Head Exploded 2

Apparently Ron White was correct.

I don't know if some of you are just clowning around, or if you really believe this cockamaimy crap. Don't really care. I can't fix stupid.

I agree that there are several semantic issues in the original statement, but the OP asserted this was a math problem and did not have anything to do with the grammatical shortcomings.

Given that, 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.

At the end, there are 2 envelopes. 1 is guaranteed to have the money. The other is guaranteed to be empty. Unless you are using additional imaginary envelopes, 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.

take all the envelopes and pistol whip the guy for berkeleying w/ your mind... done

Holy E36 M3! Is this still being argued? Its funny, the things we let bug us.

SVreX wrote: Given that, 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.

Just for fun, you should really do the bet that you proposed at some point with friends. You really will be surprised.

Not quite sure why you have lost respect for folks though.

In reply to SVreX:

Okay, let's try the "100 Doors" example from earlier. I am thinking of a number between 1 and 100. I promise that I will write down and not change what number it is.

Pick a number.

After you pick I will remove all other numbers except for the one you picked and one other number. It will be one of those two numbers. You can then choose to stay with your first guess or switch.

Salanis wrote: In reply to SVreX: Okay, let's try the "100 Doors" example from earlier. I am thinking of a number between 1 and 100. I promise that I will write down and not change what number it is. Pick a number. After you pick I will remove all other numbers except for the one you picked and one other number. It will be one of those two numbers. You can then choose to stay with your first guess or switch.

This really is the best way to describe it.

Now, do the same thing for 99 numbers. Same thing, only now instead of 1/100 and 99/100, the chances are 1/99 and 98/99. This holds true all the way down to 1/4 and 3/4, and 1/3 and 2/3.

one more way to think about this... in the original post there were 4 envelopes, one of which has the money the other 3 are empty... two of the empty ones are removed .... now think of it this way... instead of 4 envelopes there are only 2... this is what there is even at the start ( since 2 empty ones are going to be removed ) ..... since the original problem was essentially a selection between 2 envelopes not 4 then your odds would have to be 50/50

your first choice either had the $$ or the one left on the table has the $$ ... so all the statistics/probability/possibility boils down to which of the 2 envelopes has the $$

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.

True. This is a variation of the Monty Hall Problem

I'm still going with switch.

You have two samples at the first move.

The three on the table, and the one in your hand.

The three on the table represent 75%. That being that the chances that the $100 are in ONE of those three is 75%.

The one in your hand is 25%. That being that the chances that the $100 is in your hand is 25%.

Removing two guaranteed to be empties from the sample representing 75% means that the ONE left on the table still has the original 75% chance.

I'll switch every time, thanks.

Where you're all getting the 50/50 split is by ignoring the events that happened to get you to only two envelopes left. If someone came up to me and said "hey man, here's two envelopes, one has money one doesn't," i'd call that a 50% chance.

This isn't about 2 envelopes.

I don't see how the action of you picking one makes the other one "more" likely to have the money in it. Just because an envelope is in a group of envelopes, some of which have been shown to not contain the money, doesn't mean diddly about the contents of the remaining envelope in relation to the one you picked.

You either have the one with the money, or you don't. It's 50/50. Switch if you want to, but it matters not.

EastCoastMojo wrote: You either have the one with the money, or you don't. It's 50/50. Switch if you want to, but it matters not.

This is true for the remaining envelopes, but when you first picked there were three, meaning you only had a 33% chance of picking correctly, meaning that you probably picked wrong to begin with.

jg

Not to harp on, but the goat website has even run the stats:

"players who STAY have won 1511 cars out of 4547 games yielding a winning percentage of 33%; players who SWITCH have won 3705 cars out of 5604 games yielding a winning percentage of 66%"

If you needed the 100 bucks to buy the antidote for the poison you've just been injected with [you're on one of those wacky Japanese game shows], you are statistically better off switching envelopes.

Luke wrote: Not to harp on, but the goat website has even run the stats: "players who STAY have won 1511 cars out of 4547 games yielding a winning percentage of 33%; players who SWITCH have won 3705 cars out of 5604 games yielding a winning percentage of 66%" If you needed the 100 bucks to buy the antidote for the poison you've just been injected with [you're on one of those wacky Japanese game shows], you are statistically better off switching envelopes.

Yes they ran the numbers...

based on a flash game they made to prove their point. Not really fact.

EastCoastMojo wrote: I don't see how the action of you picking one makes the other one "more" likely to have the money in it. Just because an envelope is in a group of envelopes, some of which have been shown to not contain the money, doesn't mean diddly about the contents of the remaining envelope in relation to the one you picked. You either have the one with the money, or you don't. It's 50/50. Switch if you want to, but it matters not.

My offer is still open to anyone to do the 100 doors example.

If you are correct, when I remove 98 doors that are not the correct answer, you will have a 50/50 chance of being correct.

Fascinating!

wbjones wrote: one more way to think about this... in the original post there were 4 envelopes, one of which has the money the other 3 are empty... two of the empty ones are removed .... now think of it this way... instead of 4 envelopes there are only 2... this is what there is even at the start ( since 2 empty ones are going to be removed ) ..... since the original problem was essentially a selection between 2 envelopes not 4 then your odds would have to be 50/50 your first choice either had the $$ or the one left on the table has the $$ ... so all the statistics/probability/possibility boils down to which of the 2 envelopes has the $$

I have read all the posts so far... can anyone explain why my analysis is wrong... especially the last paragraph

by the way I do understand the 25/75% with the money being in one of the four... but actually it's only in one of the two since 2 ( the ones that are thrown out )are irrelevant

you either picked the correct envelope the first time or it's the one left on the table after the other 2 are removed

What if we take the "picking" part out of the equation. You are shown four envelopes on the table and told one of them has $100 inside. Then two are taken away and you are told they are empty. What are the odds that the $100 is in the one on the left?

Please just show me how it can possibly be anything other than 50%

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).

EastCoastMojo wrote: What if we take the "picking" part out of the equation. You are shown four envelopes on the table and told one of them has $100 inside. Then two are taken away and you are told they are empty. What are the odds that the $100 is in the one on the left? Please just show me how it can possibly be anything other than 50%

In that scenario the probability is 50/50 because you are not making a selection from four options. You are choosing from two.

For all you 50/50 folks, look at it this way:

Instead of removing the 2 empty envelopes, you just put all 3 envelopes in 1 big envelope. Now which envelope do you pick? The envelope in your hand, or the envelope on the table with 3 small envelopes in it?

And before you tell me it is a different problem, tell me how? You still have 2 envelopes to pick from for the final selection, and 2 empty envelopes have been removed.

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.

Switch, every time.

If someone shows me 4 envelopes, 3 empty and 1 with $100. I can either open three envelopes or just one. You'd be stupid not to open 3 - and that's basically what's happening here.

N Sperlo wrote: Holy E36 M3! Is this still being argued? Its funny, the things we let bug us.

Extreme: You buy a lottery ticket. Someone walks up to you and has one of the other 500 million tickets in HIS hand and says, with guaranteed authority, one of those two tickets will be the winner. Which one is more likely to be the winner?

And this works all the way down to 3 choices.

mtn wrote: 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've followed the logic from the beginning and have added to the thread only for the sake of debate ;)

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?

I still stand by 2