Head Exploded 2

alfadriver wrote:

mtn wrote:

alfadriver 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 3 in 4 chance of winning.

Except that when the two envelopes were removed and you were told that there was no money, the chances that your envelope was right immediately changed from 25% to 50%. You have a 1 in 2 chance whether you exchange or not. Where the odds get interesting is if the envelopes all have different values in them- which is a game show called "lets make a deal". Then the odds are very fluid based on the known values or not. Those odds are also calculated with the value on the table.

Common sense says that you are right. When this test is repeated many times in real life, the percentages come out to 25% and 75%.

The problem is that once the two envelopes are removed, not having money, the odds change for all envelopes. Those two removed become 0/4, therefore both of the remaining two are 1/2. It's true that the odds change for the remaining envelope, but one has to remember that the odds change for your's too.

Repeat the problem enough times, the percentages work out to 25% and 75%.

alfadriver wrote: The problem is that once the two envelopes are removed, not having money, the odds change for all envelopes. Those two removed become 0/4, therefore both of the remaining two are 1/2. It's true that the odds change for the remaining envelope, but one has to remember that the odds change for your's too. The other game show is called "deal or no deal". It's all about changing values to choices.

Look at it this way: If you opt to keep the envelope in your hand, you're betting that you picked the right one the first time. Odds of that would have been 1 in 4.

If you switch, you're betting you picked the WRONG one the first time, because if you did pick wrong, you're certain to get the right one when you switch. Odds of that are 3 in 4.

So, definitely, switch.

One wrinkle the Monty Hall game show that the math versions leave out was that Monty didn't have to open the door - he only opened it if he felt like it. So you'd also have to figure the odds that Monty was trying to sucker you into switching when you had the right item.

mndsm wrote:

N Sperlo wrote:

tuna55 wrote:

nderwater wrote: No, he's right - according to statistical reasoning. But by the rules of life, it makes no difference. You're never going to win the $100. You were not the 15th caller. No Nigerian official really wants you to stash his millions.

Well, I don't see how, unless by "he" you meant "Alfa".

Would you rather pick up the beer glass that is ¼ full or ½ full?

Doesn't matter. In either case, it's clearly time for more beer.

Yes it is!
And more couldn't hurt in figuring this out either.

mtn wrote: Common sense says that you are right. When this test is repeated many times in real life, the percentages come out to 25% and 75%.

Seems like that would depend on the actual procedure. Does the person removing the envelopes know which are empty and which aren't, and therefore always pick two that are empty? Or do they pick randomly, and just tell you if the two removed are empty or not?

If they know which are empty and always remove an empty 2, that is not random and does not change the probability for the first card you picked to be correct. Therefore, you should pick the remaining card, because there is still a 75% chance that it is correct.

If they just draw two randomly and tell you if those are empty, that changes the probability. There is a 50% chance that they will pick the card with the money in it. If they do not, that means that there is an equal probability of it being in either remaining envelope, so it doesn't matter which you pick.

nderwater wrote: that 'past occurences do not affect current probabilities'. At the end of the day you're stuck with two choices - 50/50.

This

tuna55 wrote:

nderwater wrote: that 'past occurences do not affect current probabilities'. At the end of the day you're stuck with two choices - 50/50.

This

But only if the other two removed are removed randomly and just happen to not have the money.

Build a tree diagram, you'll see how it works. It's a lot easier to understand than figuring probabilities. That's especially true for the Monte Hall version.

nderwater wrote: Blowing this one out a little... Let's say that there were a million envelopes to start with and you took away 999,998 and showed they were empty. By the statistical model, if you switched envelopes you'd have a 99.9998% chance of getting the $100.

ReverendDexter wrote: That actually is exactly right. The odds would be the same only if you randomly selected two more envelopes... but you don't - you specifically throw-out ONLY non-winning options.

The issue with this (Salanis alluded to it as I was typing) is that the only way to ensure that the $100 always lands in one of the two remaining envelopes is to ignore all the permutations where the envelopes (2 in the original scenario; 999,998 in mine) removed from the table contain the $100 (which would have been a 50% chance originally, 99.9998% chance in mine).

So the thought experiment is rigged - the envelope removal cannot be random, which eliminates the statistical chances that the first selection was based on. Because there is ALWAYS $100 in one of the final two envelopes, then I stand by statement that the chances of holding the envelope with the $100 is 50/50.

mtn wrote: Repeat the problem enough times, the percentages work out to 25% and 75%.

If you change the odds of the remaining envelope, tell me why the odds of the picked envelope does not change as well?

The first choice I make is 1 of 4.

The second choice I make is 1 of 2, and it's irrelevant that I made a previous choice.

Here- change what happens- the person tells you to menally take an envelope, but don't change anything. And then they take two that do not have the $100 in it, then leaving you a choice between the remaining 2. Would it matter if you change or not? It's the same sequence.

Non opponent gambling (aka, not head to head poker) is the same thing- but when it's one deck of 52 or 6 decks of 52, the odds change differently based on what has been played.

nderwater wrote: The issue with the original question (Salanis alluded to it as I was typing) is that the only way to ensure that the $100 always lands in one of the two remaining envelopes is to ignore all the permutations where the envelopes (2 in the original scenario; 999,998 in mine) removed from the table contain the $100. So the envelope removal cannot be random, the thought experiment is rigged, and you've eliminated the statistical chances that the first selection was based on. Because there is ALWAYS $100 in one of the final two envelopes, then I stand by statement that the chances of holding the envelope with the $100 is 50/50.

If the two were removed first, you have a 50/50 chance.

If you pick before any are removed, you have a 1/4 chance of being correct. There is a 3/4 chance that the $$$ is in one of the other three envelopes. When two are removed, there is still a 3/4 chance that the $$$ remains in the other set. The only difference is that the remaining set is now a set of 1 card with a 3/4 chance of having the money.

alfadriver wrote: Here- change what happens- the person tells you to menally take an envelope, but don't change anything. And then they take two that do not have the $100 in it, then leaving you a choice between the remaining 2. Would it matter if you change or not? It's the same sequence.

That is very different. If you pick the envelope mentally, there is a 50% chance that the envelope you picked will be removed.

Salanis wrote: If you pick before any are removed, you have a 1/4 chance of being correct. There is a 3/4 chance that the $$$ is in one of the other three envelopes. When two are removed, there is still a 3/4 chance that the $$$ remains in the other set. The only difference is that the remaining set is now a set of 1 card with a 3/4 chance of having the money.

That's what I was saying, just with less cake.

Salanis wrote:

alfadriver wrote: Here- change what happens- the person tells you to menally take an envelope, but don't change anything. And then they take two that do not have the $100 in it, then leaving you a choice between the remaining 2. Would it matter if you change or not? It's the same sequence.

That is very different. If you pick the envelope mentally, there is a 50% chance that the envelope you picked will be removed.

It may be different, and yes, there's a 50% chance that your mental envelope will be removed.

But you are still left with a choice between two objects- one in your had, and one on the table, or two on the table.

I still don't see how removing two envelopes changes the odds of the envelope on the table, and not the one in your hand, being that you are given a choice at a second point.

If it's just to reveal, and you are NOT given a choice, your envelope keeps it's 25% chance the whole time, since that's what it had when you picked it. But since you are given a chance, how you chose prior has no effect on the 50/50 chance of the final choice. that's the part they play with you on game shows.

You cannot stack odds. When you make the first choice, you are given 1:4. Now they remove two that do not have money. Thus, there is a choice between two envelopes, one of which has money, 1:2.

I did the same thing...struggled with it and overthought it. But it's simple. It doesn't matter whether the guy who set it up "knew" which ones to remove or removed two empty ones by chance. The odds don't change. If you picked up one and then were told you could exchange it for all three left on the table, it would be the same thing. It's trading a 25% chance for a 75% chance.

chuckles wrote: It doesn't matter whether the guy who set it up "knew" which ones to remove or removed two empty ones by chance. The odds don't change. If you picked up one and then were told you could exchange it for all three left on the table, it would be the same thing. It's trading a 25% chance for a 75% chance.

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?

chuckles wrote: The odds don't change. If you picked up one and then were told you could exchange it for all three left on the table, it would be the same thing. It's trading a 25% chance for a 75% chance.

That is probably the simplest way this could be explained.

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.

Statistically speaking, statisticians don't succeed at business. Maybe because they think there is a 75% chance of success when there is only a 25% chance.

Logic isn't a statistician's strong point.

It would be more fitting if you put that money into one of four envelopes, and then put it on the table.

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.

Salanis wrote:

mtn wrote: Common sense says that you are right. When this test is repeated many times in real life, the percentages come out to 25% and 75%.

Seems like that would depend on the actual procedure. Does the person removing the envelopes know which are empty and which aren't, and therefore always pick two that are empty? Or do they pick randomly, and just tell you if the two removed are empty or not?

The question/problem states that they remove two empty ones on the table after you have made your (first) pick.

You can't subtract the total reduced quantity from one side of the equation without subtracting it from the other.

Bad math.

SVreX wrote: You can't subtract the total reduced quantity from one side of the equation without subtracting it from the other. Bad math.

Right- the odds change every time the amount of choices are changed. First it's a one in 4 chance of winning, final is a one in 2 chance of winning.

Situation changes, odds change.

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.