Math games, puzzles, paradoxes, and questions

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So, before the serious discussion forum, there was an excellent “Math Riddles” thread in the off topic forum. I’d love to resurrect that thread, perhaps in a slightly broader sense, to allow the math geeks to babble about stuff. Post an interesting riddle you know. Ask a question that you’ve been wondering about. Post a quick description of a cool paradox you like, or even a clever false proof. Anything of the sort. Just have fun with it. :)

Monty Hall Paradox

As an example, one of my favorite simple paradoxes (i.e. counter-intuitive answer, not self-contradictory), is the Monty Hall problem. Monty Hall was the host for an old game show called “Let’s Make a Deal”. The marquee game on the show involved showing you 3 closed doors. Behind one of them is a grand prize (such as a car), and behind the other two are gag prizes (often some sort of livestock).

The game is very simple. The player selects a door. Monty Hall, in an effort to build suspense, would open one of the doors not selected and show one of the gag prizes. The player would then have the option of keeping the selected door or switching to the other still-closed door. The question is simply: should the player keep the door, switch the door, or does it not matter?

I’ll give an explanation later, but I’ll let you guys think about it first (at least, those of you who haven’t seen this problem before).

Random Number Selection

This is a question that I’ve had. This idea came from a proof that I did at one point in a math class (which I’ll post at a later date, since it still confuses me to this day), though any similar proof would work.

In simplest terms, you can prove that the chances of selecting a random real number and having it be a rational number is 0. Not close to 0, but truly 0. The reason is that there are infinitely more irrational numbers than rational numbers (countable vs. uncountable).

Let’s simplify for a second though. What are the chances of selecting the number “42” from the (infinite) set of all integers? 0, of course (i.e. 1 / infinity).

But does this mean that it’s actually impossible to select a random number from an infinite set, since the chances of selecting any particular random number are 0? Does this have any implications on our view of the universe? Must we live in a finite universe, or must we acknowledge that there is no such thing as a truly random event?

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oo I heard the Monte Hall one before… it’s really interesting to have your initial logic of what seems to be a simple question to be torn into pieces lol. I won’t spoil the answer.

I’d say.. yes it’s impossible to.. represent a random number from an infinite set, maybe? because you could say that the random number is between a range… and then you could keep reducing the range it’s between… but you will never come to an exact answer. It will have an infinite number of decimal places, because else it would be a distinct number. I guess you could call any number selected an irrational number because there will be no way that it could be represented as a fraction. Errrr… that’s as far as I go hehe. Thanks for putting that one forward!! Was interesting the think about =0 and I’d like to hear other people’s views on it. I don’t think that any events are truly random, although I don’t know less than alot of people about the universe, so I’m not really in a good position to say much about that lol. I think that quantum work seems to suggest there being randomness involved with things at a small level, and impossibilities, but I think that maybe it is all completely predictable, and just follows the laws of the universe in a way that we can’t understand or come up with a model for yet. Although then… just the concept of existance seems to suggest some randomness… (waiter at restaurant excuse me sir, did you order a big bang?)

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Let’s simplify for a second though. What are the chances of selecting the number “42” from the (infinite) set of all integers? 0, of course (i.e. 1 / infinity).

This is, technically speaking, incorrect. 1/infinity is not zero. 1/infinity is the same thing as 1/justice or 1/orange – it’s meaningless to divide a number by what is, essentially, a concept. If you allowed 1/infinity, you would by definition allow things like infinity/2, which leads to a paradox – infinity/2 = infinity/3, therefore 2 = 3.

Normally though, it’s okay to say things like that, but when you’re pondering mathematical problems you need to be specific about it. When people say “1/infinity = 0”, what they really mean is “limit n→infinity of 1/n = 0”. This is entirely different, as lim n→infinity 1/n is never actually zero. Thus, while the probability is infinitesimal, it is not actually zero.

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Well, you are somewhat correct, Einar. Yes, my shorthand of (1 / infinity) was not accurate. However, lim n→infinity of 1 / n does equal exactly zero, since you’re taking a limit. So, the argument still stands. The chances of picking the number 0 out of n integers is 1 / n, and if we’re to take the full set of integers, we would allow n to approach infinity, and if we take the limit here we do get 0.

By the way, your paradox isn’t resolved by using limits. The lim n → infinity of 2 / n = 0, and of 3 / n = 0, so does 2 = 3 now?

This Wikipedia article gives a good explanation of the difference between “surely” and “almost surely” (the difference being that if something will surely happen, then no other outcomes are possible, while if it will almost surely happen, then the probability of happening = 1). This is where math gets kinda weird (infinity has a tendency to do that). I’ll go ahead and throw another one of my counter-intuitive truths out there.

Cardinality of Rationals and Irrationals

Quick definitions: A rational number is a number that can be expressed as a fraction with integer numerator and denominator. An irrational number is any real number that is not rational.


1. The rational numbers are dense in the irrationals (i.e. between any two irrational numbers you can find a rational number).

2. Similarly, the irrationals are dense in the rationals.

3. The rationals are countable while the irrationals are uncountable (i.e. there are infinitely many more irrational number than there are rational numbers).

This always surprises me, since you’d think 1 and 2 would actual prove that they have the same cardinality (i.e. both countable or both uncountable), yet this is provably not the case.

I’ll explain Monty Hall and give my favorite false proof in a while.

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what i don’t get is in films/books/tv programs

how come people travel back in time to stop things happening then it all turns out right but surely if it all turned out alright they nether went back in time to stop whatever happened happen let me see if i can explain it simpler

Say the world got took over by elephants and someone travelled back in time to stop elephants existing then when they got to the future elephants would’ve become exctinct but then they would’ve never travelled back in time to stop it happening in the first place.

This causes it to be in an infinite loop and if some1 can explain this for me i’d be thankful.

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it"s fiction. it doesn’t have to make sense.

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I think time-travel is probably provably impossible, if for no other reason because of the inherent feedback loops. Some people describe time as the “4th dimension”. I don’t buy it.

So really, as banero said, it’s fiction and thus doesn’t have an explanation. I don’t think I’ve ever seen a time-travel story that really made sense without introducing any paradoxes (though Sound of Thunder, the short story, made a nice effort).

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some physics involves ideas of something like 11 dimentions =/ and apparently some of them are donut shaped lol.

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how dare you have all this fun while i’m in class. to add on to the cardinality of rationals and irrationals, it’s also interesting to note that the rationals and the natural numbers are the same size.

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That it is matt! And there’s a rather clever way of creating a 1:1 mapping between them (that I think Cauchy came up with it), which I imagine you’ve seen. I wonder if I can do this in text…

…on second thought, I’m find a picture. Ok, here we go – found it on Wikipedia, and got corrected (it was Cantor, not Cauchy – wrong C guy with a set named after him…)

Ok, so the way this works is you create an infinite grid with the positive integers running along both edges. For any particular point in the grid itself, we say that it is the fraction of the row to the column. So, if we go down to (2,3), we’ll write 2/3. It’ll look something like this (yes, I know it inverted, sorry, these are the only 2 pics I could find):

You trace the path shown above, and you’ll end up with all possible fractions, as well as a way to order them, meaning that we can now actually number the rational numbers (multiply by -1 and map to the negative ints to get the rest of them). Isn’t that crazy? I could say what is 3.14159 and you could tell me it’s rational number #57927592 (and yes, I made that up, please, don’t check it…unless you’re very bored…).

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Phoenix, a quick question. Even if I wanted to waste my time while the way too easy TD game I am playing wraps up, how do we count the numbers in a grid which stretches to infinity both ways?

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Oh jeez, not the Monty Hall problem. This has inspired furious debates on other forums.

Here’s the answer: If the host opens the first door and it isn’t a car, you should definitely switch doors. It will double your odds of getting the car.

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Jelly’s right (though ignored my request not to post if you had heard it :-P), though I don’t know why it would inspire debate in forums – it’s a simple and clear proof. Anyone care to try for it?

Mils – I didn’t ask you to count all of them, simply to identify where 3.14159 lies in the series. In the example above, 1/1 = 1, 2/1 = 2, 1/2 = 3, 1/3=4, 2/2=un-numbered(repeat of 1/1), 3/1 = 5, 4/1 = 6, 3/2 = 7, etc.

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ok, im not gonna solve it but im gonna state the facts ------- situation one: game host opens door with cattle Situation 1a: you say "Awesome! ill take it!" -chances of getting the car, 0/1 Situation 1b: you say "nah, ill try to get the car"- chances of getting the car, 1/2 ------- situation two: game host opens door with car Situation 2a: you say, "Awesome! ill take it!" -chances of getting the car, 1/1 Situation 2b: you say "nah, ill try to get the cattle"- chances of getting the car, 0/2 --------- There are the facts, you give me the answer buy now the question is, do you want the cattle more than the car, or the car more than the cattle?
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PyschoMonkey – I’m worried I’ve mis-stated the problem for you. Let me try again.

You have three doors. Let’s say you, the contestant, choose Door #1.

Now, Monty Hall is going to open up one of the doors that you didn’t select, but he will only open it up if there is cattle behind it (however, since two doors have cattle and you only picked one door, you know this is always possible). In this case, let’s say he opened up Door #2 and showed you cattle.

You’re now looking at Door #1 and Door #3. The question is, should you keep Door #1, or should you switch to Door #3? The answer is that you should definitely switch to Door #3, since it doubles the chances of winning the car. I’ll let you try to work out why this is the case. :)

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This is possibly the simplest and most elegant proof of the Monty Hall Paradox (without going into set notations and Bayes’ Theorem), presented so that all may understand and based on the generalization by D.L. Ferguson 1.

Where Pi = Initial Probability, Ps = Probability from switching, n = Number of initial choices and p = Number of choices revealed. Ferguson stated that:

Pi = 1/n

Ps = (n-1)/[n(n-p-1)]

So in the case of Monty we have:

Pi = 1/3

Ps = 2/3

Therefore it is better to switch.

My favorites have always been Zeno’s Paradoxes and in particular the concept of a supertask coined by James F. Thomson and Thomson’s Lamp, which pretty much goes:

Consider a lamp with a toggle switch. Flicking the switch once turns the lamp on. Another flick will turn the lamp off. Now suppose a being able to perform the following task: starting a timer, he turns the lamp on. At the end of one minute, he turns it off. At the end of another half minute, he turns it on again. At the end of another quarter of a minute, he turns it off. At the next eighth of a minute, he turns it on again, and he continues thus, flicking the switch each time after waiting exactly one-half the time he waited before flicking it previously. The sum of all these progressively smaller times is exactly two minutes.

The following questions are then considered:

Is the lamp on or off after exactly two minutes?

Is the lamp switch on or off after exactly two minutes?

Would it make any difference if the lamp had started out being on, instead of off?

Pulled directly from Wikipedia if there is a problem with the link (or you are just to lazy to follow it).

Note: OK, now that I spent all of my time attempting to get the formatting correct using textiles (which I couldn’t do) instead of just using HTML tags (which I should have just used from the start), I am to frustrated to discuss at the moment, so will post again later.

1 Selvin, S. (1975). A problem in probability. American Statistician , 29, 134.

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In order to have a being capable of performing that task, the being must be able to oscillate the switch with an infinite frequency – i.e. it takes no time at all to toggle the switch. Near the end of those two minutes, the lamp oscillates between on and off with increasing frequency. At the two-minute mark, it effectively exists in both states simultaneously.

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So is the cat alive or dead?

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Schrodinger may say otherwise though.

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OK, so I’m sitting here playing games and thought of a challenge to pose to you all. I’m sure if you are reading this thread I’m assuming that you already know that 0.9999…. = 1. This can be mathematically proven in various forms.


And even with more simple algebraic digit manipulations.

My challenge is as such; Prove this using base 2 math.

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Oh, geez – base 2? Alrighty, gonna have to ponder…

However McCoy, I have to disagree with your “simple and elegant”. While it might be nicely generalized (and elegant), I find it hard to use to explain to someone who isn’t fairly familiar with math. In the particularly small case offered with Monty Hall, it’s much easier to just do exhaustive cases:

There are two options for the first selection: either you picked the car (p = 1/3) or you picked a crap prize (p = 2/3). In case 1, if you switch, you don’t get the car. However, in case 2 if you switch you do get the car. Thus, switching generates a 2/3 chance of winning.

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I think time-travel is probably provably impossible, if for no other reason because of the inherent feedback loops. Some people describe time as the “4th dimension”. I don’t buy it.

Having time as the fourth dimension made Einsteins calculations in special relativity a whole lot easier. Since nothing could travel faster than the speed of light, and as one approached the speed of light, time goes slower (compared to a “stationary” observer [keep in mind this is an accelerated object]), one can include time as a dimension if one hypothesizes that everything travels at the speed of light through the four dimensions (x,y,z,t), meaning c=sqrt(x ^2+y ^2+z ^2+t ^2), so as spatial motion approaches the speed of light, time gets smaller (ignore negative values of x,y,z,t, because those are all relative). Now, if one wants to travel faster than the speed of light in a spatial dimension, they must have something called “imaginary time” or t=ix (where x is a real number). Some movies say it is possible to go faster than the speed of light. Hence inflicting imaginary time, which, when put into the transformation for time in special relativity, makes the object travel back in time. Some other movies have wormholes and other stuff, which basically let people exploit the safeguards put into special relativity to make it logically consistent. But, of course, this is just what I gather from being a nerd, I could be completely wrong…

My challenge is as such; Prove this using base 2 math.

Ok, I have to prove that 1= 0.111111…. (man I wish I had a bar to put over this) because 1 is the greatest digit and .111111 repeating should be analagous to .99999 repeating.
This equates to 0/1+1/2+1/4+1/8+1/16, which approaches 1 because each term halves the distance to one.
I’m now going to attempt to solve that algebraically…

n= .111 repeating

*10 *10

10n=1.111 repeating

-1n -.111 repeating (remember n=.111 repeating)



.111 repeating=1


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Hmm – I think I perhaps misunderstood McCoy’s challenge. I thought he wanted to prove that 9/10 + 9/100 + … = 1, which is very hard to represent in binary.

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mxmm: You are correct in the fact that 0.1111…2 = 0.9999…10 as in base 2 this would be defined as:

[.] 2-1 2-2 2-3 2-4

Then you go and throw in:

*10 *10

10n=1.111 repeating
which is not so because you are using 10 as a base 10 number (not allowed in my challenge) and attempting to apply 10 in terms of base 10, not base 2; so multiplication does not simply shift the radix point. Sorry; no ‘Geek of the Day’ award for you yet.

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Phoenix: OK, it may not be the simplist, but is is definately less complicated than this, as I thought you were looking for an actual matematical proof opposed to an explination or decision table of some sort.