In reply to 92CelicaHalfTrac:
(a+ (b+y+z))=100
(25+(25+0+0)=50
Just because you made two equal zero doesn't mean you changed the value of one. For all you know, it is equal to the first. 100 must stay common, so a and b must reflect three change as 50.
(50+(50+0+0))=100
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.
No. You chose one of four. Your choice is still one of four.
It's only a 50/50 chance if two empty envelopes are removed before you choose.
This, I think, is where everyone's getting hung up.
N Sperlo wrote: In reply to 92CelicaHalfTrac: (a+ (b+y+z))=100 (25+(25+0+0)=50 Just because you made two equal zero doesn't mean you changed the value of one. For all you know, it is equal to the first. 100 must stay common, so a and b must reflect three change as 50. (50+(50+0+0))=100
Nope.
You "knew" at the beginning they were all 25.
Can we agree that in the beginning that "a" is 25?
If so, then you have:
[a+(b+c+d)] = 100
a=25
You are then given that "c" and "d" both = 0.
[25+(b+0+0)] = 100.
Solve for b.
When you change "c" and "d," both of which are contained within the expression (b+c+d) and you cannot venture outside of that until the expression itself is solved. Separate expression, has no bearing on what's outside of it.
Knurled wrote: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.No. You chose one of four. Your choice is *still* one of four. It's only a 50/50 chance if two empty envelopes are removed *before* you choose. This, I think, is where everyone's getting hung up.
That us what I'm expressing.
When the two are shown to be zero, they must be removed from the equation along with their 25. You are left with a total of fifty.
That_Renault_Guy wrote: 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.
That would actually be three "players."
You'd have:
Player A with a 25% chance.
Player B with a 25% chance.
Dealer with a 50% chance. (envelopes "c" and "d")
In your scenario, the dealer is removed completely, his entire hand shown as "zero."
Players A and B are left with 50/50 chance. Doesn't matter if they switch or not.
Or, as the equation would show:
[a+b+(c+d)]=100
We can agree that A and B are equal because it's the probably of choosing the $100. So... 25.
[25+25+(25+25)]=100
However, we're shown that C and D are both "zero."
[25+25+(0)]=100
But that doesn't add up to 100. However, refer back to the fact that A and B are equal.
50+50=100.
N Sperlo wrote:Knurled wrote:That us what I'm expressing.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.No. You chose one of four. Your choice is *still* one of four. It's only a 50/50 chance if two empty envelopes are removed *before* you choose. This, I think, is where everyone's getting hung up.
The envelopes are removed AFTER you choose, though. That doesn't match what you're expressing.
N Sperlo wrote: When the two are shown to be zero, they must be removed from the equation along with their 25. You are left with a total of fifty.
Yes and no.
They're removed, but the expression (b+c+d) is still equal to 75, so you must keep it that way.
I'll ask again: Would you rather choose one envelope and take your chances? Or would you rather choose three envelopes and take those chances?
Don't know if you're actively choosing to not answer the question or if you're just missing it. I'll assume you're actively choosing to not answer it if you don't this time. 
N Sperlo wrote: In reply to 92CelicaHalfTrac: (a+ (b+y+z))=100 (25+(25+0+0)=50 Just because you made two equal zero doesn't mean you changed the value of one. For all you know, it is equal to the first. 100 must stay common, so a and b must reflect three change as 50. (50+(50+0+0))=100
The order of operations says you done messed up.
"a" is KNOWN to be 25.
The SUM of "b," "y," and "z" is KNOWN to be 75. If you don't solve the equation (b+y+z) first, you are doing it wrong. Cold hard fact of math. Until that expression is taken care of, you do not, under any circumstances, touch "a."
What you're doing is saying that after the very first step of the scenario when you have one envelope, and the dealer has three, with none of them shown as being empty.... you SOMEHOW know that you both have a 50/50 chance. And that certainly doesn't make any sense. How does your one envelope have a 50% chance, and his three have a 50% chance?
92CelicaHalfTrac wrote:N Sperlo wrote:The envelopes are removed AFTER you choose, though. That doesn't match what you're expressing.Knurled wrote:That us what I'm expressing.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.No. You chose one of four. Your choice is *still* one of four. It's only a 50/50 chance if two empty envelopes are removed *before* you choose. This, I think, is where everyone's getting hung up.
Visuals always help, right?
Envelope a
Envelope b
Envelope c
Envelope d
I choose envelope b.
Envelope a
Envelope c
Envelope d
Now lets say a and
Holy crap, the 25/75 people are right. The columns are "your pick, the money was in, stay wins, switch wins" at the bottom, the total in "n, stay wins, switch wins":
A A 1 0
B A 0 1
C A 0 1
D A 0 1
A B 0 1
B B 1 0
C B 0 1
D B 0 1
A C 0 1
B C 0 1
C C 1 0
D C 0 1
A D 0 1
B D 0 1
C D 0 1
D D 1 0
16 4 12
92CelicaHalfTrac wrote:N Sperlo wrote:The envelopes are removed AFTER you choose, though. That doesn't match what you're expressing.Knurled wrote:That us what I'm expressing.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.No. You chose one of four. Your choice is *still* one of four. It's only a 50/50 chance if two empty envelopes are removed *before* you choose. This, I think, is where everyone's getting hung up.
Visuals always help, right?
Envelope a
Envelope b
Envelope c
Envelope d
I choose envelope b.
Envelope a
Envelope c
Envelope d
Now lets say envelopes a and d are empty. Mr Matlock tells us. That basically means we can throw them out of the equation. Lets not though. They have a null chance. Note you have two envelopes left which may have 100 dollars in it. Please explain how the null value of the two envelope effects only one envelope that may contain the money.
N Sperlo wrote: Mr Matlock tells us.
Not prior to you making your random selection of 1 in 4, thus the two removed are not null, thus giving you 25/75 odds. With two states (money, no money) and one occurrence (money positive), logic tells you that the other three have no money, whether they're removed or you're told that one, two, or three have no money. As long as you don't know which is which, your odds are 1 in 4. This does not change unless the selection is randomized again.
So all that needed to be stated was that those two were not specified.
chuckles wrote: You are shown four, identical envelopes on a table. You are told that one contains a $100 bill, the rest are empty. You are told to choose one and pick it up. Then, two of the three envelopes still on the table are removed and you are told that they were empty. Now, do you keep the one in your hand, exchange it for the one still on the table or does it not make any difference?
because the original posts suggest specification.
N Sperlo wrote: Now lets say envelopes a and d are empty. Mr Matlock tells us. That basically means we can throw them out of the equation. Lets not though. They have a null chance. Note you have two envelopes left which may have 100 dollars in it. Please explain how the null value of the two envelope effects only one envelope that may contain the money.
Do you agree that on the outset, the chances of having the correct envelope are 25%?
Then in that case, the chances the correct envelope are on the table is 75%. Right?
Someone opens up two empty envelopes on the table. You see that they are empty. Why does this change the chances of the correct envelope being on the table?
The only thing that it has changed is bringing the chances of those two envelopes down to zero; you have more knowledge about the envelopes. The chance that the correct envelope is on the table is still 75%.
You are looking at this as random. The ONLY random part of this is the original choice. The removal (or opening) of two envelopes wasn't random, so the next choice (switch or stay) isn't random either.
N Sperlo wrote: because the original posts suggest specification.
Not really. They are specified only in that one has money, three do not. Logic tells you that at least two on the table do not, so they could be removed, opened, or left on the table. That doesn't change your odds. Because you don't know which is which at the time of selection, your odds are 1 in 4. Now if the two remaining envelopes were collected and shuffled and then you selected again, only then would your odds be 50/50.
In reply to mtn:
I'll take a 50% chance over 75% odds.
Alright, i'm out of this one for now. Ignoring a key question multiple times is refusal of proof.
Your mind won't be changed no matter how many times math says what's going on.
I'm cool with that. No matter how many times people tell me otherwise, P71s still suck, too. 
wrote: Apparently Ron White was correct. the odds are 50/50. Period. Anything else is a foolish misrepresentation. And frankly, if you really believe otherwise, I've lost a bit of respect for you. the odds are 50/50. But if it makes you feel better, I don't mind if you claim the win. Go for it. Won't impact me (or reality) a bit.
Irony
92CelicaHalfTrac wrote: P71s still suck, too.
Blasphemy.
mtn wrote:Apis_Mellifera wrote: The two removed aren't actually irrelevant because at the time of your selection they were viable choices. Each envelope has a 1 in 4 chance of containing the money. If you stay with your original selection, your odds are 1 in 4 of having the money. Because you have the opportunity of making a second selection from two choices, the notion is that you now have a 1 in 2 chance. This would be true if the two envelopes were collected and shuffled and the choice was now random. However, the envelopes were not shuffled. Two wrong choices were just removed from the three you didn't choose (which logic dictates to be a known whether they're removed or not). After the initial selection the choices are no longer random thus the probability does not change from 25/75. I think that's were the 50/50 folks are getting tripped up (whether they realize it not).Bingo. It isn't random, and since somebody is "in on it" the chances are not 50/50.
not sure how "somebody being in on it " changes anything... yes everyone is correct it's 75/25 when picking one out of four.... and then if you were asked if you wanted to swap the one you picked with any of the other three then there would be some thinking to do .... odds wise.... but yes two of them are irrelevant ... because they might as well not even be there.... you really are picking one envelope out of two, because the other two are going away...
if you, at 25%, actually had picked the envelope with the money then the "person in on it" would pick any two and discard them, if you had , at 75%, picked one of the empty envelopes, then the "person in on it" would toss the two remaining empty envelopes ....
so now you are right back where you started with a 50/50 chance of having the $$ just like at the beginning, even though there were four to start with... you already knew you didn't have to worry about two of them.... so not only are two of the envelopes irrelevant but to some degree so was your first pick...
now with the only two that matter you have to make your choice .... 50/50
If you still incorrectly believe the answer is 50/50, then I suggest you google Conditional Probability. This problem is an example of conditional probability.
wbjones:
Nderwater wrote: 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.
it's the same scenario with the envelopes.
wbjones wrote:mtn wrote:not sure how "somebody being in on it " changes anything.Apis_Mellifera wrote: The two removed aren't actually irrelevant because at the time of your selection they were viable choices. Each envelope has a 1 in 4 chance of containing the money. If you stay with your original selection, your odds are 1 in 4 of having the money. Because you have the opportunity of making a second selection from two choices, the notion is that you now have a 1 in 2 chance. This would be true if the two envelopes were collected and shuffled and the choice was now random. However, the envelopes were not shuffled. Two wrong choices were just removed from the three you didn't choose (which logic dictates to be a known whether they're removed or not). After the initial selection the choices are no longer random thus the probability does not change from 25/75. I think that's were the 50/50 folks are getting tripped up (whether they realize it not).Bingo. It isn't random, and since somebody is "in on it" the chances are not 50/50.
I meant that someone knows which ones are empty, and removes two of them.
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