Head Exploded 2

nderwater wrote: </cite You could just as easily argue: it doesn't matter how many envelopes there were originally, only that there are two now - one empty and one with the cash. No matter what the odds were when you picked the envelope in your hand, the odds don't change the fact that there are only two outcomes possible for the final choice. For example - A couple has three children, all girls. What are the chances that the next child will be a boy?

Your example is an entirely different problem.

Yes, there are two possible outcomes, but that doesn't mean they are equally likely. The chance that the money is in one envelope or the other is NOT random. the held envelope was one selected from a set of four, while the remaining envelope represents the rest of the set not selected. When two envelopes get removed from that set after the fact, that does not affect the selection you've already made. You're not picking again randomly from the two remaining envelopes, if you were that would indeed be a 50/50 shot. The originally selected envelope will always have a 25% chance of containing the money, that selection has been made and does not change. Which means that the chance of the money being among the unselected envelopes will always be 75%. The two envelopes removed from the remaining set are not removed randomly (if they were, then the odds of the money being in the held envelope would be equivalent to the odds of the remaining envelope, both at 25%, and there would be a 50% chance the money was in one of those taken away). This is where people get tripped up, it's intuitive to treat the second situation like you are selecting again from a new set, but you're not. You never put back that envelope you originally selected, and the odds of that selection don't magically change when the unselected set is partially revealed to you.

This thread is a great example of why I wish I owned a casino.

SVreX wrote:

mtn wrote:

SVreX wrote: I'd be happy to put a little money on the table for anyone who believes the statistics are anything other than 50/ 50. That's ridiculous.

I'll take that bet.

OK. You are suggesting that the odds of the envelope on the table having $100 in it are 75%. I am saying the odds are 50%. So, I'll put $500 in a pot, and so will you. We'll run the experiment 100 times. If you are right, the envelope on the table will have $100 75 times. If I am correct, it will only be 50 times. Whoever is closer gets the $1000. Still ready to take the bet? I'll gladly take your money.

Before you get so excited, read the experiment again. I think you are failing to understand that the envelopes removed after the initial selection are not removed randomly, but by an agent that knows which ones are empty. There are always at least two empty envelopes in the remaining set, and two OF THOSE and ONLY THOSE are removed. There is no scenario in which the envelope containing the money is removed, which is what you'd need for the odds to be equivalent after the removal.

Josh, I agree.

THAT is exactly why the odds become 50/50.

The only scenario which would make the odds 1 in 4 would be if the potential existed for the one with the money in it to be removed. It doesn't.

Josh, if you'd like to take my bet, I'll do it with you too.

You are doing the logic backwards.

Look at it this way.

While there were 4 envelopes originally, 2 were NEVER an option, and BOTH of these 2 were guaranteed to be empty.

So, the real scenario is, I'm gonna let you pick 1 envelope out of (4 minus 2). Odds are 50/50.

SVreX wrote: Look at it this way. While there were 4 envelopes originally, 2 were NEVER an option, and BOTH of these 2 were guaranteed to be empty. So, the real scenario is, I'm gonna let you pick 1 envelope out of (4 minus 2). Odds are 50/50.

He's right. Josh, you are making this too complicated. When you make the second decision, you are indeed faced with one envelope with money and one without. 50/50 regardless of previous probabilities.

My money is still on the table.

I'm hearing lots of talk, but no takers.

If the contents of the 2 removed remain unknown, the odds are 25%. As presented, 50/50, and a money maker for me.

Josh wrote: . This thread is a great example of why I wish I owned a casino.

Me, too, while hastening to add that I know better than to go in one myself.

One last shot at it: It has nothing to do with "statistics," only probability. It's about what would be the better gamble.

Looks as if my work here is done.

Not till you take my bet.

You can stay up all night here and run your own experiments: Stay or switch?

goats for everyone!

jg

SVreX, I will take your bet. Seriously. Sorry I hadn't responded, I had to go to (math) class.

Think about it like this:
Four envelopes on the table, one has money. You pick one.
Now, instead of having them remove 2 envelopes without money on the table, think of it as putting all 3 envelopes on the table into a single envelope. Now, do you switch?

In practice, it is the same problem.

I have a moral objection to taking money from people who don't understand they're being taken advantage of, so I don't really want to actually make a bet with any of you, but you're making my convictions difficult to uphold.

I really think you (SV, Tuna) just aren't grasping the scenario being presented. Forget about the envelopes. There are 4 cards face down on the table, one of them is the ace of spades (winner), the other ones are all deuces. You pick one. The dealer (who already knew which card was the ace), now flips over two deuces (there are always at least two among the three remaining cards). If you took the ace with your initial draw, it's already in your hand. If you didn't, it's the lone card remaining on the table. Are you still sure you want to keep the one in your hand? If you do, I can PM you my paypal address.

SVreX wrote: Look at it this way. While there were 4 envelopes originally, 2 were NEVER an option, and BOTH of these 2 were guaranteed to be empty.

Actually, 2 were an option. 2 are an option once, 2 are an option twice.

Maroon92 wrote: By rules of statistics, you should ALWAYS trade envelopes. The envelope in your hand has a 1 in four chance of winning, while the envelope on the table has a 1 in 2 chance of winning.

They both have $100 bills. Over the years, I have built up a tolerance to them, so they won't kill me.

SVreX wrote: Look at it this way. While there were 4 envelopes originally, 2 were NEVER an option, and BOTH of these 2 were guaranteed to be empty. So, the real scenario is, I'm gonna let you pick 1 envelope out of (4 minus 2). Odds are 50/50.

Possibly.

On the OTHER hand. You have a 25% chance of getting the money envelope. That means there's a 75% chance that you have an empty one.

Then someone takes away two empty ones. There's still a 25% chance that you have an empty envelope, and a 75% chance that the remaining envelope has the money in it...

Or, another way without numbers. You pick one. Guy takes away all of the empty ones on purpose but leaves one remaining. Now: He just told you that he knows what's in them and that he took away a whole lot of losers, leaving one left. He's telling you that the remaining one has money in it!

To be fair, part of the problem here is the wording of the setup, because it's all "you are told..." instead of being clear that there IS $100 in one of the four envelopes and that the two that are discarded ARE empty. As it's written the only unassailably logical thing to do would be to knee the other guy in the balls, run off with all the envelopes, and open them at home to see what's really in there.

RED SNAPPER!

SVreX wrote: My money is still on the table. I'm hearing lots of talk, but no takers. If the contents of the 2 removed remain unknown, the odds are 25%. As presented, 50/50, and a money maker for me.

I WILL TAKE THAT BET!

you are essentially betting against the fact that 50% of the time you will pick correctly! That is just not possible.

you will pick the correct envelope 25% of the time and the money will be ON THE TABLE 75% of the time.

Guaranteed!

Knurled wrote:

Maroon92 wrote: By rules of statistics, you should ALWAYS trade envelopes. The envelope in your hand has a 1 in four chance of winning, while the envelope on the table has a 1 in 2 chance of winning.

They both have $100 bills. Over the years, I have built up a tolerance to them, so they won't kill me.

"Fool! You fell victim to one of the classic blunders. The most famous is 'Never get involved in a land war in Asia,' but only slightly less well known is this: 'Never go in against a Sicilian when death is on the line.'"

JG's goat website offers a couple of good explanations as to why you're better off changing envelopes: http://www.stayorswitch.com/explanation.php

Luke wrote: JG's goat website offers a couple of good explanations as to why you're better off changing envelopes: http://www.stayorswitch.com/explanation.php

I always love explanations that start by referencing someones "Goat website".

A better question:

Did Monty actually let you keep the goat?

jg

In reply to JG Pasterjak:

I was thinking that same thing. Goats is good eatin': http://simplyrecipes.com/recipes/jamaican_goat_curry/

Luke wrote: JG's goat website offers a couple of good explanations as to why you're better off changing envelopes: http://www.stayorswitch.com/explanation.php

No, those are quasi-logical positions defending false predetermined assumptions designed to confuse people who don't know any better.

There is no pre-determined "stay or switch" position. The suggestion that it is a "better" strategy to approach the problem is a fallacy. It forgets the fact that you don't know the rules before you start.

It's a coin flip. One side is "stay" one side is "switch".

So, the initial 3 doors are presented and understanding all three have the same odds, a simple coin flip will suffice. Heads is stay, tails is switch.

So, in consideration of door #1, the coin is flipped. Tails. Switch to door #2. Coin flip again. Tails. The choice is to switch to door #3. No particular reason other than that I think it is pretty.

Game show host changes the rules. Provides me with a NEW choice of only 2 doors, with a guarantee that one has a car, one has a goat. Flip the coin. 50/50 odds. Very simple.

If the goat explanation was true, Truth or Consequences would have lost vast fortunes and been bankrupt before it started. Someone doesn't seem to be noticing that Monty Hall is the casino "house".

It is designed to get people excited about psychological phenomenon, not mathematical facts.

If I flip a coin 10 times and get heads every time, what are the odds it will land on heads the 11th time?