Interesting Math Facts - Page 2

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.

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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