1. If we choose between two envelopes x and 2x, or x and 1/2, switching doesnt matter but
2. if we have a envelope with money and get a offer to switch it for a envelope with 2x or 1/2 we should switch
im not sure witch off the two scenarioes you described in your initial post

1. What? It clearly does matter. I think you mean if we choose between 2X, X, and 0, then it's =EV. But that's not what the problem is saying.

2. Exactly. Again, isn't this exactly what you're saying in 1 except the conclusions are different?

You have the $5000 envelope. The other is either $2500 or $10k. If those two possibilities are equally likely, the EV of switching is $12500/2 = $6250. That's more than $5000 so unless you're extremely risk averse you should switch.

The question is, are $2500 and $10k equally likely? If this was a real world situation, I'd be trying to estimate some kind of probability distribution and it might not be evenly balanced between the two possibilities. I'd want to know the net worth of my benefactor. There's only so much money anybody could/would put in an envelope.

So for example, I'm dealing with someone whose net worth is $10 million. I'm pretty sure he's not going to give away any more than $1 million. To my surprise, I open an envelope with $1 million. In that case I'm not switching, because no way does the other envelope contain $2 million. If I thought the probability of $2 million was equal to the probability of $500k I'd do the switch, but it isn't in the situation I described.

And another thing. I was handed the envelope. It wasn't chosen at random, so now the intentions of the person who handed it to me are going to muddy the waters. Maybe this person is trying to direct me to switch to the lower value, expects me to do the math and switch, and therefore deliberately hands me the higher value envelope of the two. Or maybe the reverse, whatever, but the intentionality of the envelope selector adds another dimension (which maybe wasn't intended in the statement of the problem).

Also while I'm nitting it up, do I know in advance of the envelope opening that I'll be offered a choice to switch? If the choice of switching is only offered after the amount is revealed, I'm getting suspicious again (maybe the switch is only offered if I picked the higher amount). Could this problem be stated this way instead: 1. The person who filled the envelopes instructs me that I am to pick one of the envelopes and open it, and that after seeing the amount inside I will be allowed to switch to the other envelope, if I wish. 2. I choose one of the two envelopes at random, it contains $5000.

Sorry; hasty copy paste. Being handed an envelope is still completely random. And you'll always be offered to switch without exceptions.

Say you have 2 envelopes filled with money. The value of one envelope is half the value of the other.

1 So the envelopes are either 5k and 10k or 2.5k and 5k
we open 5k switch to 10k
we open 10k switch to 5k

we open 5k switch to 2.5k
we open 2.5k switch to 5k
switching doesnt matter

2. if we have a envelope with 5k and are offered to choose between two envelopes 1 containing 2.5k and 1 with 10 we should switch.

3. if we have a envelope with 5k and are offered to choose another envelope that might hold 2.5k or 10k now we need to figure out how likely the new envelope is to hold 2.5k or 10k

Bs mistake is thinking this is this is scenario 2 when it is scenario 1.

In the first case you choose an envelope at random. You do not open it. Your logic tells you there's no reason to switch.

In the second case you choose an envelope at random. You open it and find $5k. Your logic tells you that you should switch. In fact, the amount in the envelope is irrelevant; you'd arrive at the same conclusion (to switch) for any amount in the first envelope.

Therefore, knowing you should always switch, you don't even need to open the envelope in Scenario #2. Which puts you back in scenario #1, where there's no reason to switch.

I don't know if I can give the mathematical proof or whatever to solve that paradox, but the flaw in Person B's logic is like this:

I have an envelope worth X.
the other envelope either has 1/2X or 2X, with equal probability. Therefore my EV is 1/2(1/2)X + 1/2(2)X = 5/4x
So I should switch to the other envelope.
Now I have an envelop worth Y.
the other envelope either has 1/2Y or 2Y, with equal probability....
so on and so forth

i think the problem is intended to read like there's an auxiliary tank that each plane has, so the answer becomes only 3. though if not, this seems right

No auxiliary tank. Bostaevski's solution is correct.

Yeah, I figured that was what you meant after saying that, or that the puzzler was being an a-hole and suggesting a kamakazi mission.

I got the scale one right, right?

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Supposedly, this problem actually happened way back when. Assume you have all current knowlege to your disposal, except you do not know anything of non-Euclidean geometry, nor do you have access to electricity. I am only doing this because I don't want to confuse the issue and force you to take everything in historical knowlege on top.

A kind orders a crown from the best smith in the region. The king sends out 5lbs of gold, and recieves an exquisite crown. You are his resident genius, and he asks you if there is any way that you can confirm that the crown is 100% gold, for the smith may have pocketed some of the gold and created an alloy.

You are not allowed to smash the crown. What is your solution?

Yes, you got the scale one right.

As for the problem: "Eureka!"

If they had auxiliary tanks, I think the answer is 2.

2 planes leave at the same time in the same or opposite direction. They each get 1/2 way around the planet, refuel each other, then head home.

Provided planes cannot refuel themselves, in which case 1 plane could do it.

I see what you did there

"Eureka!" is, in fact, the correct answer.

Thanks for not spoiling the real answer.

Wtf_Fml, go here: http://en.wikipedia.org/wiki/Monty_Hall_problem. This explains it pretty well.

The main requirement is that the host HAS to open a door without a car behind it.

You have two shot glasses. One is filled to the brim with Whiskey, and the other one is filled with water just below the rim. Without using another container, you must find out how to transfer the water to the shot glass with whiskey in it, and you must transfer the whiskey into the glass with the water in it.

You are not allowed to use any sort of container that is able to hold liquid, so, rolling up a dollar bill into a funnel is not the correct answer. You are allowed to use anything else. As a quick hint, straws wouldn't work in this situation. You are not allowed to spill a single drop.

an old bar trick solution, dunno if it counts. EvilSteve's seems pretty legit. Except I have weird shaped shot glasses.

I guess evil steve's works as well. I was also thinking of another solution.



There are these logic puzzles I really used to like, but I can't remember how to set them up. I am looking for them, but...

the way most people finally get the monty hall problem is this way:

imagine its not three doors, its 100 doors. You pick door number one, and monty shows you 98 doors with no prize behind them, leaving you two choices... stay with door #1 or switch to door #x (the only one besides #1 that he hasnt shown you)

This in all likelihood is incorrect, but I gave it a shot

Bar tricks like the whiskey/water shot glass one are really awesome.

Also look up stacking an inverted empty bottle on top of another empty bottle and try to get a dollar bill out from between the two.

Also getting a dollar out from beneath an inverted bottle.

They seem really simple after you see them, but perhaps not so intuitive before.

I was going to post the famous candle and matchbox problem, but rereading wiki, it's actually too simple unless you're physically presented with the materials and the problem.

Eh. It can't hurt to post it anyway.

You're given a candle, a box of matches, and a couple tacks. You're in a room with nothing but a corkboard on the wall. How do you pin the candle to the corkboard using only the materials provided so that you can both light the candle and no wax drips onto the floor?