Math Fun

Sounds valid to me.

OK. I am satisfied that you are forced to accept that those are valid proofs in order to maintain that your solution is also valid.

So I take it that you didn’t get the whole, “I AM NOT TALKING ABOUT A SOLUTION, BUT MERELY ASKING A QUESTION” bit, huh.

Carleas, let me ask a different question;

In the canonized solution, it is proposed that a blue (you) is thinking that another blue is thinking that he might be brown, “if I am brown, then…”. Why would a blue be thinking that he is brown? He has no way to know that he is brown, especially since he really isn’t.

First, we need to be precise in our language, since this has gotten us into trouble before. It’s not what he “is thinking”, but what he logically could think.

Second, while he could assume he is brown, the relevant assumption to complete the modus tollens is that he is ‘not blue’. That is different from being brown. Being brown is one way to be ‘not blue’, but it is reasoning on the possibility of being ‘not blue’ that ultimately allows him to conclude that he’s blue.

Finally, he’s able to contemplate and reason on these assumptions because they are logically permitted. There are a million logical deductions he could make, most of which aren’t relevant to the solution of the problem, but which the givens state that every person on the island knows immediately; everything that is logically deducible. He doesn’t know his eye color, so he knows that he could be blue, not-blue, brown, not-brown, red, etc. Making a deduction that includes his being any of these does not contradict anything he knows. And he makes them all. One of those deductions is "if I am not blue, then those blues see 98 blues, and they might think, "If I am not blue, then those blues see 97 blues and they might think, "If I am not blue then those blues see 96 blues and they might think, “…” " " "

Anybody ready to try a different one?
Don’t worry, it’s far less ambiguous (I think so anyway…).

Again, no tricks/loopholes:

You’re talking to a woman, she says she has 2 children, you ask her, “Is at least one of your children a boy?” She says yes. What is the probability that the other child is a boy?

Just to be perfectly clear about no tricks/loopholes:
Assume she’s telling the truth.
Assume there’s no ambiguity about sex, everyone is either a boy or a girl in this universe.
Assume that each time a woman gives birth, it’s 50/50 boy/girl.
Assume that each child’s sex is independent of the other (no statistical anomalies caused by twins).

If you must cheat and look up the answer, at least put it in a [ tab ] thingy.

[tab]1/3[/tab]

yurp

A 1-metre stick is broken into two pieces at random. What is the length of the shorter piece, on average?

[tab].25 meters?
My reasoning is this: It will always be between .5 and 0 meters long, and the average of all the lengths between .5 and 0 should be .25[/tab]

Correct

You decide to play a game with your friend where your friend places a coin under one of three cups. Your friend would then switch the positions of two of the cups several times so that the coin under one of the cups moves with the cup it is under. You would then select the cup that you think the coin is under. If you won, you would receive the coin, but if you lost, you would have to pay.

As the game starts, you realize that you are really tired, and you don’t focus very well on the moving of the cups. When your friend stops moving the cups and asks you where the coin is, you only remember a few things:

He put the coin in the rightmost cup at the start.

He switched two of the cups 3 times, as follows:

The first time he switched two of the cups, the rightmost one was switched with another.

The second time he switched two of the cups, the rightmost one was not touched.

The third time he switched two of the cups, the rightmost one was switched with another. This was the last switch

Which cup is most likely to hold the coin?

Cup and coin:[tab]1/2 cup on the right[/tab]

Yes. It is what he COULD think.
And he COULD think, “if everyone were to simply start with the same number then…”
And just as he wasn’t actually brown when he thought of what COULD be the case, to see where it would lead, he wasn’t thinking of an actual number when he thought, “what if everyone simply started with the same number…” so as to see where that would lead. Just as he COULD possibly be brown, he COULD possibly think of a common number.

And very importantly, he MUST eliminate such a possibility before he can ASSUME that everyone isn’t thinking such a thing rather than thinking of whether they are brown (or “not blue”).

Perfect logicians don’t ASSUME, they ELIMINATE alternatives by EXAMINING them (not simply jumping to the first possible assumption of what 200 others might be thinking).

James, I’m happy to keep talking about this, but it seems like everyone else has moved on and I don’t want to monopolize the thread. I’ll send you a PM with my response.

New Math Logic Fun:

Say that a chess board has two diagonally opposite corners removed, so that it contains 62 squares. Is it possible to place 2x1 rectangular tiles on the board to cover every square without the rectangles hanging off the board?

I know!
[tab]No
By my calculations, no it’s not. I’ve seen a similar problem before, and the logic works like this:

First of all, let’s call these ‘2x1 tiles’ dominos, and let’s assume that the dimensions of the dominos are exactly 2x1 relative to the size of each equally sized tile on the chessboard.
Also, we’re assuming that the only way to place them is for each to fully cover 2 tiles, of course, and the only way to do that is to align them with the chess grid (no partial covers, no diagonal pieces, etc).

Now, notice this: whichever way you place the domino, such that it fully covers two squares and is perfectly aligned with the grid, it covers 1 black and 1 white square. You cannot place it anywhere where it covers 2 squares fully, it’s not diagonal, and the squares are of the same color.

So if you think about it, every domino must be covering 1 black and 1 white square, so the number of blacks must be equal to the number of whites covered.

Now think back to the problem statement, in which opposite corners of the chess board were removed. You start out with a chessboard of 64 squares, 32 black and 32 white. Equal black and white. When you remove 2 corners that are opposite each other, you remove 2 corners of the same color. This means that you’re left with 30black and 32white, or 32black and 30 white. In either case, you’re left with an unequal number of black and white squares.

Since each domino, to be placed correctly, must be covering 1 black and 1 white square, you can only fully cover 60 of the squares with even the best strategy, as the remaining 2 will be of the same color.[/tab]

Correct, and thorough!

Awesome explanation. I couldn’t put my finger on why I couldn’t do it.

I know the explanation to the cup and coin:
[tab]The cup that initially covers the coin is swapped with one of the other two, giving a probability of 1/2 that it’s in either. The probability that it’s in the initial position is zero at this point.
Next, swapping the two not in the initial position does not change these probabilities.
Finally, swapping one of the two not in the intial position, with the cup in the initial position, means there’s a probabilty of 1/2 for each of them that they were swapped back to the initial position, and 1/2 that they were not.
For each of these two, the probability of 1/2 that the coin was swapped into that position AND the probability of 1/2 that it was NOT swapped back out makes a probability of 1/4 that it’s still there.
Two probabilities of 1/4 (for the coin being in a non initial position leaves a probability of 1/2 that it’s back in the original position.[/tab]
And the explanation of the stick one:
[tab]The average position of the first snap is bang in the middle.
The average position of the second snap is bang in the middle of either of the 2 pieces resulting from the first snap.
So the average length of (either of the 2) the shorter piece(s) is a quarter of the original length of 1m.[/tab]
I can intuitively work out the baby one from long-unused conditional probability knowledge that sits somewhere in my subconscious, but as for the proper explanation I can’t quite explain it right.

I think you misunderstood that one.
There aren’t 2 snaps. There’s 1 snap, into 2 pieces.