Counter-Intuitive Science

Try this:Instead of thinking through the various outcomes, consider 2 different contestants who adopt opposing strategies: Messers Stick & Change. Under what circumstances will each win the car?If Mr Stick is playing, he would only win if he choose correctly at the first time of asking - on average 1/3 of the timeIf Mr Change is playing, then he wins the car if he chooses the wrong door first time around because Monty is forced to open the other goat-door leaving only the car for him to change to (as he always does) - on average 2/3 of the time.Clearly better to be Mr Change.For the avoidance of doubt, this is the original version of the game where Monty does know where the car is & deliberately avoids it (and the contestant understands this). Also anal, I know, but you absolutely have to be clear about who knows what and when to understand puzzles like these.I think it's fairly obvious that where Monty chooses blindly & his door isn't opened before you re-choose, then it doesn't matter whether or not you change doors. This is similar to drawing lots: it doesn't matter which order everyone chooses.If Monty chooses blindly and his door is opened before you get to change, then again it doesn't matter. Mr Stick is still going to win only when he chooses the right door to start with - 1/3 of the time.Mr Change will win when he chooses an incorrect door to start with (prob = 2/3) AND Monty chooses the other goat's door (prob = 1/2). Overall probability = 1/3.We can ignore the cases where Monty opens the car's door because either strategy has an equal probability of success (0). If one strategy were better than the other, it would be because it succeeds more often in the cases when Monty does not choose the car.

I saw a couple of good examples in my undergraduate days:One involves 3 plain cards with faces coloured red/red, red/green & green/green. Now if one card is chosen at random and placed on a table so that you can see one side (red say), what it the probability that the other side is also red?Secondly, perhaps more of a paradox, but anyway... draw a circle (doesn't matter how big) and then inscribe an equilateral triangle (inside the circle with its corners touch the circle). What is the probability that, for a straight line drawn through the circle, the part of the line which is inside the circle is longer than the length of the equilateral triangle's sides? This one is not obvious, but the point is that there are perfectly reasonable arguements which appear to demonstrate that the probability is both 1/2 & 1/3.Finally, and it's quite complicated - also >20 years ago so I can't fully remember, but an undergraduate problem I was set was to come up with two (mathematically) continuous lines which would connect diagonally opposite corners of a square without leaving the square and without crossing over one another! To be specific, there is no point which is common to both lines. It involved a function like sinx/x which has a thing called a removable singularity at x=0 - a singularity which you can "remove" by simply redefining it at the singularity. The point being that the function can take a range of values at the singularity and remain continuous. Such redefining allows the 2 lines to cross over at x=0 without touching.

Cavediver, Well, I suppose. No, there's no trap. But I find a lot of people expect it to be 1/2. Sorry it was too intuitive! Me too. You are "correct" in your approaches to getting 1/2 & 1/3. Clearly this can be extended to any value in between. But 1/4 is new. What? You mean I haven't convinced you that 1/2 = 1/3?Myself, I'd question the whole basis of a probability of choosing any one element of an infinite subset, from a larger set, and finding the answer is "1/3". I thought about an argument based on assuming finitely wide strips, and then limiting, but again, does a probability carry across a limit (or an integral)? Functional analysis is not my area.