Logic/Math Puzzle

[bobert]

Originally posted by: 91TTZ
This sounds like they're going for a Monte Hall type situation, but the way it's worded doesn't allow the same type of math.

In the Monte Hall problem, you're initially faced with a 1/3 chance, then he shows you an empty window and asks if you want to change your guess. You should, since now you'll have a 1/2 chance of being right, as opposed to 1/3 before.

But in this problem, you start with a 1/2 chance of it being the 2x amount, and you still have a 1/2 chance after he asks if you want to change your guess. I can see people trying to argue that you didn't know what your chances were when he asked you the first time, but now you do know the chances. But that doesn't change the probability since you'd always end up with a 50/50 shot.

The wiki link above states you have 2/3 chance when switching, not 1/2.

 

[Gunslinger08]

I'd switch if my initial envelope had an amount of money in it that is not divisible by 2.

 

[BrianH1]

oh no, not again. not the stupid goat behind a door...

 

[91TTZ]

Originally posted by: bobert

Originally posted by: 91TTZ
This sounds like they're going for a Monte Hall type situation, but the way it's worded doesn't allow the same type of math.

In the Monte Hall problem, you're initially faced with a 1/3 chance, then he shows you an empty window and asks if you want to change your guess. You should, since now you'll have a 1/2 chance of being right, as opposed to 1/3 before.

But in this problem, you start with a 1/2 chance of it being the 2x amount, and you still have a 1/2 chance after he asks if you want to change your guess. I can see people trying to argue that you didn't know what your chances were when he asked you the first time, but now you do know the chances. But that doesn't change the probability since you'd always end up with a 50/50 shot.

The wiki link above states you have 2/3 chance when switching,

My mistake, you're right. There are 3 places, and one is revealed. That brings it from 1/3 to 2/3.

 

[BrownTown]

this isn't the gosh darned Monty Hall problem, if you wanna talk about that one then do so in the threads already provided. This is TBH a much more subtle and difficult problem.

EDIT: i honestly suggest people take an hour or two one day to read Wiki's paradox listing and familiarize yourself with those. There are alot of apparently very simple problems that you can learn alot from. Not just about math either, there are alot of other philosophical ones too that are great for making you think. Some of the more interesting ones are the ones that deal with humans ability to manage risk, for example why the nuclear plants are considered much more dangerous to be around than coal or hydro plants when the probability of dying from one is far less (very infrequent large problems are weighed much more heavily than they should when compared to very frequent smaller problems like the hundreds of fatal coal minning accidents each year).

 

[Zenmervolt]

Very interesting... In effect it is a variant of the Monty Hall problem. (Bear with me here.)

Assume we have two people playing against each other using the Monty Hall problem. There are three doors, each of the two players must choose a door and they cannot choose the same door as the other player. One of the two players has chosen the correct door, so Monty opens the one remaining door revealing nothing. Is it in the players' best interest to switch?

Traditional logic says that they should switch, that each player has a 2/3 chance of winning when he or she switches to the other remaining door. However, when there are two players, this would then mean that if they switch the combined probability is 4/3, which isn't possible. It would likewise mean that if neither switched, the probability that one of the two would win is only 2/3, which again isn't possible since Monty has revealed that the prize is not behind the thrid door and must therefore be behind the door of one of the players (that is, a 1/1 probability that one of the players will win).

The base level paradox is the same.

In the OP, imagine two people playing this envelope game against each other. The expected value equation leads us to believe that it is in both players' best interest to switch. However, that means that the combined probabilities for the players add up to more than 1, which cannot happen.

My own response to the OP regarding the paradox:

I friggin hate you. This question is going to haunt me now.

ZV

 

[desteffy]

Originally posted by: CaesaR

Originally posted by: desteffy
This is not the monty hall problem, this one is harder.

Think about a "double or nothing" bet, if you bet X dollars you could either get 2X or 0 back, and that is a "fair bet", if you do it over time you will come out even.

But this you could get either 2X or X/2, which is better than the double or nothing bet.


If you dont understand the solution to monty hall problem or why .9999.... = 1 you probably wont even understand what the question is here...

I know that Einstien. I thought your problem may have a similar solution. But thanks for being condescending.

I dont mean to be condescending, I'm just saying there were big arguments on this forum where people didnt believe .999... = 1, and this is harder to understand than that...

But you are right, this problem is similar in statement to the monty hall problem.

 

[msparish]

Originally posted by: Zenmervolt
Very interesting... In effect it is a variant of the Monty Hall problem. (Bear with me here.)

Assume we have two people playing against each other using the Monty Hall problem. There are three doors, each of the two players must choose a door and they cannot choose the same door as the other player. One of the two players has chosen the correct door, so Monty opens the one remaining door revealing nothing. Is it in the players' best interest to switch?

Traditional logic says that they should switch, that each player has a 2/3 chance of winning when he or she switches to the other remaining door. However, when there are two players, this would then mean that if they switch the combined probability is 4/3, which isn't possible. It would likewise mean that if neither switched, the probability that one of the two would win is only 2/3, which again isn't possible since Monty has revealed that the prize is not behind the thrid door and must therefore be behind the door of one of the players (that is, a 1/1 probability that one of the players will win).

The base level paradox is the same.

In the OP, imagine two people playing this envelope game against each other. The expected value equation leads us to believe that it is in both players' best interest to switch. However, that means that the combined probabilities for the players add up to more than 1, which cannot happen.

My own response to the OP regarding the paradox:

I friggin hate you. This question is going to haunt me now.

ZV

Your whole analogy breaks down when you say that one of the players chose the correct door.

 

[Oceandevi]

Its free money either way.

 

[nonameo]

Originally posted by: Oceandevi
Its free money either way.

QFMFT

 

[Mark R]

Originally posted by: Slew Foot
Assume your envelope has $10. This means that the other has either $5 or $20.
If you always keep your envelope, you'll end up with $10 on average.
If you switch, assuming a fair game, youll average (5+20)/2, 12.50 each time on average.

Except, if you think about it - this is clear nonsense. The two envelopes are indistinguishable, so let's say you are playing with a friend, and the friend opens it and finds $20. Should you swap. Your friend performs the same calculation and swaps. The method you use to calculate the amount you are likely to win following a switch is wrong.

This is explained in depth in the links. However, the best explanation was provided by thesurge at Link. The problem is that this is heavy on mathematical theory and isn't easy to understand if this is the first time you have come across this concept.

Why is your method of calculation wrong? It's because it assumes an impossible set of potential prizes.

Let's say that there is a maximum prize of $100. If you open the envelope and find $60, should you switch in this circumstance? The game is fair - you have a 50:50 chance of having picked the higher or lower envelope. So you should switch right? Of course not. You *know* that you can't have picked the lower prize - and this knowlege is essential to perform a useful calculation. Without this knowledge you don't have enough information to estimate the gain from switching. (This concept of using additional knowledge to refine probabilities is called Bayesian mathematics).

The mistake in your method is subtle - essentially, it assumes that the probability of having picked the lower prize, given that knowledge of what prize you picked, is the same as having picked the higher prize, given the knowledge of what the prize you picked. This is not the same as the probability of picking the higher prize as opposed to the lower prize, in the absence of this knowledge. And, as demonstrated in the example above - knowledge of the prize you picked can substantially affect the odds. In more formal language, the calculation requires the posterior odds, but you only the know the prior odds - and have incorrectly assumed that they are the same

So, is there a situation where you could assume that the odds given the knowledge of the prize are 50:50? The answer is no. The only way in which knowing the value of the prize you picked can have no effect is if the potential prize money is infinite, and any possible dollar value (or fraction) can be chosen. In other words the probability of the host chosing a losing prize of .002 cents would have to be the same as that of chosing a losing prize of $8,819,277,425,299.78. This is unlikely to be a valid assumption.

 

[91TTZ]

Sometimes it's easier to rely on common sense in situations like this, since the math involved gets so complicated and abstract that you lose a grip on it and lose the ability to pick out pretty basic logical errors.

On the other hand, trying to use common sense without knowing the math can get you in trouble, too. I guess the "keep it simple, stupid" principle applies. Always try to use the simplest method that works.

 

[Zenmervolt]

Originally posted by: msparish

Originally posted by: Zenmervolt
Very interesting... In effect it is a variant of the Monty Hall problem. (Bear with me here.)

Assume we have two people playing against each other using the Monty Hall problem. There are three doors, each of the two players must choose a door and they cannot choose the same door as the other player. One of the two players has chosen the correct door, so Monty opens the one remaining door revealing nothing. Is it in the players' best interest to switch?

Traditional logic says that they should switch, that each player has a 2/3 chance of winning when he or she switches to the other remaining door. However, when there are two players, this would then mean that if they switch the combined probability is 4/3, which isn't possible. It would likewise mean that if neither switched, the probability that one of the two would win is only 2/3, which again isn't possible since Monty has revealed that the prize is not behind the thrid door and must therefore be behind the door of one of the players (that is, a 1/1 probability that one of the players will win).

The base level paradox is the same.

In the OP, imagine two people playing this envelope game against each other. The expected value equation leads us to believe that it is in both players' best interest to switch. However, that means that the combined probabilities for the players add up to more than 1, which cannot happen.

My own response to the OP regarding the paradox:

I friggin hate you. This question is going to haunt me now.

ZV

Your whole analogy breaks down when you say that one of the players chose the correct door.

Only if you assume that the player knows he or she chose the correct door.

In the question posed by the OP, one of the "players" does choose the "correct" envelope. It's the same paradox.

ZV

 

[dullard]

Originally posted by: 91TTZ
Sometimes it's easier to rely on common sense in situations like this, since the math involved gets so complicated and abstract that you lose a grip on it and lose the ability to pick out pretty basic logical errors.

On the other hand, trying to use common sense without knowing the math can get you in trouble, too. I guess the "keep it simple, stupid" principle applies. Always try to use the simplest method that works.

In this case, keeping it simple AND using math both result in the correct answer.

Lets say the envelopes have a $10 bill or a $20 bill.

Case 1: you pick the $10 bill. If you switch, you gain $10.
Case 2: you pick the $20 bill. If you switch, you lose $10.

Thus, you either gain $10 or lose $10. The odds are even 50% chance of gaining $10 and 50% chance of losing $10. There is no reason to switch. There is no fallacy. There is no paradox. There is no confusion. The math says you gain or lose $10. The math is simple. The simple conclusion is correct: there is no reason to switch.