2 boxes, you open one and find it has $100

[OverVolt]

How do you always pick the $100 box? I get the impression you guys don't understand the hypothetical. You randomly pick between 2 boxes. In this first attempt, you happen to pick the $100 box - meaning the other box is either half that or double it. In the next attempt you have a 50/50 chance of picking the $100 again or whatever is in the other box.

On gameshows you always switch tho.

 

[Mr.IncrediblyBored]

On gameshows you always switch tho.

Hell yes you switch. That's called gambling and it's TV. Switching boxes on a game show is fun and wouldn't hurt your chances (or help them), so why not? Makes for good TV.

The question is whether there a strategy to make the most amount of money. 50% of the time you've already picked the box with more money. And a coin flip is not a strategy.

 

[Rakehellion]

And, on the wikipedia page, the major rules are neglected - thus there's a huge flaw on that page. The only way you can do the probability is if you know that either the host randomly chooses when to reveal the extra door, else the host always reveals the extra door. That knowledge is absolutely required for the problem, but it's often left out.

To be honest, I didn't read the Wikipedia page or your post. I just skimmed them. But Wikipedia isn't the only source of information for that problem and I'm sure it's explained better elsewhere.

And my answer was correct: You have a 50/50 shot.

 

[Rakehellion]

How do you always pick the $100 box? I get the impression you guys don't understand the hypothetical. You randomly pick between 2 boxes. In this first attempt, you happen to pick the $100 box - meaning the other box is either half that or double it. In the next attempt you have a 50/50 chance of picking the $100 again or whatever is in the other box.

The exact amount is irrelevant. Replace $100 on the chart with 100% and the results are the same.

 

[Rakehellion]

No, on the contrary I'm completely ignoring the values in the box/envelope.

So you're wrong. You can't possibly say one choice is better than the other if you don't understand the concept of money.

There are numerous possible solutions to this problem. Take the first envelope, take the second envelope, keep your envelope, the first two choices are the same, or it's irrelevant, etc. In your strange world where $50 is equal to $100, these are all the same. In the real world, they aren't.

 

[Carson Dyle]

Whoa! I don't have to wait until lunch tomorrow! We have a winner! I am not disappoint. For the Monty Hall problem to be correctly stated, it must be known that the host isn't acting malevolently. You need to know that either the host always offers the contestant the opportunity to change their mind (which Monty Hall did), else that the host does so randomly.

Your pet peeve is way off the mark. There is no flaw in how the wikipedia article presents the problem. The problem is stated as a singe trial in which the contestant is given the opportunity to switch their selection after one of the other two doors has been revealed. So there is no need to state whether the host is or isn't acting malevolently, or may not have offered the opportunity to switch. It's "always" the same scenario.

 

[BxgJ]

To be honest, I didn't read the Wikipedia page or your post. I just skimmed them. But Wikipedia isn't the only source of information for that problem and I'm sure it's explained better elsewhere.

And my answer was correct: You have a 50/50 shot.

For the monty hall problem? No, try reading his post, and think of it as two sets. The first is the door you pick, and the second is the entire set of other doors (he used 99 in his example to make it easier to understand).

 

[DrPizza]

Your pet peeve is way off the mark. There is no flaw in how the wikipedia article presents the problem. The problem is stated as a singe trial in which the contestant is given the opportunity to switch their selection after one of the other two doors has been revealed. So there is no need to state whether the host is or isn't acting malevolently, or may not have offered the opportunity to switch. It's "always" the same scenario.

You need to know that the host isn't acting malevolently. If he is, your probability is zero if offered the opportunity to switch. If he isn't, your probability is .67 by switching. That is, there is a factor that the probability is conditional upon - the host's intentions.


That is, if there were 100 doors, and one prize of a million dollars, after picking a door, the host could say, "I'm sorry, you lost." Or, the first and only time the game is played, with the host knowing which door is the winning door, upon realizing that you picked the winning door, the host could show you 98 of the 99 losing doors, and then ask if you want to switch to the last door remaining. Dumb people might get lucky by not knowing the Monty Hall problem and having poor statistical skills. But, smug smart people would switch every time, because no one explained to them that the host could be acting malevolently. Monty Hall offered the opportunity every time, because this made his show more exciting.

 

[sontakke]

I do not understand you. Assuming host is malevolent, how does the problem change? We all know host never opens the door containing the prize. He is always able to open a door NOT containing the prize. That actually means he is always acting in malevolently. Or is your definition of malevolent different?

By the way, isn't probability theory itself based upon multiple random events? You can't really apply probability odds to single situation. Only when multiples of a situation *can* occur, probability can be computed.

 

[Rakehellion]

For the monty hall problem? No, try reading his post, and think of it as two sets. The first is the door you pick, and the second is the entire set of other doors (he used 99 in his example to make it easier to understand).

But the fact that he's trying to make you lose isn't known so it can't be factored into the solution.

 

[Rakehellion]

I do not understand you. Assuming host is malevolent, how does the problem change? We all know host never opens the door containing the prize. He is always able to open a door NOT containing the prize. That actually means he is always acting in malevolently. Or is your definition of malevolent different?

By the way, isn't probability theory itself based upon multiple random events? You can't really apply probability odds to single situation. Only when multiples of a situation *can* occur, probability can be computed.

In Monty Hall's game, you get a second chance after choosing a door. In DrPizza's game, you don't. It's a completely different game.

 

[Schfifty Five]

You need to know that the host isn't acting malevolently. If he is, your probability is zero if offered the opportunity to switch. If he isn't, your probability is .67 by switching. That is, there is a factor that the probability is conditional upon - the host's intentions.


That is, if there were 100 doors, and one prize of a million dollars, after picking a door, the host could say, "I'm sorry, you lost." Or, the first and only time the game is played, with the host knowing which door is the winning door, upon realizing that you picked the winning door, the host could show you 98 of the 99 losing doors, and then ask if you want to switch to the last door remaining. Dumb people might get lucky by not knowing the Monty Hall problem and having poor statistical skills. But, smug smart people would switch every time, because no one explained to them that the host could be acting malevolently. Monty Hall offered the opportunity every time, because this made his show more exciting.

But...Monty Hall asking if the contestant wants to switch their selected door IS part of the game. There are no "what ifs" with it. It's not something you need to ponder as it's part of the original premise.

What you seem to be doing is making up a new game all together, though it looks closely like the original Monty Hall problem.

Those are just my thoughts however, and I might not be fully understanding your posts.

 

[Mr.IncrediblyBored]

So you're wrong. You can't possibly say one choice is better than the other if you don't understand the concept of money.

No, on the contrary, I said it does not matter whether you keep the first envelope or switch to the second. It's a coin flip whether you get the larger amount or smaller amount. The actual amounts in the envelope don't matter. Not sure I can put this any simpler.

There are numerous possible solutions to this problem. Take the first envelope, take the second envelope, keep your envelope, the first two choices are the same, or it's irrelevant, etc. In your strange world where $50 is equal to $100, these are all the same. In the real world, they aren't.

I don't know what any of this means. I don't mind playing along, but you need to at least keep it coherent.

 

[Carson Dyle]

You need to know that the host isn't acting malevolently.

Again: You do know that. Exactly as the problem is presented in the wikipedia article that you think is flawed.

Why you think that's even a possibility is the real head-scratcher.

 

[Rakehellion]

I don't know what any of this means.

Clearly.

 

[Schfifty Five]

Again: You do know that. Exactly as the problem is presented in the wikipedia article that you think is flawed.

Why you think that's even a possibility is the real head-scratcher.

I'm with you Carson Dyle. It's possible I'm not fully grasping what Dr. Pizza is trying to say, but I feel like he's removing a premise or assumption of the "monty hall problem" and creating something different.

When you talk about the Monty Hall problem or game, it's understood that you have the option to stay with your door choose the other door. There is no "up to Monty to offer that choice" factor.

 

[DrPizza]

I'm with you Carson Dyle. It's possible I'm not fully grasping what Dr. Pizza is trying to say, but I feel like he's removing a premise or assumption of the "monty hall problem" and creating something different.

When you talk about the Monty Hall problem or game, it's understood that you have the option to stay with your door choose the other door. There is no "up to Monty to offer that choice" factor.

On Monty Hall, as someone stated above, it's understood that the host always offers the chance to switch - it's part of the game. The wikipedia makes no mention of this part; people assume that it's understood. In the version I proposed, I followed the way Wikipedia proposed the problem. However, you odds if you switch are exactly zero, because the only time I offer to let you switch is if swirching results in you losing.

Two possibilities - you pick the correct door with your first pick, or you pick the wrong door. if you pick the correct door, I open another door and allow you to switch. if you switch, you will lose. If you pick the incorrect door, I show you what you won - oops, I'm sorry, you lost. There's no opportunity to switch doors if your choice was a losing door.

In other words, you need to know what motivates the host to offer the switch - is it *always* a part of the game? Or is the host more clever than you are and realizes the offer will make you lose.

 

[DrPizza]

I'm with you Carson Dyle. It's possible I'm not fully grasping what Dr. Pizza is trying to say, but I feel like he's removing a premise or assumption of the "monty hall problem" and creating something different.

When you talk about the Monty Hall problem or game, it's understood that you have the option to stay with your door choose the other door. There is no "up to Monty to offer that choice" factor.

Very close to correct - I'm pointing out that an assumption has been omitted by wikipedia and in my first post about this problem. You guys think that assumption is implicit - it is not.

 

[Mr.IncrediblyBored]

Clearly.

Sorry, I can't make sense of whatever gibberish that comes from you mashing on the keyboard.

 

[Carson Dyle]

Very close to correct - I'm pointing out that an assumption has been omitted by wikipedia and in my first post about this problem. You guys think that assumption is implicit - it is not.

There IS no assumption.

The problem as stated on widipedia:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

How in the world you think anyone is "assuming" this is always the case is a baffling. Why you even brought it up is a mystery.

 

[Tweak155]

There IS no assumption.

The problem as stated on widipedia:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

How in the world you think anyone is "assuming" this is always the case is a baffling. Why you even brought it up is a mystery.

His point is that knowledge changes your probability. Suppose the door the host opened contained the prize and you were offered a chance to switch to another door, what is your chance of picking the prize now?

 

[Tweak155]

Speaking of the Monty Hall problem (assuming host always offer the switch and picks an empty door to show), none of the explanations have made sense to me in the past. I have come up with my own explanation which is as follows:

You have a 1/3 chance of picking the correct door with no knowledge of what is behind each door, which means you are 2/3 likely to pick the incorrect door (which means the prize is behind the doors not chosen have 2/3 probability of containing the prize). When the host opens a door with no prize and you choose to open the last door remaining, you've effectively opened both doors which held the 2/3 probability in your original problem. Thus your chances are greater of winning if you switch.

Not sure if that even follows good reasoning, but it makes sense in my head!

 

[kranky]

Exactly right, Tweak155. Same reasoning I have used to explain it to people.

Regarding some comments above, I never considered the MH problem to allow for the host NOT opening one of the losing doors after the initial door is chosen, or to not allow a switching option thereafter. That's how Monty Hall always did it on the show. It's called the Monty Hall Problem, that's how it works. But it's the Internet, and you can find lots of examples where the description of the problem was garbled.

 

[sontakke]

If you know that the host is always malevolent, then you can change your strategy and improve your odds.

But if the host knows that you know that the host is always malevolent you are back to the original problem.

Bit if you know that the host knows that you know that the host is always malevolent ...

Do you see where this is going? I have utmost respect for DrPizza but I think is unnecessarily complicating the essence of the puzzle.

 

[Tweak155]

If you know that the host is always malevolent, then you can change your strategy and improve your odds.

But if the host knows that you know that the host is always malevolent you are back to the original problem.

Bit if you know that the host knows that you know that the host is always malevolent ...

Do you see where this is going? I have utmost respect for DrPizza but I think is unnecessarily complicating the essence of the puzzle.

I think he just prefers it is clearly stated. I can see it either way, but you should at least be able to understand why it could be a factor. Clearly you do.