Head Exploded 2

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. I saw this problem explained very well in a fascinating book called "There are two errors in the
the title of this book." Highly recommended.

I'm willing to admit I may not be right, but it would be nice if I understood why. Thank you all for your patience.

EastCoastMojo wrote:

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%.

But in that example, where did the other 50% go?

If initially you have your 25%, and the other three representing 75% as (25%+25%+25%) soo....

[25%+(25%+25%+25%)] = 100%

You can only change values as you open an envelope within the expression.

So you're left with:

[25%+(37.5%+37.5%+0%)] = 100%
[25%+(75%+0%+0%)] = 100%

The issue is that much as your brain wants to tell you otherwise (and i struggled with this when my high school math teacher thought he would be cute and give us this), you HAVE to keep the envelope in your hand separate from the other three. The other three are their own entity. The guy trying to play this trick on you opening envelopes has no affect on the chances of your original envelope containing the dollhairs.

three of the envelopes are always empty - you know that before you even choose. 51 of the cards are not the ace of spades.

the act of seeing the evidence that most of the unchosen options are null (a fact you already knew anyway) does not change the odds for the item in your hand... unless you put it back on the table and choose at random.

I'm stuck on the initial problem of four envelopes.

• (25)
• (25)
• (25)
• (25)

Now 2 empty are removed.

• (0)
• (0)
• (50)
• (50)

I know the empty two, I don't the other two. Therefore, 50/50.

92CelicaHalfTrac wrote: 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?

That was for the game which has 3 doors, not 4 envelopes. With three choices, you have a 33.3% chance of picking correctly the first try.

Salanis wrote:

92CelicaHalfTrac wrote: 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?

That was for the game which has 3 doors, not 4 envelopes. With three choices, you have a 33.3% chance of picking correctly the first try.

Ah ok. Now i don't feel so dumb, thanks! (Maybe i should have waded through all 6 pages of arguing? )

EastCoastMojo wrote: 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.

Opening a door does change the odds for all the remaining doors. But the door you first selected has been removed from the set already. Your door is not a "remaining door".

N Sperlo wrote: I'm stuck on the initial problem of four envelopes. * (25) * (25) * (25) * (25) Now 2 empty are removed. * (0) * (0) * (50) * (50) I know the empty two, I don't the other two. Therefore, 50/50.

The other guy isn't opening envelopes from a pool of all 4. He's opening from "his" envelopes, which are three of the original four, and represent 75% combined.

It's not 25%+25%+25%+25%=100%

It's [25%+(25%+25%+25%)]=100%

The issue is that you aren't seeing how the equation should be written, and each time envelope opens, you're creating a new equation/instance with a different sample size.

Salanis wrote:

EastCoastMojo wrote: 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.

Opening a door *does* change the odds for all the remaining doors. But the door you first selected has been removed from the set already. Your door is not a "remaining door".

D'OH!

That's what i couldn't formulate into words... That's the answer right there.

Its fun arguing two different equations simultaneously. Lets do it again sometime!

N Sperlo wrote: Its fun arguing two different equations simultaneously. Lets do it again sometime!

Ok but seriously...

If you were looking for that $100... Would you want to choose one of the four envelopes? Or would you want to choose three of the four envelopes? This is really all the original question is. You can ignore the rest.

Hopefully I'm being helpful and not just muddying, but another spin on how to explain why the "swap" isn't 50/50 just occurred to me...

You are choosing between two envelopes, but the two envelopes are not random; that is, they have not just been shuffled.

There is one in your hand, and there is one on the table. The important thing to remember here is that it's not chance which decided which one is in your hand and which is on the table.

To arrive at the situation we're at now, there was a 25% chance the winning envelope would be in your hand, and a 75% chance it would be on the table, because it was a one in four choice to pick one up.

So, although you are now choosing between two items, the decision about which one was left on the table and which one went in your hand was not 50/50, and as a result the last decision is "stacked" in favor of the envelope on the table.

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.

Okay, think of it as [a+(b+x+y)]=100%

When you first choose, the probability of it being "a" is 25%.
25%+(b+x+y)=100%

x and y are removed, not at random, and shown to be empty. Their values are revealed.
25%+(b+0+0)=100%

b=75%

Salanis 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.

Okay, think of it as [a+(b+x+y)]=100% When you first choose, the probability of it being "a" is 25%. 25%+(b+x+y)=100% x and y are removed, not at random, and shown to be empty. Their values are revealed. 25%+(b+0+0)=100% b=75%

Bingo!

That's what i was missing when i used all "x." D'oh! again!

ransom wrote: Hopefully I'm being helpful and not just muddying, but another spin on how to explain why the "swap" isn't 50/50 just occurred to me... You are choosing between two envelopes, but the two envelopes are not random; that is, they have not just been shuffled. There is one in your hand, and there is one on the table. The important thing to remember here is that it's not chance which decided which one is in your hand and which is on the table. To arrive at the situation we're at now, there was a 25% chance the winning envelope would be in your hand, and a 75% chance it would be on the table, because it was a one in four choice to pick one up. So, although you are now choosing between two items, the decision about which one was left on the table and which one went in your hand was not 50/50, and as a result the last decision is "stacked" in favor of the envelope on the table.

Noooooooooooooooooooooooooooooo!

Salanis 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.

Okay, think of it as [a+(b+x+y)]=100% When you first choose, the probability of it being "a" is 25%. 25%+(b+x+y)=100% x and y are removed, not at random, and shown to be empty. Their values are revealed. 25%+(b+0+0)=100% b=75%

Once two are removed, the equation changes. In all reality, the chances were always 50/50 because someone was going to remove two empty ones anyway.

N Sperlo wrote:

Salanis 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.

Okay, think of it as [a+(b+x+y)]=100% When you first choose, the probability of it being "a" is 25%. 25%+(b+x+y)=100% x and y are removed, not at random, and shown to be empty. Their values are revealed. 25%+(b+0+0)=100% b=75%

Once two are removed, the equation changes. In all reality, the chances were always 50/50 because someone was going to remove two empty ones anyway.

They aren't truly removed though. Just shown to be zero, and his equation changed to reflect that.

OK. Thank you all. I am just going to go find something shiny to chew on for a while.

92CelicaHalfTrac wrote:

N Sperlo wrote:

Salanis 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.

Okay, think of it as [a+(b+x+y)]=100% When you first choose, the probability of it being "a" is 25%. 25%+(b+x+y)=100% x and y are removed, not at random, and shown to be empty. Their values are revealed. 25%+(b+0+0)=100% b=75%

Once two are removed, the equation changes. In all reality, the chances were always 50/50 because someone was going to remove two empty ones anyway.

They aren't truly removed though. Just shown to be zero, and his equation changed to reflect that.

The value of x and y changed, and a as well as b need to change to reflect that.

N Sperlo wrote:

92CelicaHalfTrac wrote:

N Sperlo wrote:

Salanis 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.

Okay, think of it as [a+(b+x+y)]=100% When you first choose, the probability of it being "a" is 25%. 25%+(b+x+y)=100% x and y are removed, not at random, and shown to be empty. Their values are revealed. 25%+(b+0+0)=100% b=75%

Once two are removed, the equation changes. In all reality, the chances were always 50/50 because someone was going to remove two empty ones anyway.

They aren't truly removed though. Just shown to be zero, and his equation changed to reflect that.

The value of x and y changed, and a as well as b need to change to reflect that.

No... just "b." The brackets are there for a reason. ;) And that is, to solve what's inside, first.

(Please excuse my dear aunt sally, order of operations, blah blah blah)

N Sperlo wrote: Once two are removed, the equation changes. In all reality, the chances were always 50/50 because someone was going to remove two empty ones anyway.

No, because the two removed changes each time and is not done randomly. The dealer is not going to remove B and C every time because you might pick one, or one might be the winner. What you pick and what wins can change every round.

Say there are envelopes A, B, C, D. Envelope C is the winning envelope.

You choose A.
B and D are removed. If you stay, you lose. If you switch, you win. (Stay 0 : Switch 1)

You choose B.
A and D are removed.
If you stay, you lose. If you switch, you win.
(Stay 0 : Switch 2)

You choose C.
The dealer has 3 ways to remove 2 of the remaining envelopes (AB, AD, BD).
If you stay, you win. If you switch, you lose.
(Stay 1 : Switch 2)

You choose D.
The dealer removes A and B.
If you stay, you lose. If you switch, you win.
(Stay 1 : Switch 3)

There are 16 permutations of this. If you write them out the Stay 1:Switch 3 ratio will remain the same.

Seriously, if you think it's 50/50, guess a number 1-100. I'll pick a winning number randomly. Pick a number. I'll remove 98 choices after you pick and guarantee one is the winner. If you are right, it will be a 50/50 chance each time.

No money on the line. I won't cheat.

If you switch, you'll have a 99% chance of winning.

N Sperlo wrote: Simulations cannot accurately predict chance.

There is no chance. It's only an illusion of chance.

Ok, how about this:

Same scenario, except with two players.

Player A has a 25% chance of having the correct envelope and a 75% chance that it’s one of the other 3.

Player B also has a 25% chance of having the correct envelope and a 75% chance that it’s one of the other 3.

If a dealer reveals that the remaining two envelopes are empty, what is each player’s chance of having the $100?

Based on the logic presented previously, each player would increase their odds by trading with the other.