or No Deal the contestant has no more information about their box than any other, so the Monty Hall problem doesn't apply, even though the banker does sometimes offer to let the contestant switch the final two boxes.

Actually, it does.

And I didn't say that the Monty Hall problem applies. I said the solution applies.

Being shown 28 cases that don't have the million dollars in it doesn't, as they hosts claim, make it a 1 in 2 chance of you having the million. You still have only a 1 in 30 chance getting the million.

Dammit... I said I was going to avoid getting involved in the math portion of this discussion.

FoolKiller:

Being shown 28 cases that don't have the million dollars in it doesn't, as they hosts claim, make it a 1 in 2 chance of you having the million. You still have only a 1 in 30 chance getting the million.

Yes it does make a difference. Check here, where they've already done the maths to tackle this exact debate, and have run simulations to distinguish between the random selection of Deal or No Deal and the forced selection of Monty Hall.

I can try to explain the difference by thought experiments. Lets stick with Deal or No Deal. You selected a box at the start - let's call it box A. It's got a 1 in 30 chance of being the million.

Random Selection (how it works on Deal or No Deal):
Either box A has the million, or it doesn't. If it does (1 in 30), then it doesn't matter which 28 other boxes randomly get opened - in 1 out of every 30 games, then whatever happens you'd be better off sticking with box A.

If box A doesn't have the million, that leaves 29 boxes. You'll randomly select 28 boxes to open - this is equivalent to randomly picking 1 box to stay closed. Let's call the closed box B.

In 28 out of every 30 games, the million will be in one of the boxes you just opened. In this case, it doesn't matter whether you switch between A and B - the million is unavailable.

And in 1 in every 30 games, the million will be in box B, at which point you're better off switching.

For 28 in 30 games, the million got accidentally revealed. For 1 in 30 games, you're better off sticking. And for 1 in 30 games you're better off switching. So if the million hasn't shown up yet, you've got a 50:50 chance regardless of what you do.

Forced selection (how it works in the Monty Hall problem):
Either box A has the million, or it doesn't. If it does (1 in 30), then the host is free to open any 28 boxes of their choosing. It doesn't matter which 28 they choose - in 1 out of every 30 plays, then you'd be better off sticking with your box.

If box A doesn't have the million (29 in 30), then the host is forced to open the other 28 boxes that don't have the million either. This leaves one box behind that does contain the million, so in 29 out of every 30 games, you'd be better off switching.

***

For forced selection, all 28 of the "million accidentally revealed" games that happened previously have instead been converted into "Box B has the million". This is why switching gives you that 29 in 30 chance. (Or the 2 in 3 chance that switching in traditional Monty Hall gives.)

But Deal or No Deal has random selection instead, so the Monty Hall solution doesn't apply. Those 28 "million accidentally revealed" games are important.

So the box problem had stumped me. I wanted to believe the answer was 50%. You either chose the box with GG or you chose the box with GS in your initial selection. So, for all you "boxers" out there, I was in you camp at 50%.

Then there was all this "just look at the coins!" talk and I still couldn't wrap my head around it. So, for all of you "boxers" like me, look at it like this:
1. Forget the SS box. It's dead to all of us. You have two boxes to choose from: GG ad GS.
2. You found a gold coin. Good for you. Any option of drawing silver is eliminated. Let's break it down by boxes...
-You had two chances to be successful at finding gold if you had stuck your hand into Box GG.
-You had one chance to find gold in Box GS. Getting to this point is statistically more difficult.

So, you had two chances on your "stab in" to find the GG box in which case the other coin will indeed be a G. And you had 1 chance to find the G coin in the GS box in which case you will lose (other coin is S). 2/3=win.

I was figuring that once you found gold in a given box, which of the two boxes it was in didn't matter. But I completely dismissed the fact that finding that gold coin in the GS box is less likely than finding a gold coin in the GG box.

Bad Player:

2xDouble:
Because this is a "thought exercise" and math problem, we have to concede that anything which didn't happen in the story didn't happen.

No we don't.

Oh, but we do. Without that restriction, the infinite possibility of human existence must be considered. How do we know A didn't just stab the warden, take his keys, and leave (or immediately get shot)? because it didn't happen. How do we know B didn't receive a pardon immediately before or after the conversation, possibly without the warden's knowledge? because it didn't happen. How do we know C wasn't digging a tunnel as they speak and A was engaging the warden to distract him? because it didn't happen. How do we know the warden wasn't lying to mess with A's head? because it didn't happen. How do we know aliens didn't show up and beam the prisoners to a spaceship? because it didn't happen... and is statistically insignificant.

This is indeed a probability problem, and like all probability problems, we must account for possible outcomes. As mathematicians, we don't get to simply decide which outcomes are and are not possible unless we place fixed limits (or calculate it out and disregard as insignificant). The Monty Hall problem has built-in restrictions; the data set is fixed, there are three cards with printed pictures. This problem, however, humanizes the data points, and in doing so adds humanity (and therefore choice and decision-making) into the equation. The cards can't choose not to be goats, but prisoners can choose not to follow a plan. To correctly answer the question, we must either account for this or assume it didn't happen.

wulfy42:
Not true at all, in fact, it's actually the opposite in some ways.

Lets look at the 3 door problem. You have a 1/3 chance of getting the car to start. That is, in the end, your exact chance of getting the car no matter what else happens along the way. Why?

Well lets look at what is really happening here real quick.

You initially choose a door, there is a 33% chance you chose the care. The game announcer knows if you have the car or not, and knows which other door (or doors) do not have the car.

Either way, there is a door without a car for him to open.

Now, at that exact point in time, you have a 50% chance to have the right door. In no way does the actions of the annoucer opening a door make any difference in your probability of initially choosing the right door. If you do choose again, you now have a 50% chance of being right....but...you have that SAME 50% chance of being right if you don't choose again (it's still the exact same chance).

Choosing again, does not change your odds. Saying it does it just plain silly. The annoucer always is going to offer you another pick, and one of the other doors is always doing to have a goat. Your probability of choosing the correct door is not going to change, but....if you choose again at this point, your NEW probability of choosing the correct door will be 50% (instead of the original 33%). That is the same if you choose the door you currently have, or choose the other door.

In effect you are starting off with a 33% chance (your initial choice)...which for all intents and purposes is pointless (as long as the announcer always has to give you the option to choose again after getting rid of one door), and then ending up with a 50% chance to win. Your net over all chance to choose the car in this case is 50% (as one goat is always removed). Keep the same door, choose the other door, it doesn't matter, you have a 50% chance to win.

The reason it's 66% and not 50% is because they're all part of a set of three options even though one option is now revealed.

Initially you had 33% chance to win and a 66% chance to lose. This means that your single door represents 33% of the options and the other two doors represents 66% of the options. [D1]/[D2+D3] = 33%/66% (Where D1 represents which ever one you chose and D2+D3 represents whichever two you did not).

When one door is revealed to be empty, they're all still part of the same set. The two other doors, the one open and the one closed doors that you didn't pick still represent the other 66% from when you started and you knowing that one door is empty is just you having special insight into the other 66%. With that door open, it is not merely the door that you picked vs the door that you did not but it is instead the door you opened vs the two doors you did not but you now know one isn't it.

This is basically why all of these are considered apparent paradoxes. It seems like it would be one thing but actually, the additional knowledge changes the odds. Like in the box equation, the likelihood of getting a coin from the boxes with gold and silver or with gold and gold are equivalent. But the moment you pull out one gold coin the odds that you got the double golden box over the single golden box changes with that knowledge because the 1gold 1silver box had a lower likelihood of you getting the gold. Basically, the odds that it is the box with Gold and Silver diminishes while the odds that it is the double golden box remains the same. Hence it going from 1/2 to 2/3rds due to the gold-silver box having a half chance of you getting gold.

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