Interesting Math Facts - Page 3

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