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

[Rakehellion]

I'm sorry, you're also wrong. Who wants to be the third person who is wrong? Read my post again, and assume that I'm pretty good at math. The Monty Hall problem is one of my biggest pet peeves (in mathematics), because so many people get it wrong. I'll consider this successful if even a few more people gain a better understanding of the Monty Hall problem. The computer simulations you're speaking of get the problem I proposed wrong - I'll explain why tomorrow at lunch time.

Or, maybe I'm wrong? Nah, I don't think so.

You're wrong. There's a 50% chance of getting the right door.

 

[cubby1223]

You're wrong. There's a 50% chance of getting the right door.

Everyone can agree that your initial choice has a 33% chance of being correct. Thus 67% chance of being wrong, one of the other doors is the correct door.

The fact that the host then opens a non-winning door, does not provide any new information as to whether your original choice is right or wrong. If you stick, you win 33% of the time.

Therefore if you switch, you will win 67% of the time.

 

[DealODay]

Take the 100, pay someone else 25 to take both boxes and give them to you
/thread

 

[DrPizza]

You're wrong. There's a 50% chance of getting the right door.

Huh?! It's not a 50% chance. You should re-read the Monty Hall problem. To understand this with the Monty Hall problem, if there were 100 doors, you only have a 1% chance of choosing the correct door, and a 99% chance that you chose the wrong door. If Monty Hall - the Monty Hall of the Monty Hall problem - said at that moment, would you like to keep your door? Or would you like the sum of all the prizes from behind the other 99 doors (99 of the 100 doors have a prize worth zero dollars), then in that scenario, by switching, you'd have a 99% chance of winning the big prize. That he knows where the big prize is and reveals 98 of the zeros immediately after allowing you that chance doesn't change the probability that you were correct with your initial guess.

 

[DrPizza]

You're wrong. There's a 50% chance of getting the right door.

Everyone can agree that your initial choice has a 33% chance of being correct. Thus 67% chance of being wrong, that one of the other doors is the correct door.

The fact that the host then opens a non-winning door, does not provide any new information as to whether your original choice is right or wrong. If you stick, you win 33% of the time.

Therefore if you switch, you will win 67% of the time.

Assuming it was Monty Hall, yes.

 

[Charmonium]

Assuming it was Monty Hall, yes.

So how was your problem any different? The game host still reveals one of the other doors which has a non-prize, just like in MH. The only thing I can see from your wording of it is that Monty picks the other door at random.

 

[DrPizza]

How the problem is different is probably the most important part of the Monty Hall problem. Yet, that part is overlooked the vast majority of the time. In fact, I just glanced through the wikipedia article and feel that the article is, unfortunately, wrong in the statement of the problem. It overlooks one important aspect of the Monty Hall problem - Monty Hall. If I recall correctly from an earlier analysis, Marilyn vos Savant was also wrong.

 

[Rakehellion]

Huh?! It's not a 50% chance. You should re-read the Monty Hall problem. To understand this with the Monty Hall problem, if there were 100 doors, you only have a 1% chance of choosing the correct door, and a 99% chance that you chose the wrong door. If Monty Hall - the Monty Hall of the Monty Hall problem - said at that moment, would you like to keep your door? Or would you like the sum of all the prizes from behind the other 99 doors (99 of the 100 doors have a prize worth zero dollars), then in that scenario, by switching, you'd have a 99% chance of winning the big prize. That he knows where the big prize is and reveals 98 of the zeros immediately after allowing you that chance doesn't change the probability that you were correct with your initial guess.

If there are 100 doors, you don't know what's in 99 of the doors, so you switch. If there are 3 doors, you don't know what's in 2 of the doors, so it's a 50/50 shot making your choice meaningless.

 

[Rakehellion]

The fact that the host then opens a non-winning door, does not provide any new information as to whether your original choice is right or wrong. If you stick, you win 33% of the time.

It provides information as to which of the other doors is wrong, and by process of elimination, yours.

 

[disappoint]

Let's say that I'm the host of my own game show. There are 3 doors, and behind one of those doors, there's a big prize. Behind the other two doors, there's a non-prize. You pick a door. I then reveal what's behind one of the other two doors - a non-prize. And then, I offer you the opportunity to change your pick. Would you change your pick?

The question I have is: Are you offering the opportunity to change my pick because you know I have the right door and want me to switch in order to make me lose? (all this assuming you know which is the winning door)

Conversely would you NOT offer the opportunity to change my pick if you knew I had the losing door to begin with?

If it's all just random that's a different problem and has already been answered.

 

[Rakehellion]

How the problem is different is probably the most important part of the Monty Hall problem. Yet, that part is overlooked the vast majority of the time. In fact, I just glanced through the wikipedia article and feel that the article is, unfortunately, wrong in the statement of the problem. It overlooks one important aspect of the Monty Hall problem - Monty Hall. If I recall correctly from an earlier analysis, Marilyn vos Savant was also wrong.

"Everyone in the world is wrong but me."


k.

 

[Rakehellion]

The question I have is: Are you offering the opportunity to change my pick because you know I have the right door and want me to switch in order to make me lose? (all this assuming you know which is the winning door)

Conversely would you NOT offer the opportunity to change my pick if you knew I had the losing door to begin with?

If it's all just random that's a different problem and has already been answered.

Everyone is offered to change. It adds suspense and all that BS to make the show more entertaining.

 

[DrPizza]

The question I have is: Are you offering the opportunity to change my pick because you know I have the right door and want me to switch in order to make me lose? (all this assuming you know which is the winning door)

Conversely would you NOT offer the opportunity to change my pick if you knew I had the losing door to begin with?

If it's all just random that's a different problem and has already been answered.

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. If you don't know that, then you hope the host isn't malevolent, which gives the 67% chance of winning by switching. But, if the host is malevolent, then 100% of the time you switch, you lose.

If you chose wrong with the first choice, you lose, period. I don't offer you the chance to change doors. If you choose correct with the first door, then, and only then will I offer the opportunity to switch doors and show you one of the two losing doors that you didn't choose. By switching, you lose 100% of the time. That was my plan from the start. Not knowing my intentions doesn't change your probability of winning if you switch doors - this is why knowing that the host always (or randomly) allows the switch is a necessary part of the statement of the problem.

 

[Rakehellion]

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. If you don't know that, then you hope the host isn't malevolent, which gives the 67% chance of winning by switching. But, if the host is malevolent, then 100% of the time you switch, you lose.

If you chose wrong with the first choice, you lose, period. I don't offer you the chance to change doors. If you choose correct with the first door, then, and only then will I offer the opportunity to switch doors and show you one of the two losing doors that you didn't choose. By switching, you lose 100% of the time. That was my plan from the start. Not knowing my intentions doesn't change your probability of winning if you switch doors - this is why knowing that the host always (or randomly) allows the switch is a necessary part of the statement of the problem.

It isn't really a "logic problem" if you neglect to reveal major rules of the game.

 

[Charmonium]

It isn't really a "logic problem" if you neglect to reveal major rules of the game.

I was thinking the same thing. The presentation is so close to the original Monty Hall problem that it's hard to avoid making the same assumptions.

 

[HamburgerBoy]

The way he presented it, there is no game. In his scenario, you start with the wrong choice 100% of the time.

 

[ImpulsE69]

This would be much more compelling if it was body parts at risk.

When talking about money that you didn't have in the first place you aren't really risking anything because you won regardless. At that point it just comes down to if you'd rather have money or don't care one way or the other.

Let's say the first box says you will lose 2 legs and 2 arms, but that the other box may be 1 leg, or it may be your head. It makes the decision much more intense. We might be onto a movie here. ATOT SAW

 

[OverVolt]

..

 

[DrPizza]

It isn't really a "logic problem" if you neglect to reveal major rules of the game.

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.

 

[SlitheryDee]

I'd just take my $100 and leave. I don't care to game the system. I just got $100 I didn't have before. Good enough.

 

[TwiceOver]

To me, the potential $50 loss is worth the potential $100 gain. So I'd take the chance. Someone in a more desperate situation might think otherwise.

I'm sure the probability is there, but going on gut/risk...

 

[OverVolt]

To me, the potential $50 loss is worth the potential $100 gain. So I'd take the chance. Someone in a more desperate situation might think otherwise.

I'm sure the probability is there, but going on gut/risk...

If I were the one presenting the boxes you'd get $50 every time man. There would be no $200 box .

Should've taken the first one!

 

[disappoint]

Whoa! I don't have to wait until lunch tomorrow! We have a winner!...

:sneaky::thumbsup:

 

[Mr.IncrediblyBored]

Chances to do what? Pick the "correct" envelope or make the most money?

Your argument completely ignores the amount in each envelope and assigns them arbitrary values of "positive" and "negative." This is bad math. By your logic, you shouldn't switch even if your envelope contains $1 because you still have only a 50% chance of winning.

No, on the contrary I'm completely ignoring the values in the box/envelope. The amount doesn't matter as it's a coin flip either way. If the question is how can you get the most amount of money possible? The answer is: Nothing.


Edit: That's the gamblers fallacy. Sure, you have the chance to double your money this one attempt. Over the long term, giving up the first box/envelope for the second nets you nothing (doesn't hurt you either).

 

[Mr.IncrediblyBored]

On a piece of paper make two columns, and any amount of rows (at least more than 3 rows). On the left-hand side write $100 in each row. On the right-hand side alternate between $50 and $200 for each row. With two columns of equal length, sum up all the values of the left-hand column and all the values of the right-hand column.

Which summation is greater?

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.