That Monty Hall problem has always bothered me.

i thought it was clear if you swap then you keep the amount of money you swapped for and the point of having a set amount is that you always want to swap

This isn't the Monty Hall problem.

In the Monty Hall problem there are 3 doors. Two have no prizes, one has some valuable prize. You choose one, and the host removes an empty door from the two unchosen ones. You then have the option to swap, and you should always swap.

I understood it pretty quickly after reading about it and thinking it through, but apparently lots of people can't fathom it at all.

The one in this thread is simpler, because even though the chance of gaining or losing is 50-50, the potential loss is $25 but the potential gain is $50. It's more like a psychological test: if you can't get it, you just aren't much of a risk-taker.

Even though it works out logically, the answer is so unintuitive.

nich's problem was the monty hall problem, not mine

:( herp

For the Robertson Hotel doesn’t merely have hundreds of rooms — it has an infinite number of them. Whenever a new guest arrives, the manager shifts the occupant of room 1 to room 2, room 2 to room 3, and so on. That frees up room 1 for the newcomer, and accommodates everyone else as well (though inconveniencing them by the move).

Now suppose infinitely many new guests arrive, sweaty and short-tempered. No problem.

How does he do it?

Later that night, an endless convoy of buses rumbles up to reception. There are infinitely many buses, and worse still, each one is loaded with an infinity of crabby people demanding that the hotel live up to its motto, “There’s always room at the Hilbert Hotel.”

The manager has faced this challenge before and takes it in stride.

First he does the doubling trick. That reassigns the current guests to the even-numbered rooms and clears out all the odd-numbered ones — a good start, because he now has an infinite number of rooms available.

But is that enough?

Note: learend this in a cantor lecture, you can look it up on google but don't ruin the fun

For the Robertson Hotel [...]

demanding that the hotel live up to its motto, “There’s always room at the Hilbert Hotel.”

Robertson should rethink their motto, it will drive away business.

The Monty Hall Problem - you should swap, its intuitive and easy to prove.

I have no idea x how you can say that the two envelope problem is easier. Surely you are not suggesting that it is mathematically provable that you should swap. Swapping is just the same as picking the other one first a sticking. It is for sure of no benefit to swap, the difficulty is in proving it. Very tricky.

http://www.maa.org/devlin/devlin_0708_04.html

A nice explanation.

can't belive i got duped, i've even read that before

:( double-herp

For the envelope problem, I think that it was easier than the Monty Hall Problem. Considering that one envelope is double the other and one of them contains $50, then switching works to your benefit. Looking at it one way, you can either earn fifty more dollars or lose twenty-five. Another way to look at it is that you have twenty-five dollars and have to choose between getting nothing more, twenty-five more dollars, or seventy-five more dollars. The only way it would be better to stay is if there is nothing in an envelop, which results in a 50-50 split, or if you must pay to open the envelop, which would be an overall net loss.

SGV I assure you that there is no advantage in swapping. No disadvantage either but definitely no advantage. It is not like the Monty Hall problem. The great thing about this problem is that it will only trick people who are quite good at maths and logical. Someone who doesn't get the Monty Hall problem would probably stick but for the wrong reason.

Please read the explanation I posted. Or better this one that I just googled. http://works.bepress.com/cgi/viewcontent.cgi?article=1006&context=raam_gokhale

This is your reasoning to swap in mathematical form. It is independent of the amount found in the first envelope (A).

0.5*2A + 0.5*0.5A = 1.25A (i.e it looks like you get a greater expected return)

or with 50 inserted instead of A as someone above stated.

0.5*(2*50) + 0.5*(0.5*50) = 62.5

It doesn't matter the value of A. Using this reasoning you always get a greater expected return by swapping. Therefore if you didn't look in the envelope it is still worth swapping and if you are then asked again, using the same reasoning you would swap again and again ad infinitum. This can't be right so there must be a flaw in the above calculation.

Its all about the fact that you have already chosen either the bigger or smaller envelope which already have the money in. If you have already chosen the bigger one you cant expect it to get bigger by swapping. The above expected formulae calculation is therefore wrong and the two instances must be separated. Your expected return is decided by which envelope you chose first.

If you chose the bigger one, X first your expected return by swapping is 0.5 X.
If you chose the smaller one Y first you expected return by swapping is 2Y.

There are lots of much cleverer and probably clearer mathematical explanations out there. Perhaps someone could explain this more simply. I am interested to see.

Finding Sir F's explanation rather unsatisfactory, let me give it a try from another angle. (No offense intended, Sir F, just saying that your explanation didn't make it "click" for me.)

Before, we called the amount shown A. However, this didn't lead to tangible results, because (1) 50% chance A is the lower amount and (2) The higher amount is 2 times the lower amount did not lead us to be able to say what the size of the higher amount is. In the attempted solution, we concluded that the higher amount would be 2 * A, but this is mistaken because the amounts were set before the exercise and knowing their relation in combination with whether we guess the amount we have seen is the higher or the lower one changes nothing about that.

Instead, we'll call the lower amount B.

There's a 50% chance the amount we have seen is the higher amount. In that case, the expected value when not changing is 2*B and the expected value when changing is B. There's also a 50% chance that the amount we have seen is the lower amount. In that case the expected value when not changing is B and when changing is 2*B.

The total expected value when not changing is 0.5 * 2 * B + 0.5 * B = .75 * B and the total expected value when changing 0.5 * B + 0.5 * 2 * B = .75 * B. As you see, both are equally high so it does not matter at all whether you switch or not.

---

As for the original question of a proof for the lack of fours, let me give that a try as well.

Row r is the first row that contains a number of 4 or higher number and is not row 1.
Row 1 does not contain a 4 or higher number, thus r > 1
Row r - 1 must contain either a 4 or higher or a sequence of 4 or more times the same number.
The only row before r that can have a 4 or higher number is row 1
Row 1 does not contain a 4 or higher number
As such, row r - 1 must have a sequence of four or more times the same number.

Assume there is a row with four or more times the same number and call this row s
Row 1 does not have a sequence four or more times the same number, thus s > 2
Row s - 1 must have two subsequent sequences sequences of the same number (because the longest sequence of number x you can have without having (n times the number X twice) is three.
However, whenever that happens it is regarded as a single sequence instead
Because of this, no row s exists

Row r-1 is (a) row s, so row r - 1 doesn't exist
This means row r does not exist.
This means there are no rows with the number 4 or higher


This sequence must have one of those configurations:
- n times number a, x number of times (a times the number a), a times number m
-> This will result in the sequence (n + a * x, a, a, m), rather than four or more times a
- (x + 1) number of times (a times number a)
-> This will result in the sequence (a * (x + 1), a) rather than four or more times a
- n times number a, x number of times (a times the number a)

Q.E.D.

Note: Skip that last part (from "This sequence must have..." to "n times number a, x number of times (a times the number a)". It was an earlier attempt at the proof I abandoned (not the least of all because it was unclear) and the rest should be read without it to form my proof)

Nice one Jasper. That is the sort of thing I was looking for.

I have a problem for you all :)

A box of chocolates contains white, milk and plain varieties. The number of plain chocolates is at least half the number of white and at most half the number of milk. Given the total number of plain and white chocolates is over 55, what is the minimum number of milk chocolates in the box?

Please don't just post the answer, I'd like to see how you got your answer.

Okay, let's try that one out.
Here's the given data (W=white, M=Milk, P=plain)
P>=1/2W
P<=1/2M
P+W>55

Since we are looking for the lowest value of M, we start with the lowest P possible, which is 19 (making W 37, thus P+W=19+37=56).
Then the lowest number of M is 38, which is 2×P.
Did I get it right?

What the hell is plain chocolate?

I don't know it either. Maybe he means dark chocolate?