Ok, now what i'm having a hard time wrapping my head around is that the results seem to be 66% FOR switching, and 33% against in terms of the simulation.
With that large of a sample, it's a little bit too far outside of acceptable deviance.
Is it because you really end up with 3 samples in the scenario? Two removed, one you picked, and the remaining? But if the two that were removed are guaranteed to not have the money in it, why does it matter?
OK, seriously people. Its been cut down to a choice of two. Thats a 50/50 chance. IT DOESN'T berkeleyING MATTER WHAT YOU PICK!
N Sperlo wrote: OK, seriously people. Its been cut down to a choice of two. Thats a 50/50 chance. IT DOESN'T berkeleyING MATTER WHAT YOU PICK!
Seriously buddy.... read this. 
http://en.wikipedia.org/wiki/Monty_Hall_problem
Sperlo - the final choice is indeed a choice of two. But the odds are greater that the $100 is in the other envelope.
Let's say I put two envelopes on a table in front if you, one containing $100. Yes, there's a 50% chance of getting the $100 if you chose at random, but what if I tell you that I put the $100 in the blue one? Would you still pick the red one?
I think that MTN's illustration is easy to visualize and pretty compelling, so let's blow it out a little. Let's say there were 100 envelopes on the table. You pick one. Then the remaining 99 envelopes are combined into one big envelope. Now do you want to trade?
Get it? The 50% chance is there only if you chose at random between the final two envelopes. But what went into the final two envelopes is not completely random - it's statistically more likely that the envelope not in your hand contains the $100.
I think I am following things up until the envelopes are considered a group. My brain only wants to consider them as individual envelopes. In the first step, when there are four envelopes, each one has a one in four chance of containing money, or 25%. Regardless of me picking one. Then, lets say one envelope is removed and shown to not have any money, that leaves three on the table, each one now having improved odds of having money inside, now it's a one in three chance, or 33%. Now, let's remove one more envelope, and it is shown to not contain money either. Now we are left with two envelopes, with their new and improved chances being one in two, or 50%.
My picking one and removing it from the group does not change that percentage. I've read the thing on the Monty Hall Equation, but it just sounds like BS.
^Look at the "removing" of the envelopes as simply opening a couple of them in the other group.
Your first choice leaves you with two samples. The envelope in your hand, and the others. As you said, there's a 25% chance that the one in your hand has the money, and the others is a 75% chance.
OPEN one, but don't remove it, and it's still a 75% chance that one of the remaining three have the $100. Open a second one, and it's still a 75% chance that one of the remaining three have $100. Switch and you jump from your lowly 25% to essentially getting whatever happens to be in the other THREE envelopes. 
The confusion comes with the word "remove."
With the way the "game" works, you're essentially getting the chance to see the contents of 3 of the 4 envelopes. With that knowledge, it's not a 50% chance.
Its a puzzle to find out chances, which give you absolutely no edge, because your chances are still 50/50.
92CelicaHalfTrac wrote: ^Look at the "removing" of the envelopes as simply opening a couple of them in the other group. Your first choice leaves you with two samples. The envelope in your hand, and the others. As you said, there's a 25% chance that the one in your hand has the money, and the others is a 75% chance. OPEN one, but don't remove it, and it's still a 75% chance that one of the remaining three have the $100. Open a second one, and it's still a 75% chance that one of the remaining three have $100. Switch and you jump from your lowly 25% to essentially getting whatever happens to be in the other THREE envelopes.
How in the world does opening a door NOT change the odds for all the remaining doors? That's what I don't get at all. Sorry to be such a complete dumbass, but that makes no sense to me.
Sperlo - If I told you that the envelope in your hand was empty, and I knew because i marked the correct envelope - then would you trade?
That would change the odds considerably, don't you think?
In reply to EastCoastMojo:
It doesn't. It remains that there are two envelopes and one has a bill in it. Your chances do not stray from 50/50 just because they were not 50/50 before. Chances are still 50/50.
EastCoastMojo wrote:92CelicaHalfTrac wrote: ^Look at the "removing" of the envelopes as simply opening a couple of them in the other group. Your first choice leaves you with two samples. The envelope in your hand, and the others. As you said, there's a 25% chance that the one in your hand has the money, and the others is a 75% chance. OPEN one, but don't remove it, and it's still a 75% chance that one of the remaining three have the $100. Open a second one, and it's still a 75% chance that one of the remaining three have $100. Switch and you jump from your lowly 25% to essentially getting whatever happens to be in the other THREE envelopes.How in the world does opening a door NOT change the odds for all the remaining doors? That's what I don't get at all.
Think of it as brackets/parantheses within an equation.
You have [x + (x + x + x)] = 100
Each "x" is 25, and stays 25 no matter how you solve it.
Which is more? The single "x"? Or the expression of (x+x+x)?
N Sperlo wrote: In reply to EastCoastMojo: It doesn't. It remains that there are two envelopes and one has a bill in it. Your chances do not stray from 50/50 just because they were not 50/50 before. Chances are still 50/50.
Out of curiousity... how do you explain the simulation results not giving a 50/50 split with a very high sample count?
92CelicaHalfTrac wrote:N Sperlo wrote: In reply to EastCoastMojo: It doesn't. It remains that there are two envelopes and one has a bill in it. Your chances do not stray from 50/50 just because they were not 50/50 before. Chances are still 50/50.Out of curiousity... how do you explain the simulation results not giving a 50/50 split with a very high sample count?
Simulations cannot accurately predict chance.
92CelicaHalfTrac wrote:EastCoastMojo wrote:Think of it as brackets/parantheses within an equation. You have [x + (x + x + x)] = 100 Each "x" is 25, and stays 25 no matter how you solve it. Which is more? The single "x"? Or the expression of (x+x+x)?92CelicaHalfTrac wrote: ^Look at the "removing" of the envelopes as simply opening a couple of them in the other group. Your first choice leaves you with two samples. The envelope in your hand, and the others. As you said, there's a 25% chance that the one in your hand has the money, and the others is a 75% chance. OPEN one, but don't remove it, and it's still a 75% chance that one of the remaining three have the $100. Open a second one, and it's still a 75% chance that one of the remaining three have $100. Switch and you jump from your lowly 25% to essentially getting whatever happens to be in the other THREE envelopes.How in the world does opening a door NOT change the odds for all the remaining doors? That's what I don't get at all.
Certainly, (x+x+x) is more than x, but how can (X+0+0) be greater than x? That is the question I struggle with. The two empty envelopes are known quantities, and their quantity is 0. Therefore, the odds change.
If you chose a playing card from a deck, there's a 1/52 chance that it's the ace of spades...
Without turning your card over, are the odds greater that yours is the ace, or that the rest of the deck contains the ace? The odds are 98% (51/52) that the ace is still in the deck.
So then the dealer privately looks through the rest of the deck and chooses 50 cards and turns them over, showing that they are not the Ace.
The last card in his hand has a 51/52 chance of being the ace. Your card has a 1/52 chance. Sure, if you set them both down, shuffle and chose one at random, you chance of getting the ace is 50%. But you haven't done that - you're still holding the card with the 1/52 chance.
N Sperlo wrote:92CelicaHalfTrac wrote:Simulations cannot accurately predict chance.N Sperlo wrote: In reply to EastCoastMojo: It doesn't. It remains that there are two envelopes and one has a bill in it. Your chances do not stray from 50/50 just because they were not 50/50 before. Chances are still 50/50.Out of curiousity... how do you explain the simulation results not giving a 50/50 split with a very high sample count?
With a large enough sample, they absolutely can. I'm not saying it'll get to exactly 50/50 like you say... but it's wayyyyyy off of that, with thousands of recreations.
EastCoastMojo wrote:92CelicaHalfTrac wrote:Certainly, (x+x+x) is more than x, but how can (X+0+0) be greater than x? That is the question I struggle with. The two empty envelopes are known quantities, and their quantity is 0. Therefore, the odds change.EastCoastMojo wrote:Think of it as brackets/parantheses within an equation. You have [x + (x + x + x)] = 100 Each "x" is 25, and stays 25 no matter how you solve it. Which is more? The single "x"? Or the expression of (x+x+x)?92CelicaHalfTrac wrote: ^Look at the "removing" of the envelopes as simply opening a couple of them in the other group. Your first choice leaves you with two samples. The envelope in your hand, and the others. As you said, there's a 25% chance that the one in your hand has the money, and the others is a 75% chance. OPEN one, but don't remove it, and it's still a 75% chance that one of the remaining three have the $100. Open a second one, and it's still a 75% chance that one of the remaining three have $100. Switch and you jump from your lowly 25% to essentially getting whatever happens to be in the other THREE envelopes.How in the world does opening a door NOT change the odds for all the remaining doors? That's what I don't get at all.
I should have spent more time with that example, that's my fault.
What you're struggling with is that once you see the "zero," you're throwing those envelopes out, along with their chances. You've gotta keep them in. Just because they're a "zero" doesn't mean that their original chance doesn't still apply to the situation.
There's 4 envelopes in play, throughout the entire scenario. Not just at the beginning. All you're doing is making the choice as to whether or not you want to take your chances on one envelope, or three.
If there is an option of two, the chance will be 50% that the simulator gets it, just like everyone else. Its just as plausible that if you lined up the same amount of people, the results would end up completely opposite.
Three doors. Random door wins. Random selection. Its randomization and nothing more.
nderwater wrote:EastCoastMojo wrote: Certainly, (x+x+x) is more than x, but how can (X+0+0) be greater than x? That is the question I struggle with. The two empty envelopes are known quantities, and their quantity is 0. Therefore, the odds change.If you chose a playing card from a deck, there's a 1/52 chance that it's the ace of spades... Without turning your card over, are the odds greater that yours is the ace, or that the rest of the deck contains the ace? The odds are 98% (51/52) that the ace is still in the deck. So then the dealer privately looks through the rest of the deck and chooses 50 cards and turns them over, showing that they are not the Ace. The last card in his hand has a 51/52 chance of being the ace. Your card has a 1/52 chance. Sure, if you set them both down at random, you chance of getting the ace is 50%. But you haven't done that - you're still holding the card with the 1/52 chance.
What I am saying is that the former unknowns, the 50/52, have been removed from the set of current unknowns, because their value is now known. The set of current unknowns now consists of 2; your card and the one on the table. Each one carries the same likelyhood to be the ace, as all of the "former unknowns" have been revealed to not be the ace.
EastCoastMojo wrote:nderwater wrote:What I am saying is that the former unknowns, the 50/52, have been removed from the set of current unknowns, because their value is now known. The set of current unknowns now consists of 2; your card and the one on the table. Each one carries the same likelyhood to be the ace, as all of the "former unknowns" have been revealed to not be the ace.EastCoastMojo wrote: Certainly, (x+x+x) is more than x, but how can (X+0+0) be greater than x? That is the question I struggle with. The two empty envelopes are known quantities, and their quantity is 0. Therefore, the odds change.If you chose a playing card from a deck, there's a 1/52 chance that it's the ace of spades... Without turning your card over, are the odds greater that yours is the ace, or that the rest of the deck contains the ace? The odds are 98% (51/52) that the ace is still in the deck. So then the dealer privately looks through the rest of the deck and chooses 50 cards and turns them over, showing that they are not the Ace. The last card in his hand has a 51/52 chance of being the ace. Your card has a 1/52 chance. Sure, if you set them both down at random, you chance of getting the ace is 50%. But you haven't done that - you're still holding the card with the 1/52 chance.
All things considered, if you were looking for that ace.... would you rather have your one card? Or would you rather be sitting in the dealer's spot with the other 51 cards as your initial choice if that's what was presented to you?
N Sperlo wrote: If there is an option of two, the chance will be 50% that the simulator gets it, just like everyone else. Its just as plausible that if you lined up the same amount of people, the results would end up completely opposite. Three doors. Random door wins. Random selection. Its randomization and nothing more.
I guess i'll ask you the same question. If you want that $100... would you rather pick one envelope? Or would you rather pick three?
92CelicaHalfTrac wrote: I should have spent more time with that example, that's my fault. What you're struggling with is that once you see the "zero," you're throwing those envelopes out, along with their chances. You've gotta keep them in. Just because they're a "zero" doesn't mean that their original chance doesn't still apply to the situation. There's 4 envelopes in play, throughout the entire scenario. Not just at the beginning. All you're doing is making the choice as to whether or not you want to take your chances on one envelope, or three.
Yes, this is definitely the part I am struggling with, as what the hell is the point of showing me the contents of those envelopes? I have no problem accepting that the set of three envelopes on the table have collectively a larger chance of being the one with the money in it, as their values are unknown and together they represent (25%+25%+25%), which is considerably more than my 25%, but once we know more about the group, my consideration of what that group has to offer HAS to change. I now know that the empty envelopes have a 0% chance of containing the money, so that changes the equation to (25%+0%+0%), which is the same as my 25%.
My limited internet connection is not fast enough to get involved in this argument over, but I just thought I'd chime in and say that Ben is right.
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