Interesting Math Facts

http://www.businessinsider.com/the-most-controversial-math-problems-2013-3#-1

anyone who says the monty hall problem is an easy, simple problem is a pompous ass

not rly if you practice thinking things through sequentially.. it isn't an easy problem if you just rely on your fast response system 1 tho

Well there you have it mnj

I still struggle with the Monty Hall problem... blows the mind

Here's a problem. Car A vs Car B in a 1/4 mile drag race. Car A gets a running start and travels at a consistent speed of 225 mph. Car B is a top of the line dragster and is not allowed to take off until Car A is even with it. In other words, Car B is sitting on the track, Car A is behind it and comes driving past at 225 mph and Car B can start once the two cars are even. Who wins the race?

wut

The 225 mph car will make a 4 second pass which is faster than any dragster.

On March 29 2013 17:03 blackjacki2 wrote: Here's a problem. Car A vs Car B in a 1/4 mile drag race. Car A gets a running start and travels at a consistent speed of 225 mph. Car B is a top of the line dragster and is not allowed to take off until Car A is even with it. In other words, Car B is sitting on the track, Car A is behind it and comes driving past at 225 mph and Car B can start once the two cars are even. Who wins the race?

+ Show Spoiler +

On March 29 2013 17:03 blackjacki2 wrote: Here's a problem. Car A vs Car B in a 1/4 mile drag race. Car A gets a running start and travels at a consistent speed of 225 mph. Car B is a top of the line dragster and is not allowed to take off until Car A is even with it. In other words, Car B is sitting on the track, Car A is behind it and comes driving past at 225 mph and Car B can start once the two cars are even. Who wins the race?

Miles*((Seconds/hour)/(Miles/hour)
(1/4)(3600/225) = 4

It takes A 4 seconds to complete the entire track.

Top Fuel Dragster (TF/D). The rail dragsters, or "diggers", are the fastest class. Among the fastest-accelerating machines in the world, these cars can cover the dragstrip in less than 3.8 ­seconds and record trap speeds over 325 mph. Top Fuel cars are 25 feet long and weigh 2,320 pounds in race-ready trim. Methanol fuel mixed with up to 90% nitromethane is used.

The top level dragster if it is allowed proper traction and no wind.

my favorite math fact (formula) is e^(pi*i)+1=0 uses 5 of the most important numbers (or complex numbers ) in all of mathematics.


some other facts/proofs/formulas/random stuff i love:
-shapes can have finite volume and infinite surface area: fractals or gabriels horn are good examples

-cantor sets (http://en.wikipedia.org/wiki/Cantor_set): perfect sets that are nowhere dense, pretty amazing

-the proof of square root of 2 being irrational:
suppose root(2) is rational. that means it can be represented as integers a and b where a/b are not reducible. then (a/b)^2 = 2 or a^2=2b^2. this means that a^2 is even which implies that a is even. say a=2k. (2k)^2 = 4k^2 = 2b^2 or 2k^2=b^2, so b^2 is even which implies that b is even. so both a and b are divisible by 2 but we assumed they were not reducible. => <= contradiction, thus root(2) is irrational. very cool proof by contradiction showing that there existed numbers that could not be represented by fractions.

-transcendental numbers are numbers that are not roots of any polynomial equations with rational coefficients. aka they are not algebraic. its very hard to prove something is transcendental and we dont know of many transcendental numbers, but almost all of the real numbers are transcendental since they are an uncountable set while the algebraic numbers are countable.

-non transitive dice: this one is a bit tricky but really cool.
suppose there are 3 dice

die A: 1,1,4,4,9,9
die B: 2,2,6,6,8,8
die C: 3,3,5,5,7,7

there is a 20/36 chance that die A outrolls die B.
there is a 20/36 chance that die B outrolls die C.
there is a 20/36 chance that die C outrolls die A.


-Russell's paradox:

from wikipedia:

Let us call a set "abnormal" if it is a member of itself, and "normal" otherwise. For example, take the set of all geometrical squares. That set is not itself a square, and therefore is not a member of the set of all squares. So it is "normal". On the other hand, if we take the complementary set that contains all non-squares, that set is itself not a square and so should be one of its own members. It is "abnormal". Now we consider the set of all normal sets, R. Determining whether R is normal or abnormal is impossible: If R were a normal set, it would be contained in the set of normal sets (itself), and therefore be abnormal; and if R were abnormal, it would not be contained in the set of all normal sets (itself), and therefore be normal. This leads to the conclusion that R is neither normal nor abnormal: Russell's paradox.

did lots of set theory in college so i always found them really interesting

oops, yeah top fuel dragsters can do a 1/4 mile faster than 4 seconds. Pretty fuckin sick

Why the conversation in this thread is still all about math and not about me and how I made an amazing contribution to this forum by finding this presentation and starting this thread here?

None of these are controversial. Imagine someone said it was controversial whether aces were the best nlhe hand to have and you can have a good idea how silly calling these controversial is

You can say what you want but I don't think the "monty python" problem is that abstract... I am by no means a mathematical genius and I solved the problem in highschool. I'm really not trying to brag either because i am very far from being genius.

DAUT, I ALWAYS WANTED TO KNOW MORE ABOUT E AND WHERE IT SHOWS UP IN NATURE.

NEVER COULD FIND EXAMPLES THOUGH

e = 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + ... = approx 2.7818

e is also used in finance. For instance, suppose I were to give you an interest rate of 100% (never going to happen, but for practical purposes). You put $1 in the bank and you will get $2 if compounded yearly. if we compound biyearly, we get
$1 x (1 + 1/2)^2 = $2.25
compounded quarterly
$1 x (1 + 1/4)^4 = $2.4414 approx
compounded monthly
$1 x (1 + 1/12)^12 = $2.613035 (do you see the pattern?)
compounded "n" periods
$1 x (1 + 1/n)^n
compounded continuously
lim n->infinity $1 x (1 + 1/n)^n = $2.7818 approx = e

This is why when you hear the term compounding continuously we use the formula e^rt where r = interest rate and t = time period to calculate how much money you will have after a certain period of time at a certain rate.

On March 30 2013 01:56 KoeBawlt wrote: None of these are controversial. Imagine someone said it was controversial whether aces were the best nlhe hand to have and you can have a good idea how silly calling these controversial is

That's not true. When the monty python example was first released to the public there were multiple debates from Ph.D mathematicians regarding the true solution to the problem. While the idea seems simple enough, the difficulty at the upper echelons of mathematics was to prove the answer as 2/3 always. You can repeat 10000 trials and realize it's always 2/3 but this is not a mathematical proof.

And as a person that has worked on many many problems similar to all of these out of entertainment, there as definitely been high scale debates over these topics at one point or another. Only thing in today's society is we have all the answers right in front of us. When you go back to when these problems were first contrived, there was much more heated debate.

When you multiply any number by 9, the sum of the resulting numbers added up will always be 9. Cool huh ?

On March 31 2013 09:05 brambolius wrote: When you multiply any number by 9, the sum of the resulting numbers added up will always be 9. Cool huh ?

This post led me to play with the calculator for a little bit. You sir, might be on to something

if the digits of a number add up to a multiple of 3, the original number is divisible by 3.

On March 29 2013 17:36 Daut wrote: -non transitive dice: this one is a bit tricky but really cool. suppose there are 3 dice die A: 1,1,4,4,9,9 die B: 2,2,6,6,8,8 die C: 3,3,5,5,7,7 there is a 16/36 chance that die A outrolls die B. there is a 20/36 chance that die B outrolls die C. there is a 20/36 chance that die C outrolls die A. did lots of set theory in college so i always found them really interesting

Am I right?

Just a little thing I wanted to add to the Monty Hall problem. There is often an assumption that is implied, but not stated. For the conclusion of the Monty Hall problem to be correct, it has to be part of the rules of the game game that after you pick, he is not allowed to open up the door with the prize behind it.
Let's say that Monty doesn't know where the prize is either, and could pick the door with the prize (in which case you just lose right away when he picks the prize door). If you pick a door, then he picks a door at random and behind it is a goat, then you do NOT need to switch, it is the same probability either way. The probability only changes if his door opening is not at random, and he always picks a door that doesn't have a prize.

I am not sure if I ever saw the Monty Hall problem presented with that fact explicitly, which it really needs to.

On March 31 2013 14:33 HungarianGOD wrote: Just a little thing I wanted to add to the Monty Hall problem. There is often an assumption that is implied, but not stated. For the conclusion of the Monty Hall problem to be correct, it has to be part of the rules of the game game that after you pick, he is not allowed to open up the door with the prize behind it. Let's say that Monty doesn't know where the prize is either, and could pick the door with the prize (in which case you just lose right away when he picks the prize door). If you pick a door, then he picks a door at random and behind it is a goat, then you do NOT need to switch, it is the same probability either way. The probability only changes if his door opening is not at random, and he always picks a door that doesn't have a prize. I am not sure if I ever saw the Monty Hall problem presented with that fact explicitly, which it really needs to.

This is wrong. Even if monty opens it at random, its its a donkey u should still switch. there is no difference between if he purposely picks a donkey or randomly picks it, the info u get is the same.

[/QUOTE]
This is wrong. Even if monty opens it at random, its its a donkey u should still switch. there is no difference between if he purposely picks a donkey or randomly picks it, the info u get is the same.[/QUOTE]

It's not, I can write out a proof later if you would like.

Take the problem expanded into a ton of doors (like 100). If you pick a door, and then he randomly opens 98 doors and doesn't pick the prize, the probability of him doing that if you DIDN'T pick the door with the prize is extremely low, 1/99. In fact from this point of view it looks like you would actually do much worse if you swapped than if you didn't (but this isn't actually the case, because the probabilities balance out when you look at the similarly low probability that you would pick the door in the first place).

I don't think it matters since you are only looking at instances where he successfully opens all donkeys, so the "probability" is 1 i.e. not very low.

EDIT:
After some thought I'm convinced the fact that he chooses randomly indeed changes the calculation and it's 50% chance, but your explanation is off. Also, it's implied that he will not open the prize, so it can't be random.

I haven't even explained it yet, how can my explanation be off ;-p

But yeah, isn't it interesting how such an essential piece of info to the problem is just never mentioned

HungarianGOD, I don't see how there is any difference in probability between the host knowing which doors have goats and choosing a door with a goat, or restricting the outcomes to the host randomly choosing a door and revealing a goat.

On March 31 2013 14:50 HungarianGOD wrote: en if monty opens it at random, its its a donkey u should still switch. there is no difference between if he purposely picks a donkey or randomly picks it, the info u get is the same It's not, I can write out a proof later if you would like. Take the problem expanded into a ton of doors (like 100). If you pick a door, and then he randomly opens 98 doors and doesn't pick the prize, the probability of him doing that if you DIDN'T pick the door with the prize is extremely low, 1/99. In fact from this point of view it looks like you would actually do much worse if you swapped than if you didn't (but this isn't actually the case, because the probabilities balance out when you look at the similarly low probability that you would pick the door in the first place).

With your 100 doors example, what will happen is 98 times monty will open a door with the prize, which makes you loose (in this case there is no decision to make, you just lost), 2 times he will open all donkeys, and those 2 times you should def switch since ur door is 1/100 to be prize while the other door is 99/100 to be prize.

Edit: Actually.... fuck you're right. Apologies lol. Leaving my wrong explanation here cuz rereading myself made me realize why you are right.

On March 31 2013 15:00 GsOne wrote: I don't think it matters since you are only looking at instances where he successfully opens all donkeys, so the "probability" is 1 i.e. not very low. EDIT: After some thought I'm convinced the fact that he chooses randomly indeed changes the calculation and it's 50% chance, but your explanation is off. Also, it's implied that he will not open the prize, so it can't be random.

You are right. The other guy's explanation is kinda off hence why i took a while to realize he's right.

Bertrand's box helped me finally see the monty hall problem. Until now my thought process was: well if I choose 1 in 3 then 1 choice not picked is eliminated and they ask me to switch it is now a 1 in 2 chance. Better right? But why should it matter what my initial choice was if I could now REchoose either box?

On March 31 2013 09:05 brambolius wrote: When you multiply any number by 9, the sum of the resulting numbers added up will always be 9. Cool huh ?

Could some math buff explain why this happens anyway ? Having trouble googling it

To clarify - you sum up the digits, and you continue to do so until you get 9, so for example 11 * 9 = 99 sum up to 18, sum up to 9.

The multiplication part ensures that when you start to add up the digits you do so for a number that is divisible by 9. Now, for any number divisible by 9 you can show that the sum of its digits is also divisible by 9 (and vice versa - if the sum of the digits is divisible by 9, the number is divisible by 9). As you can see it's a recursion - you start with a number divisible by 9, then you get a smaller (unless it's already 9) number divisible by 9 and so on.

The interesting fact about the number 9 that makes this true is (modular arithmetic ahead):
For any number n:
n = 10 * n (mod 9)
Which means that if you divide n by 9 you get some remainder r, and if you divide 10 * n by 9 you get the same remainder r. For example:
17 = 1 * 9 + 8 (remainder 8)
170 = 18 * 9 + 8 (same remainder 8)

Now all you need to see is that numbers are expressed as sums of multiplications, 126 = 10 * 10 * 1 + 10 * 2 + 1 * 6. If you do the math modulo 9, you can just drop all those 10s, which basically is adding up the digits, so 126 = 1 + 2 + 6 = 9 (mod 9), which is what we wanted to prove.

In Group Theory (an area which in some respects can be seen as a deeper abstraction of common arithmetic), all the groups which are essentially indivisible, in a sense the primes of group theory, are known and there are 26 of them. This was only finally proved 5 years ago and is mega, like sequencing the genome of mathematics.

I think this fact, and it's lack of obvious symmetry about something so fundamental, totally mind-blowing.

And if any mathematicians want to be pernickity about my layman's description above then please don't bother, I don't think there's anything wrong with my (hedged) analogies.

If you draw any great circle around earth, then there exist some two opposite points on that circle with exactly the same temperature. It's a not so obvious result of the seemingly trivially obvious Intermediate Value Theorem, which I think is cool.

And finally, in Projective Geometry (which can be thought of as a Euclidean Geometry viewed with perspective, so parallel lines meet in the infinite distance) any theorem which can be expressed in terms of points and lines is still true when you exchange the words 'point' and 'line'! This applies quite often to 'normal' geometry as an easy to understand example that you can try yourself with a pencil and paper is Pascal's Theorem, which becomes Brianchon's Theorem when you do the exchange.

On March 31 2013 11:00 SolarM wrote: Show nested quote + On March 29 2013 17:36 Daut wrote: -non transitive dice: this one is a bit tricky but really cool. suppose there are 3 dice die A: 1,1,4,4,9,9 die B: 2,2,6,6,8,8 die C: 3,3,5,5,7,7 there is a 16/36 chance that die A outrolls die B. there is a 20/36 chance that die B outrolls die C. there is a 20/36 chance that die C outrolls die A. did lots of set theory in college so i always found them really interesting Am I right?

my bad its actually:

die A: 22,44,99
die B: 11,66,88
die C: 33,55,77

Back to Monty Hall.

It's easy to show that if Monty selects the gate to open at random, your chances are improved to 1/2 for every gate if he opens a donkey (I think it's what makes people believe your chance is always 1/2):

Let's say you always pick gate 1 at first and Monty opens gate 2. The prize can be in any of those gates, there are 3 equally possible situations. If you get the information that gate 2 was empty, one of those possibilities disappears, so the other two are improved to 1/2.

What's different about the actual game is that no scenario can be discarded, as Monty will always open an empty gate. This makes it 1/3 for your first pick even when you know Monty's gate is empty, and 2/3 for the third gate.

Now, the interesting part: in contrast to blatantly obvious (imho) fact that Monty is not picking at random, there's one more piece of information that's actually hidden: the gate that Monty chooses to open in case you pick the gate with the prize.

For the solution to be 1/3 vs 2/3 you actually need to assume that he opens either gate with exactly same probability. This can also be achieved by making the gates you did not pick indiferentiable, and that's assumed when enumerating only 3 possibilities as in here http://en.wikipedia.org/wiki/Monty_Hall_problem#Solutions

But you could argue, that no one bothered to actually randomize the pick (coin toss would suffice) and Monty (or some technician), armed with knowledge of where the prize is, decided which gate to open. In this case, since humans make terrible random number generators, Monty could be biased, so for example he'd open a gate with a bigger number with probability p > 1/2.

Let's again assume that we pick gate 1. If we assume an extreme bias of ALWAYS opening gate 3 (p = 1) when possible, a person knowing this p could calculate that:

A) If Monty opens gate 2, the prize is in gate 3 with probability 1
B) If Monty opens gate 3, the prize is in gate 1 or 2 with same probability (not 1/3 vs 2/3).

As p gets closer to 1/2, the solution gets closer to being 1/3 vs 2/3 for either gate.

I find this confusing as hell.

monty hall problem
car donkey donkey
donkey car donkey
donkey donkey car
shows up 66% in other door
33% in chosen door

On April 02 2013 06:52 GsOne wrote: To clarify - you sum up the digits, and you continue to do so until you get 9, so for example 11 * 9 = 99 sum up to 18, sum up to 9. The multiplication part ensures that when you start to add up the digits you do so for a number that is divisible by 9. Now, for any number divisible by 9 you can show that the sum of its digits is also divisible by 9 (and vice versa - if the sum of the digits is divisible by 9, the number is divisible by 9). As you can see it's a recursion - you start with a number divisible by 9, then you get a smaller (unless it's already 9) number divisible by 9 and so on. The interesting fact about the number 9 that makes this true is (modular arithmetic ahead): For any number n: n = 10 * n (mod 9) Which means that if you divide n by 9 you get some remainder r, and if you divide 10 * n by 9 you get the same remainder r. For example: 17 = 1 * 9 + 8 (remainder 8) 170 = 18 * 9 + 8 (same remainder 8) Now all you need to see is that numbers are expressed as sums of multiplications, 126 = 10 * 10 * 1 + 10 * 2 + 1 * 6. If you do the math modulo 9, you can just drop all those 10s, which basically is adding up the digits, so 126 = 1 + 2 + 6 = 9 (mod 9), which is what we wanted to prove.

Don't wanna derail or anything, but you seem to know math where as I'm a math retard, why does it only work with 9 ? Are you explaining why it happens, or just that it happens ? Once again sorry, but not versed in math at all .

Reading a bit on wiki:
If a calculation was correct before casting out, casting out on both sides will preserve correctness. However, it is possible that two previously unequal integers will be identical modulo 9 (on average, a ninth of the time). lol

On April 02 2013 14:26 GsOne wrote: Back to Monty Hall. It's easy to show that if Monty selects the gate to open at random, your chances are improved to 1/2 for every gate if he opens a donkey (I think it's what makes people believe your chance is always 1/2): Let's say you always pick gate 1 at first and Monty opens gate 2. The prize can be in any of those gates, there are 3 equally possible situations. If you get the information that gate 2 was empty, one of those possibilities disappears, so the other two are improved to 1/2. What's different about the actual game is that no scenario can be discarded, as Monty will always open an empty gate. This makes it 1/3 for your first pick even when you know Monty's gate is empty, and 2/3 for the third gate. Now, the interesting part: in contrast to blatantly obvious (imho) fact that Monty is not picking at random, there's one more piece of information that's actually hidden: the gate that Monty chooses to open in case you pick the gate with the prize. For the solution to be 1/3 vs 2/3 you actually need to assume that he opens either gate with exactly same probability. This can also be achieved by making the gates you did not pick indiferentiable, and that's assumed when enumerating only 3 possibilities as in here http://en.wikipedia.org/wiki/Monty_Hall_problem#Solutions But you could argue, that no one bothered to actually randomize the pick (coin toss would suffice) and Monty (or some technician), armed with knowledge of where the prize is, decided which gate to open. In this case, since humans make terrible random number generators, Monty could be biased, so for example he'd open a gate with a bigger number with probability p > 1/2. Let's again assume that we pick gate 1. If we assume an extreme bias of ALWAYS opening gate 3 (p = 1) when possible, a person knowing this p could calculate that: A) If Monty opens gate 2, the prize is in gate 3 with probability 1 B) If Monty opens gate 3, the prize is in gate 1 or 2 with same probability (not 1/3 vs 2/3). As p gets closer to 1/2, the solution gets closer to being 1/3 vs 2/3 for either gate. I find this confusing as hell.

it also doesn't work if they always put the prize behind door 2

**Some edits

On April 02 2013 18:02 blackjacki2 wrote: it also doesn't work if they always put the prize behind door 2

They can't. By this I mean - they can put the prize however they want, but we get to choose a gate at first, any one. So instead of choosing a gate a random we can model always picking gate 1 and randomize position of the prize.

On April 02 2013 17:20 brambolius wrote: Don't wanna derail or anything, but you seem to know math where as I'm a math retard, why does it only work with 9 ? Are you explaining why it happens, or just that it happens ? Once again sorry, but not versed in math at all .

It actually works with 3 also. I guess you could say it happens because we use 10 as a default base for our numbers, and that defines what summing up digits means. And you can easily see that 10 = 1 (mod 9) as well as 10 = 1 (mod 3). That interesting fact I mentioned earlier can be rather easily proven with induction:

1) We can see that 10 * 1 = 10 = 1 (mod 9).
2) Suppose that 10 * k = k (mod 9). We have
10 * (k + 1) = 10 * k + 10 (mod 9)
At this point we use our assumption, so
10 * k + 10 = k + 1 (mod 9)

We have proven this for 1, and proven that if this works for k it also works for k + 1, so it works for any positive integer (you can prove for 0 and negatives too). You can try to prove this for 3 yourself, very similar.

So in short it's an effect of relation of 3 and 9 to the base of decimal number system.

On April 02 2013 17:20 brambolius wrote: Reading a bit on wiki: If a calculation was correct before casting out, casting out on both sides will preserve correctness. However, it is possible that two previously unequal integers will be identical modulo 9 (on average, a ninth of the time). lol

I don't think anyone should mention this anywhere but in bit obscure riddle or property books, and I would associate this with probability, not modular arithmetic.
What this means is mathemathicians defined a new way to write equations (modular arithmetic, a = b (mod c) etc.), and we want this to hold: if k = k then k = k (mod n) (this equality sign should be three line for reasons I don't really know/care about). They call this casting, so from a normal equation we can get an equation in modular arithmetic. And for this to not be nonsensical, that property is a must.
The second part deals with the fact that in this new notation, all operations like addition and multiplication are done on just few numbers (layman terms all around, look up group theory if you want to know more). For example modulo 3 you have just 3 numbers: 0 1 and 2. But we can make this much more useful, if we tell how map bigger numbers to those. At once again, only useful option seems to be:
0 = 0
1 = 1
2 = 2
3 = 0
4 = 1
5 = 2
6 = 0
etc.

Which you can define as reminder of diving k by 3.

As you can see, this reminder is the same for different numbers, so it can happen that a =/= b but a = b (mod n) with 3 =/= 6 and 3 = 6 = 0 (mod 3) as an example. And now some bright fellow decided to point out what the probability of this happening to two random numbers modulo 9 is and share. No value in understanding modular arithmetic.

This might not be that interestesting, but I remembered this puzzle and I always liked one solution for its elegance and simplicity.

Fact: you can't fill 12 x 15 board (180 squares) with 45 T-shaped blocks (each block is 4 squares).

Puzzle: prove it

Hint
+ Show Spoiler +

Also, I always wanted to give this to a person with something OCD-like and tell them to try to fill it up, then walk.

I have a point to be made about The monty hall problem.
I notice that I would select the first door more often than the 2nd or 3rd door in guessing which door the car is in. Because its completely random might as well just go with the first door.
What that means is the host should put the car in the first door always
then put the donkeys in the 2nd and 3rd door.
So when the stupid contestant picks the first door, he will get screwed when he goes for the two donkey doors.

we should select the 3rd or 2nd door to avoid the 1st door trick.

monty trolLOL

On April 02 2013 07:30 Oly wrote: In Group Theory (an area which in some respects can be seen as a deeper abstraction of common arithmetic), all the groups which are essentially indivisible, in a sense the primes of group theory, are known and there are 26 of them. This was only finally proved 5 years ago and is mega, like sequencing the genome of mathematics. I think this fact, and it's lack of obvious symmetry about something so fundamental, totally mind-blowing. And if any mathematicians want to be pernickity about my layman's description above then please don't bother, I don't think there's anything wrong with my (hedged) analogies.

edit for double post.

On April 02 2013 07:30 Oly wrote: In Group Theory (an area which in some respects can be seen as a deeper abstraction of common arithmetic), all the groups which are essentially indivisible, in a sense the primes of group theory, are known and there are 26 of them. This was only finally proved 5 years ago and is mega, like sequencing the genome of mathematics. I think this fact, and it's lack of obvious symmetry about something so fundamental, totally mind-blowing. And if any mathematicians want to be pernickity about my layman's description above then please don't bother, I don't think there's anything wrong with my (hedged) analogies.

Group theory is huge, and sometimes can be visual. I like when math meet art. And an example is the Alhambra in Spain. There you can see the 17 different way to tile in 2D (wallpapers group), which have been found before the year 1000, but the demonstration of the limited number came way later. http://en.wikipedia.org/wiki/Wallpaper_group

Also, the four color theorem is also linked to this. http://en.wikipedia.org/wiki/Four_color_theorem

On April 02 2013 20:09 GsOne wrote: **Some edits

It doesn't work with 3 :/.

? If a number is divisible by 3, it's digits will sum up to something divisible by 3 (and vice versa).

+ Show Spoiler +

Oh you can end up in 6 and 9 also when summing up the digits for 3, so this works for 9 only

i cant believe so many posts in this thread are about monty hall.

reminds me of the days when people argued for hundreds of posts about that stupidly worded question: Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?


oh no what have i just done?

On April 03 2013 09:23 Daut wrote: i cant believe so many posts in this thread are about monty hall. reminds me of the days when people argued for hundreds of posts about that stupidly worded question: Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys? oh no what have i just done?

50%?

On April 03 2013 09:23 Daut wrote: i cant believe so many posts in this thread are about monty hall. reminds me of the days when people argued for hundreds of posts about that stupidly worded question: Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys? oh no what have i just done?

at least one of them is a boy... = 1/3:
possible outcomes:
girl girl (can't be this because there is no boy)
girl boy
boy girl
boy boy - this is the desired result of the three possible outcomes

if the question is worded like "one of them is a boy..." ( as opposed to "at least one of them is a boy" ), the chance is 1/2 because the probability of the second child being a boy is independent from the first child.

On April 03 2013 07:49 GsOne wrote: Oh you can end up in 6 and 9 also when summing up the digits for 3, so this works for 9 only

Indeed