"Simple" Math Problem

[Scythiroth]

What about the 4 in 134 though? Does that not count as a loop?

[Wulfgang]

153, no closed loops in the numbers. All 3 other options have a closed off loop in the numbers (6, 9, 0)

If this is the correct answer, many baby seals will be clubbed.

What about the 4 in 134 though? Does that not count as a loop?

Not if you write 4's like I do.

[Drex]

Not if you handwrite it, it translates badly onto the internet though. The site I found it on used a font that looked like handwriting for the question, so the 4 was the goal-post looking deal

Idk, maybe this is a different question, but it looks a lot like the closed loop thing at least

I guess you could say circle instead of closed loop and be more correct, since 4 doesn't have a circle in it but 0 6 and 9 do

[Eaglestrike]

1+4+0=5

Haha, I always make a stupid mistake during math...it's usually only once per test, sad it had to occur during a 1 question one. But I have far less patience for attempting math than when I was in school and it "meant something" so that probably hurts me even more.

[Charla]

153, no closed loops in the numbers. All 3 other options have a closed off loop in the numbers (6, 9, 0)

I've heard this somewhere before and had to be told, it's stupid

A 1st grader can figure it out more easily because they aren't trying to make a pattern out of the numbers, they read the numbers as pictures and find the picture that looks the most like the others...in theory. I think it's BS but hey, whatever.

I don't know about you, but I knew how to read before third grade, and once I could read I stopped seeing written glyphs as pictures.

Maybe it's a test for asbergers or something?

[Jotaru]

While we're on the topic, I'd like to bring up something I saw on lol4chan last night, and really got me thinking, much in the same fashion as the 'Will the plane on a treadmill fly?' question...

Bag 1 has a gold coin, and a silver coin
Bag 2 has two silver coins
Bag 3 has two gold coins.

You pick one bag, and toss the other two out. You pick out a coin; its gold. What are the chances the other coin in the bag is gold?

The reason it interested me is because it (like the airplane problem) seemed to have a very simple answer (50%, it could be bag 1 or 3, thus, it could be gold(3) or silver(1)).

Then again, I recalled back to the movie 21, where they discuss variable odds... and my answer swayed a bit different. I figured, well, you pulled a gold coin out, so you can make some inference as to the likelihood of which bag you have, thereby changing the odds...

Woozie / some other smart math person?

[Tyche]

Berttrand's paradox

http://www.youtube.com/watch?v=BqXToEU5eDY

[Tulun]

I guess the 4 technically isn't a loop... but that's just silly...

[Jotaru]

I feel smarter now Tyche. I also can never look at that problem again without cringing at all the people falling for the 'simple' answer and being so sure that they're right, similar to the airplane problem. Thanks

[Norellicus]

While we're on the topic, I'd like to bring up something I saw on lol4chan last night, and really got me thinking, much in the same fashion as the 'Will the plane on a treadmill fly?' question...

Bag 1 has a gold coin, and a silver coin
Bag 2 has two silver coins
Bag 3 has two gold coins.

You pick one bag, and toss the other two out. You pick out a coin; its gold. What are the chances the other coin in the bag is gold?

The reason it interested me is because it (like the airplane problem) seemed to have a very simple answer (50%, it could be bag 1 or 3, thus, it could be gold(3) or silver(1)).

Then again, I recalled back to the movie 21, where they discuss variable odds... and my answer swayed a bit different. I figured, well, you pulled a gold coin out, so you can make some inference as to the likelihood of which bag you have, thereby changing the odds...

Woozie / some other smart math person?

33%.

You have six total coins, 3 of each; the fact that you discard the bags is irrelevant. Since you know you've already pulled out one gold one, the chance of your next pull also being one of the gold ones is 2/6.

[Charla]

After watching the video, I think that this is a case of a misleading question. Being a programmer (theoretically) I'll describe the process of finding this probability using some pseudocode:

select bag
pull coin
if (coin == silver)
break
if (coin == gold)
{
pull coin
if (coin == gold)
return true
if (coin == silver)
return false
}

Each time this code returns would then represent one trial (in the statistical sense). For anyone not familiar with the meaning of the word break, it means that if a silver is pulled first, that trial is not counted, and we proceed to the next trial. Now, out of an infinite number (N) of trials, this code should return N/2 true and N/2 false.

According to the description of the problem given in the video, however, it appears the real code should be more like this:

select bag
pull coin
if (coin == silver)
return false
if (coin == gold)
{
pull coin
if (coin == gold)
return true
if (coin == silver)
return false
}

In this case, it would return true N/3 times, and false 2N/3 times.

[Daydreamer]

33%.

You have six total coins, 3 of each; the fact that you discard the bags is irrelevant. Since you know you've already pulled out one gold one, the chance of your next pull also being one of the gold ones is 2/6.

You should watch that vid that was posted, it makes total sense and says you're wrong.

[Blarg]

33%.

You have six total coins, 3 of each; the fact that you discard the bags is irrelevant. Since you know you've already pulled out one gold one, the chance of your next pull also being one of the gold ones is 2/6.

That's what I get. If your first pick has to be a gold then there are no more golds left in 2 of the 3 boxes so the only way you get another gold is if you picked the box with 2 golds. The odds of picking the 2 gold box is 1/3

[Ramor]

Berttrand's paradox

http://www.youtube.com/watch?v=BqXToEU5eDY

Farnsworth's Parabox!
(episode of futurama for those of you that don't immediately get it)

[Norellicus]

Bah, you're right, I'm viewing all of the coins as one matrix, when your odds are fixed based on the moment you pick the bag. Since we know the arrangement of coins, the coin we've already viewed, and the fact that you'll be pulling from the same bag AGAIN, your odds in this case remain 2/3. The silver coins aren't a factor at all, since we're only concerned with how likely you are to have picked a second coin identical to your first. Since they're all paired up, it's elementary, but deceptive.

I've been doing so many similar puzzles to this via Professor Layton (DS puzzle game) I got em all mixed up D:

Edit: The wikipedia page on it and the reference to the drawers helps make the most sense in my opinion, it delineates the fact that you're examining the individual odds of two coins relative to eachother, not the odds of which box or bag you picked.

[archibaldcrane]

Basically, 2/3rds of the gold coins are in the 2 gold coin bag, so 2 out of every 3 times you grab a bag and pull a gold coin out of it, it's going to be the 2 gold coin bag. So once you get to a position where you have that first gold in your hand, 66% of the time you're holding the 2 gold bag.

[Jotaru]

That's the way I answered it on 4chan, and got laughed out (not surprised, however) by the LOL 50% CHANCE PWNT noobs.

[Khamsin]

A balloon with just enough helium to offset its weight (so it floats in place without rising or falling) is placed in an elevator. Will the balloon move with the elevator or hit the ceiling/floor, assuming it doesn't accelerate too quickly?

What about a balloon on a non-enclosed platform open to the air?

A bird in an enclosed elevator?

A bird on a non-enclosed platform?

[Lysithea]

answer is 39 only other number in multiple choice that is a secondary prime number so sequence would be 57+20 semi prime numbers 134-25 semi prime numbers 39 +30 semi prime numbers145 -35 semi prime numbers 26...

39 57 134 39 145

[Niiro]

Basically, 2/3rds of the gold coins are in the 2 gold coin bag, so 2 out of every 3 times you grab a bag and pull a gold coin out of it, it's going to be the 2 gold coin bag. So once you get to a position where you have that first gold in your hand, 66% of the time you're holding the 2 gold bag.

This is wrong and I'll show you why.

The fact that there are 3 gold coins total is irrelevant.

Bag A has 99 gold coins.
Bag B has 1 Gold coin and 1 Silver coin.
Bag C has 99 silver coins.

By your logic:

99 of the 100 gold coins are in bag A, so out of every 100 times you grab a bag and pull a gold coin out of it, it's going to be from bag A.

You see how that doesn't make sense?

-----

My opinion: it's 50%, and here's why I think that

Bag A has 99 gold coins.
Bag B has 1 gold coin and 1 silver coin.
Bag C has 99 silver coins.

You pick a bag, and pull out a coin, it's gold.

You now know it's not Bag C, so it can be taken completely out of the problem.

So, we are left with 2 possibilities:

You are holding Bag A.
You are holding Bag B.

If you are holding Bag A, the coins will match.
If you are holding Bag B, the coins will not match.

You are no more likely to be holding Bag A than you are to be holding Bag B.

There is a 50% chance you are holding Bag A and the coins will match.
There is a 50% chance you are holding Bag B and the coins will not match.

It's a 50% chance.

That's how I see it at least.