Why I Love Mathematics (the real thing)

by Vlad Dolezal on October 25, 2008

Aaaargh! He said the M-word! Run for your lives!

Wait up! If your reaction to the words mathematics is “Oh, it’s that really annoying boring thing”, or “Oh, it’s that really hard and complicated thing”, we’re not talking about the same thing. In fact, if you think like that, you probably never experienced REAL mathematics.

Sure, you did a thing called mathematics at school. But I could just easily make a subject at school called “sex” where you would memorize sexual positions and human anatomy and pass exams and get a degree. And yet it wouldn’t be anything like REAL sex.

So let me tell you what I love about real mathematics. Maybe, just maybe, I’ll make you curious enough to give it a try.

What real mathematics is like

If you only experienced mathematics at school, you might think it starts with some boring formal definitions, then some boring formal algebra, followed by a boring formal proof, from which you get a mildly interesting result.

In reality, mathematics starts from the other end. It always starts with a riddle.

Here’s a riddle you might have heard before:

A farmer has a fox, a goose, and some cabbage. He wants to cross a river on his boat, but the boat is small, so he can only take one item with him at a time. He can’t leave the fox with the goose, or the goose with the cabbage, because the former would eat the latter. How can he cross the river?

The answer is quite simple, so I won’t even tell you. And the answer is where it would end for any normal human. But not for a mathematician. As mathematicians, after we enjoy the glowing feeling of having solved the riddle, we wonder what would happen in similar cases. Let’s say the genetically modified cabbage suddenly becomes conscious and decides it loves eating foxes. Is there still a solution?

(The answer is yes. Screw the goose and the fox! You’ve got frickin fox-eating cabbage on your hands! Get on national TV, become famous, then sell the cabbage on e-bay and never have to work again. Or, alternately, build up an army of fox-eating cabbages and try to take over the world. Or kill the cabbage before it turns against you, because you’re obviously starring in a  third rate horror flick.)

Then, if we feel like it, we might go on to make some generalizations. How many animals could we transport like this if we had two places on the boat? How about three places? What about the general case of having n places? What if the relationships between what animal eats what were more complicated than a simple top-down chain?

Here’s the fun thing. No one is forcing us to do this. If we get bored, we can just leave all the gooses and fox-eating cabbages behind, and go weigh balls on a balance scale, or square a circle with just a compass and an unmarked ruler (which is impossible btw).

But what are the applications?

I can hear some of you thinking… “But what are the applications of knowing how the farmer can transport his stuff across the river?” Let me give you a long and complicated answer:

None.

Ok, now on to the short and simple answer…

What are the applications of paintings? Of playing music? Of playing chess?

Sure, painting skills can be used to make advertisements more effective, music can be used to add soundtracks to make movies sell better, chess can be used… well it can’t :). But anyway. The point is, we don’t paint, or play music, or play chess, because it can give some actual results. We just do it for fun. We enjoy it. To paraphrase Richard Feynman:

Mathematics is like sex. It can give practical results, but that’s not why we do it.

Get over the idea that mathematics is just something used by engineers and physicists to solve problems. Mathematics is an art in itself, just like music or drawing.

Some more examples of real mathematics

Here are a few more examples of mathematics. I’ll give you an easy example, a moderately hard example, and an evil example.

1. “The two kids” riddle

Imagine you’re chatting with a friend about one of your common acquaintances.

“I heard she has two kids,” you say.

“Yeah, that’s right. By the way, I met her yesterday at the supermarket, and she was with a small boy. We started chatting, and she told me it was her son. So at least one of her kids is a boy.”

What’s the probability that both of her kids are boys?

(Hint: Think hard. It’s not as straightforward as it seems.)

(Update 28/10/2008: Well, it turns out my original answer to this riddle was wrong. Check out the comments for more detail. Hey, you get to see me being proven wrong! :D )


2. The coin tossing riddle

Imagine you’re tossing a coin, and recording the results in a row like this: TTHTHHT…

I’d like you to consider two cases. In the first case, you keep tossing the coin until you get HTH, and then you stop. In the second case, you keep tossing the coin until you get HTT and then stop.

If you tried both of these cases a few thousand times, you would find that, on average, in one of the cases you will stop sooner than in the other. Which one and why?

3. The famous ball weighing riddle

This riddle is satisfyingly hard. (no, satisfyingly wasn’t the first word I thought of). It took me personally a couple of days to solve it.

Imagine you have twelve balls. One of them is either heavier or lighter than the rest, but you don’t know which ball, and whether it’s heavier or lighter. Can you find out which ball it is, and whether it’s heavier or lighter, by doing just three weighings with a standard balance scale? (one that simply tells you whether the balls you put on one half are heavier than the ones on the other half, but nothing else). You can number the balls for reference.
I won’t give you the answers to the above riddles. In fact, my favorite site for riddles is wu riddles. And you know why? Because it doesn’t tell you the answers :) (Yes, that’s a good thing. Look out for an article about “activation energy”, in which I will explain that in more detail.)

I’ll just let you enjoy thinking about the riddles.

Where are the numbers?

Wait a second. I’m writing an article about mathematics… and yet I haven’t written a single bit about numbers yet? (apart from the two kids riddle). Surely mathematics is all about numbers?

That’s because I don’t think numbers are the essential part of mathematics. It’s logic.

“When a problem has a correct solution, and the solution can be PROVEN to be correct, that’s mathematics.”
- Me :D

In other words, what I like about mathematics is the 100% certainty that a solution is correct. And if there is no solution… then there is a way to prove that with 100% certainty.

Sure, I still enjoy riddles that rely on real-world intuition. Where the answer makes a lot of sense, and any other answers are either unlikely, or too complicated. But it’s the 100% certainty that I really love about mathematics.

Is mathematics for you?

Did the riddles above make you think? Did you enjoy it? If yes, you would enjoy mathematics (the real mathematics, anyway). If not… then mathematics is not for you. And that’s fine too.

I hope I cleared up some misconceptions about what makes mathematics mathematics. (god, that was a pain to spell :p “mathematics mathematics mathe… argh! maths! why can’t I just call it maths!!!”)

If you think mathematics sounds like fun, and want to try some more, just google around a bit. I’m sure you will find plenty of interesting math questions out there. Or, for more fun riddles (that aren’t necessarily related to maths, or have a single solution), visit wu riddles. Cheers!

Afterword

If you’re a math geek, you might be shouting at the computer screen by now.

“What the hell is this guy talking about? Mathematics is about numbers, and their sequences, and beautiful patterns, and wonderful geometrical theorems. Not some silly riddles! Aaargh!”

And you’d be right. Mathematics IS about all those things. But when they were first discovered, they started as a riddle. As a nagging question inside some mathematician’s head. Is the sum of angles inside a triangle constant? How long is the circumference of a circle, compared to its radius? Can the diagonal of a square be written as a fraction?

I believe mathematics isn’t about the knowledge. It’s about enjoying the thinking. And I just wanted to introduce the concept to non-math folks. Cheers!

{ 53 comments… read them below or add one }

Nick October 26, 2008 at 03:19

I think putting the mathematics in a context really helps make it more interesting. I took trigonometry in high school, and absolutely loathed it, but then when I took statistics (which isn’t just raw number crunching, at the very least there’s a context) I enjoyed math immensely, and the same goes for a formal logic class I took last year. Formal logic is absolutely amazing, and I was wondering how much of it you’ve been exposed to? I get the feeling you’d enjoy it a lot, and by formal logic I mean propositional calculus in case I’m being inadvertently unclear. You should look into it.
A book that provides a pretty good jaunt into it and ties it in to other more humanities-esque things is Godel, Escher, Bach by Douglas Hofstadter. His focus is on how cognitive activity arises in the human brain, but you get a good dose of logic (as well as Bach and Escher). And it’s a pretty straightforward read, not exactly a breeze, but it’s not hard and not horribly bogged down by technicalities. And it’s a really insanely awesome book. Hell, you may have already read it, Vlad, but if I can get at least another commenter to look into it, I’ll be happy. Damn fine book.

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fairyhedgehog October 26, 2008 at 11:20

I used to be frightened of maths but I’m learning not to be. My first real understanding that maths might be fun was when I read Fermat’s Last Theorem.

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shAdOwArt October 26, 2008 at 14:13

That’s one hell of a goldmine you’ve dug up there ;)

The twelve balls problem was was a hard nut to crack. I’ve been pondering it for many of my waking hours since I first read it yesterday evening and I think I’ve at last found a water-proof algorithm.

In high school we had an university lector who came over once a week to give extra classes in ‘real’ mathematics. He used to argue the same points as you have here. I’ve only gone to college for half a semester but so far there’s only been engineering maths (guess I’ve got myself to blame for going to an engineering school). Pleasure maths (if you don’t mind the expression) are much more amusing indeed.

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bytenirvana October 26, 2008 at 14:27

Hi – I’m long time silent reader – nice post about maths and why it isn’t that boring (And you even mentioned Feynman briefly ;) ), but can you give the answer for the riddles somewhere (or atleast the links to them)? kthxbai

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Vlad Dolezal October 26, 2008 at 21:40

@nick:
Yeah, maths is so much more fun when you see where you’re coming from and where you’re heading, instead of just dry number crunching. And no, I haven’t read that book, but I’ll definitely look into it.

I also have two friends who are doing their PhDs in some area of logic. So you’re the third person telling me this logic thing is fun! I’ll have to look into it more :)

@fairyhedgehog:
I also loved Fermat’s Last Theorem! In fact, I think it was the first maths book I ever picked up at a bookstore. (Oh, no wait. It was the second one. The first one was about mental arithmetics. Never mind.)

@shadowart:
I love the expression “pleasure maths” :)

@bytenirvana:
Well, giving you the answers would kind of ruin the point :). But if you’re stuck, just drop me an e-mail at vlad@anamazingmind.com and explain your thoughts. I might give you a hint without giving away the answer :)

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shAdOwArt October 27, 2008 at 16:16

I’ve been pondering the two kids riddle some more, and come to the conclusion that it indeed is as straightforward as it first seems.

The riddle, as it is often presented (like the “Two Coin Flip” one on wuriddles), is quite different from the one in this text. Of course this could be intentional too, and I just had a weird first perception. Or maybe my maths are just wrong… Anyway, I’d like to settle this. Mind me spoiling the riddle in order to do so?

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Vlad Dolezal October 27, 2008 at 19:14

@shadowart:
Okay, permission granted :). Feel free to write your thoughts here. Then I can respond with my thoughts and we’ll see if we can agree on an answer.

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shAdOwArt October 27, 2008 at 21:08

The birth of two kids may generate four different outcomes gender-wise: boygirl, girlboy, girlgirl and boyboy. The knowledge that at least one is a boy eliminates the girlgirl possibility. If this had been all we’d known the chance of dual boys would have been one in three. This is the answer to the shorter version of the riddle and what I suspect you’ve been hinting at, forgive me if I’m wrong.

However, if we look closer we’ll discover the magic words: “she was with a small boy” and “she told me it was her son”. Thanks to these sentences we do not only know that at least one kid is a boy, we do also know exactly which one of the kids it is – and this makes all the difference. As we now can tell them apart we can eliminate all possibilities where the kid that was presented to our friend is a girl. Which leaves us with two different possibilities, of which one is dual boys.

I can also imagine a totally other interpretation of the question where the answer would be one in four. Though I think the word “are” in the last sentence eliminates that possibility.

So let’s hear your thoughts =) I haven’t done probability calculations in some time and is a little unsure whether my identity argument is sound.

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Mattsville October 28, 2008 at 13:13

Quite right Shadowart. Although we can only assume that 1/2 was seen as the obvious answer. It may be that 1/3 was assumed to be the obvious answer, and when you think futher you realise it is actually 1/2.
I think in this case though that you’re right, and the writer was thinking that 1/2 is incorrect.
The key here is that the two statement “She was with a small boy and she told me it was her son” is NOT the same as
“She has at least one son” as you explained above. We are also however ignoring the fact that the womain may be lying – but that’s just being silly :-)

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Vlad Dolezal October 28, 2008 at 13:16

I’ve thought about it a bit, and here’s what I’ve come up. (And no, it’s definitely not what I expected I would come up with.) I see two possible trails of logic:

1. She was with a boy. That’s one kid. The other kid can be either a girl or a boy, which gives us a probability of 1/2 of her having two boys.

2. Imagine we did this as an experiment. Let’s assume she has two kids, who are each of random gender. Then, each day, she randomly picks one child to take to the store with her. (reasonable assumptions so far?)

The cases are, as you said, boyboy, boygirl, girlboy and girlgirl. Now clearly if she had two girls, she would never take a boy to the store, so that didn’t happen. When our friend encountered her, she had a boy with her, so she has to have boygirl, girlboy, or boyboy. So… boyboy is only one of three cases. BUT… there are two boys in boyboy, so meeting her with a boy at the store would happen twice as often as with the other two possibilities. In other words, the chance is 2/4, or 1/2.

Some final thoughts…

So if I say “I tossed two coins and at least one came up heads. What’s the chance both came up heads?”, the chance is 1/3. But if I say “I tossed two coins, I looked at one of them and it’s heads. That means at least one of them came up heads. What’s the chance both of them came up heads?” – the answer is 1/2.

Did we just change the probability just by observing the coins? Quantum probability :D

Let me know if you have any more thoughts on this.

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shAdOwArt October 28, 2008 at 19:33

With one mighty look we’ve changed the “initial” conditions, and hence also the probability. What a power we possess ;)

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fairyhedgehog October 30, 2008 at 17:32

I’m not sure if you do tags but you’ve one of my favourite blogs so I’ve tagged you.

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axel g October 31, 2008 at 17:10

Great site!

Found you thru StumbleUpon +_+

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Brian November 7, 2008 at 01:36

That was an absolutely fantastic post. I try to convey the same thing on my blog, yet have never put is as succinctly. Kudos!

Brian (a.k.a. Professor Homunculus at MathMojo.com )

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Dustin November 8, 2008 at 14:58

I almost hate to leave a comment like this because of how it’s going to sound…

Regarding the riddle of the 2 kids…it seems to me that the solutions that are being presented are WAY over-complicating the matter. Wouldn’t it be logical to make the assessment this way:

Since we know one child is a boy and we are trying to figure out the probability that the other child is also boy, the (in my mind) correct way to pursue this is. What are the odds that the ONE unknown child is a boy? It comes up with the same solution (1 in 2), but without all the extra fodder. Really, we’re not trying to determine the probability that both children are boys (because we already know one is). We’re trying to determine the probability of ONE child being a boy.

Secondly (and this is where it may sound like I’m being boastful, but really that’s not my intention), but am I seriously missing something in the riddle of the 12 balls? I had a solution in under a minute and it seems so unbelievably simple that I am wondering if I have misinterpreted something or my solution (somehow) just couldn’t work…but again…it seems so simple. Could I email Vlad Dolezal to discuss this solution?

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Dustin November 9, 2008 at 14:38

In the spirit of being honest – I figured out during the course of the day yesterday why my solution for the riddle with the 12 balls would not work. It took me actually starting to say the solution out loud to someone I know for me to realize that there was a BIG flaw in my solution.

To keep a long post short – turned out my solution would only work if the ball was heavier.

Oh well, at least I figured out that I didn’t figure it out! Now I’m a little afraid to try to actually figure it out! lol

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caleblicata November 9, 2008 at 19:35

oh, so 25% is wrong >.> i was terrible at probablity

i’m in algebra 2 currently, but i don’t find it terrible at all =p (at least, once we got into quadratic expressions… i thought they were fun)

i guess i’ve got trig/precalc and calc to start hating school math again though

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Aish A November 24, 2008 at 21:49

This post is as important for parents as it is for kids. I emphasize to parents that work with us constantly that they should not give a negative spin to math for their kids. They should try to make math a fun activity and information provided in articles like this can really help parents to do that.
It’s easier for students to relate with things that have a practical application as opposed to just a topic in the books that they have to learn for the upcoming test.

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Matt November 26, 2008 at 16:28

I think that 1/3 is the right answer all along. First of all, when Vlad says ‘So if I say “I tossed two coins and at least one came up heads. What’s the chance both came up heads?”, the chance is 1/3. But if I say “I tossed two coins, I looked at one of them and it’s heads. That means at least one of them came up heads. What’s the chance both of them came up heads?” – the answer is 1/2.’

Vlad is wrong. Those statements are in fact the same. You failed to say that you saw the FIRST coin was heads or the SECOND coin was heads. If you looked at 1 of them and saw that it came up heads…that means that at least 1 of them is heads. DIFFERENT statements would be, I flipped THE FIRST COIN and it came up heads. For example: I peeked at the coins and saw 1 of them is heads. First coin could be heads or tails AND the second one could be heads or tails (but there must be at least 1 heads). If you say that you saw the FIRST COIN as heads, that means that THE FIRST COIN MUST BE HEADS and the second coin could be heads or tails. If you still don’t understand what I’m saying, label the coins: Coin A and Coin B. If you say you saw one of them heads, it could be A or B. Just as if you said there is at least one head, it could be A or B. The sentences ARE in fact the same. The flaw in shadowart’s reasoning is the same. You don’t know if her boy is the first or second child (child A or child B). YOU DO KNOW that one of them MUST be a boy.

Since you don’t know if the boy is the first or second child, but there must be a boy, the possibilities are:
BoyGirl
GirlBoy
BoyBoy

Hence, there is a 1/3 chance that the OTHER child is a boy. The owner of the site was right to begin with, it is 1/3.

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Ava November 29, 2008 at 21:18

Thanks for the post. I loved the Feynman quote. I teach 7th grade though, so I’m not sure I can use it with them (well I could but it would require more explaining than they would be comfortable with…)

Anyway, I enjoyed working through your riddles, and I get what you’re saying about math being fun (I’ve always loved math myself), but I’m still stuck on one point here. Many of my students don’t feel any personal connection to geese, cabbage, coins, etc. Solving riddles seems meaningless to them and not fun (they would rather read, write, draw, whatever…). Some of these students are even quite adept at math. I know there are lots of cool math projects (you brought up fractals, tesselations, etc.) But at some point they all need to learn how to solve an equation. And this as well can seem meaningless, especially in the context that many textbooks and riddles provide. This is just my experience in my classroom… It’s an issue I am continually struggling with in my classroom. How do you make math meaningful to students who don’t innately find it fun?

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Vlad Dolezal December 1, 2008 at 02:10

@fairyhedgehog:
In the end, I got around to answering your tag. Check my post “Random Ideas” :)

@Dustin:
Not as simple is it looks, huh? :)

If you haven’t found the answer yet, maybe first try answering the same riddle with only 8 balls. And then with 10. The answers you come up with may give you some pointers.

@Aish:
Yeah, well-meaning parents that actually make maths sound really boring and annoying are unfortunate. We can’t really fix their perception of maths. But we can still influence today’s kids!

@Matt:
Ah, you fell into the same fallacy is as me :)

If you look at your explanation:

BoyGirl
GirlBoy
BoyBoy

Notice the last case has two boys. In other words, that case will happen twice as often as the other two (first boy being picked, and the second boy being picked). If you still don’t get it, maybe we could try hacking a quick python program to simulate this, run it a million times, and see what happens :)

@Ava:
Actually, Feynman’s original quote was about physics, not mathematics. I was paraphrasing :)

And how to handle people not caring about mathematics? I have a solution in mind:

Just STOP FORCING THEM. An ordinary person only needs addition, multiplication, percentages, and maybe some basic understanding of graphs. They don’t need Pythagoras, or solving quadratic equations. They DEFINITELY don’t need integration.

I would make all the rest of maths optional. The maths would be more fun, the pupils would be more motivated, and the teachers would have more freedom and a more positive environment. Those who would choose not to continue maths beyond the basics won’t get anything useful out of the current system anyway. Everybody wins!

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Sky June 1, 2009 at 01:29

With the fox+goose+cabbage riddle, is the answer supposed to be that it’s unsolvable? Since he can’t leave the goose with the cabbage, or the fox with the goose, then his only option for his first trip is to take the goose across the river and leave the fox with the cabbage. But then for his second trip he must either take the fox or the cabbage across the river, therefore leaving either the goose and fox or the goose and cabbage alone at the destination when he goes back to make his third trip.

Or am I just missing something big?

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Vlad Dolezal June 1, 2009 at 01:43

@Sky:

Yes, you’re missing something, the riddle can be solved. Don’t feel too bad though.

That’s why I chose this riddle. It’s so simple you can explain it to a 4-year old… but the solution isn’t immediately obvious, and it takes a stroke of insight to solve. Keep at it, and you’ll find the solution.

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Sky June 1, 2009 at 05:25

D’oh. Got it now.
*Facepalm*

That’s one of those kinds of riddles that a 4-year-old would probably get right away. Too much in-the-box thinking for me.

I did manage to solve the other ones though (the balls one I first heard a while back).

The children one can be confusing, but is essentially just an application of Bayes’ Theorem. If we can rephrase the question as asking: “What is the probability of both kids being boys given that at least one kid is a boy?”, then by Bayes’ Theorem, this is equal to:

[the probability she has two boys and she has at least one boy]
————————————————————————
[the probability that she has at least one boy]

The probability of the top is simply 1/4 (i.e. the general probability of her having 2 boys, given no prior knowledge), and the probability of the bottom is 3/4 (i.e. either BG, GB, or BB). (1/4) / (3/4) = 1/3. So, the probability of her having two boys given that one child is a boy is 1/3.

And for those who haven’t seen it yet, here’s a really fun puzzle:
http://xkcd.com/blue_eyes.html

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BPS June 1, 2009 at 06:21

For the coin-flipping riddle, shouldn’t the second permutation be TTH? I think HTH and HTT have equal probability of being reached, while TTH has a clear advantage.

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Vlad Dolezal June 1, 2009 at 06:37

@Sky:

Congrats on the fox-goose-cabbage riddle!

With the two-kids riddle… your solution is exactly what I had in mind when writing it. But it turns out the way I wrote it, the solution is different :p

Esentially it boils down to the difference between “I looked at one kid, and it’s a boy” and “I looked at both kids, and at least one is a boy”.

I’ve heard the blue eyes riddle before, but I haven’t figured out the solution yet. Thanks for reminding me!

@BPS:

I definitely meant HTH and HTT as the two possibilities.

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E June 1, 2009 at 11:15

I am still not sure about the two kids riddle. It seems to depend on the assumption that she randomly chooses one kid to take to the store with her each time, so that any given time you see her there is a 50% chance she will have either kid. But, I don’t think that is really a fair assumption. I have two kids, and I often go to the grocery store with only one of them. But, I don’t take each equally often – I actually take one much more than the other. (Really, I don’t know if I have even taken the other to the store alone at all in the last year or so.) It just so happens that the one who comes to the store with me is a boy. If you happened to meet me in the store with my son, what would that tell you about the probability of my other child also being a boy?

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mex June 1, 2009 at 11:43

Well, my solution for the ball game relies on the fact that you don’t weigh your complete set of balls but always leave two out. thus if your weighed balls are equal you’re just one weighing away from the real ball by weigheing the two you left out. does that work? so you weigh 5vs5 leave 2 out. then you’re left with 5 candidates and you weigh 2vs1 and leave 2 out and the next weighing you definitely got your balls right. took me only half an hour so i have my doubts if its correct b/c the standard solution on google seems a little bit different. pretty ballsy riddle

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Vlad Dolezal June 1, 2009 at 12:07

@E:

Good question, I’m trying to think if that influences your probability.

Of course, you could also point out that the girl-to-boy ratio worldwide is 53-47 or whatever, but let’s not get too exact, ok? :)

@Mex:

The problem is, you don’t know if the ball is heavier or lighter than the others. So if you weigh the groups of 5, you will (usually) find out that one is heavier that the other. But you still won’t know which one contains the special ball. Hope that helps.

And you’re right, it’s a very ballsy riddle :D

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Jacob June 1, 2009 at 17:55

Well I was bored and waiting for hot water, so I wrote a program to test at least one of the cases we’re talking about. As far as the coin flip, if you know one of them is heads then the chances that both of them are heads is 1/3. If you don’t believe me compile this ;).
#include
#include
#include
using namespace std;

int main()
{
srand(time(0));
int coin1,coin2,oneHeads=0,bothHeads=0;
for(int i=0;i<100000;i++)
{
coin1=rand()%2;
coin2=rand()%2;
if(coin1==1||coin2==1)
oneHeads++;
if(coin1==1&&coin2==1)
bothHeads++;
}
cout << (double) bothHeads / (double) oneHeads << endl;
return 1;
}

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Sky June 1, 2009 at 22:51

@Vlad: Ah, I think I see where you’re coming from now. The probability of *seeing* one boy at the store is actually different from the probability that she has (at least) one boy, so the way I originally rephrased the problem was wrong. So I think we are really looking for is this: P(she has two boys | we *saw* one boy).
And that’s equal to this:

P(she has two boys and we saw one boy)
———————————————-
P(we saw one boy)

The top is (1/4 * 1), since the probability of her having two boys is 1/4, and the probability of us seeing a boy in that case is 1.

The bottom is P(she has two boys and we saw a boy) + P(her youngest is a girl and we saw a boy) + P(her oldest is a girl and we saw a boy) + P(she has two girls and we saw a boy) = (1/4 * 1) + (1/4 * 1/2) + (1/4 * 1/2) + (1/4 * 0) = 1/4 * (1+.5+.5+0) = 2/4 = 1/2

So the total probability we’re looking for is (1/4) / (1/2) = 1/2.

Man, I don’t think I’ve made use of my whiteboard this much since I bought it. :-)

And as for the blue eyes riddle, a good hint is to try to think inductively. Build it up from base cases. ;-)

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Vlad Dolezal June 2, 2009 at 00:03

@Sky:

Ya, that’s what I got for the two-boy riddle too, after a reader corrected me! (I also originally thought it was 1/3.)

You have a whiteboard at home? Damn you! ;)

And about the blue eyes… it completely didn’t occur to me to start with the basic cases. Even though that’s my usual approach with these riddles involving big numbers. I’ll get right on it!

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Vlad Dolezal June 5, 2009 at 09:13

@Sky:

Ahhhh, I finally got it!

Damn that was an evil riddle.

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Katie August 2, 2009 at 01:15

As far as the two kids riddle is concerned, I don’t think whether it’s boygirl or girlboy matters at all. The chance in either of those cases that both are boys is the same, zero. So you can just pick one.

Her children are either boyboy or boygirl, meaning the chance that both are boys is 1/2.

Her children are either boyboy or girlboy, meaning the change that both are boys is 1/2.

I don’t see the difference logically in these two statements, or the reasoning behind numbering or ordering the kids, for me that just makes it too complicated, but maybe I’m missing something important there.

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Vlad Dolezal August 2, 2009 at 09:19

@Katie:

I don’t think you’re missing anything. It’s just that we all need a different way of coming at the right answer (and avoiding some mistakes along the way).

The reason we’re messing with the whole boygirl vs. girlboy difference… let’s say you toss two coins. Quick, what’s the chance of getting one heads and one tails, as opposed to getting two heads?

Hopefully you answered that one heads and one tails is twice as likely. Because it is.

And the reason we were arguing about the girlboy vs. boygirl is the semantic difference between saying “given that at least one coin lands heads” and “given that the first coin lands heads”. Because that makes the difference between your chance of getting two heads being 1/3 and 1/2.

(I think I’m not explaining it very well. Oh well. As an excuse, here’s a picture of a kitten: http://tinyurl.com/dcfwga )

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Aurora August 31, 2009 at 21:55

This is more psychological than mathematical, but for the two kids problem…if both of her children were boys, wouldn’t she most likely have said that was her oldest/youngest son, not just her son? So it’s most likely that her other child is a girl, since she just says the boy is her son, with no clarifying adjectives.
Still completely stumped on the blue eyes riddle, though.

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Vlad Dolezal September 1, 2009 at 12:18

@Aurora:

I like the way you’re thinking :p. I guess this should teach me to start riddles with “assuming an idealized spherical mother in vacuum…”

Or maybe not.

As for the blue eyes, have you tried starting from the basic cases and then working up the numbers? That always helps with riddles involving “1000 prisoners” etc.

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Aurora September 1, 2009 at 15:46

Discussed it with a friend, and we finally got it. Still working on the balls one, though. I’m not as good at this sort of thinking as I’d like to be…

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shAdOwArt October 1, 2009 at 16:49

A lovely riddle, that ‘Blue Eyes’ one. Somehow I always find it much more rewarding to solve riddles that have been asked by someone rather than riddles found among a list of 100 other riddles.

Here’s two for future readers chew on!

Take the ball riddle, but modify it to n balls and k allowed weightings, when can the odd ball be found and it’s nature identified?

What is the product of all the natural numbers? All the whole numbers? All the rational numbers? All the real numbers? All the complex numbers? Always excluding zero of course.

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fadhilah July 30, 2010 at 15:14

i love math…i start learn math when i was 5 years old…now i learn precalculus…i feel so excited to know more about math

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David Bandel May 23, 2012 at 15:21

Though this view of mathematics is somewhat true and could possibly help the math-phobic to appreciate the finer points, it doesn’t really escape the real issue.

Less intelligent people are generally not drawn to mathematics. More intelligent people generally are.

You talk about the boring bits leading up to the interesting bits.

Those initial boring bits were fascinating to me as much as the so called “interesting bits”

I loved arithmetic and elementary analysis long before I got into the theoretical side with algebra and geometry. (those who think I’m talking about high school algebra or geometry have absolutely no clue what I’m talking about) I didn’t need to look at it backwards to understand it. I was taught in the conventional style and it did no harm because I am a genius with a gift for mathematics.

If this is about enabling people who don’t have a talent for math to enjoy it.. what’s the point?

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JE EF October 25, 2012 at 01:22

This guy is retarded. He is so bad at math its funny.

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Toiler in Darkness November 23, 2012 at 23:34

I’m afraid the answer IS 1/2.

Before we know about any of the kids, our problem space is [Gg,Gb,Bg,Bb], once we see the one B/b we know it’s not Gg. Your error is in not assuming which it is between Bg and Gb, then assuming which it is between the 2 in Bb. There are 4 boys in the problem space: [Gb, Bg, Bb], of those B/bs, half have brothers, and half have sisters.

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Dereck March 18, 2013 at 04:35

@david bandel

You got bullied out of one forum, so you felt like boosting your confidence here? Can you please enlighten me on your genius? We are talking about mathematics, here, not yourself. Please, perhaps you would now like to share some of your mathematics?

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David Bandel March 18, 2013 at 11:12

Hmm. I guess you probably lack the intelligence to grasp this concept but you are more or less the enemy of rational discourse.

All you’ve done is launch a series of assertions containing false premises at me.

Try again. One question that contains zero assumptions and we’ll see if you warrant a response.

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Ejazz April 13, 2013 at 19:23

The twelve balls problem:

1. The following problem can be rephrased as per follows:
There are 12 positive integers which are denoted by a,b,c,d,e,f,g,h,i,j,k and l
One of them is either greater than or smaller than the other integers. The rest of the integers are equal. Using only three known relationships between them, can you determine which number is smaller or greater than the others and whether it is smaller or greater?

2. How to solve it is beyond me as of now.

3. A few tries:
a) weighing 4 balls with 4 balls
assume they are equal
b) then the other 4 ball contains one greater or one smaller ball than the other
ball.
c) Now choose one ball and weigh it with another ball (from the 4 balls, of
course!) assume they are equal. Using this one ball, weigh it with another
ball. If it is equal, then the last ball is either heavier or lighter than the other
balls.
Problem is even after so many assumptions, I can’t tell you whether the ball is heavier or lighter( and that is with so many assumptions made!)

I will try again.

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Ejazz April 13, 2013 at 19:37

To all…(Spoiler of twelve balls question!)

How about this other approach?

1. Weighing 4 balls with 4 balls.
Assume they are equal.
2. Now, there are 4 balls left. Out of this 4, choose 3 balls. Measure with 3 balls
from the 8 balls above. If they are equal, then measure last ball with any balls
see if it is heavier or lighter. If they are not equal, then choose 2 balls out of
this 3 balls and weigh them up. If they are equal, then last ball is bad ball.
If they are not equal, then from the measurement with the 3 balls vs the 3
balls, the ball is either heavier or lighter.
3. Question: if balls measured are not equal( from 1)??
then I would have to mix the balls, although I am not exactly sure how?( the
fiber details have yet to be worked out).

Is this method acceptable or is the first assumption I made wrong and that this general method of attack is unacceptable??

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Ejazz April 13, 2013 at 19:38

Finer**

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Vlad Dolezal April 14, 2013 at 10:46

That depends on whether or not you can figure out these “finer details” in a way that always reliably gives you the answer ;-)

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Dan June 15, 2013 at 13:17

Hi Vlad,

I have a question about the toin cossing riddle. I hope you don’t mind me sharing my thoughts on this. The way I see the procedure is that you keep flipping coins until either HTH or HTT appears, then you record which one of the two it was, and you repeat. After many games, you compare which outcome appeared more often.

So on any one run, you keep flipping coins until the first H appears, then you flip until T reappears, for example
TTTTTTTHHHHHHHHHHT
or just
HT
The game won’t have stopped by then, but now you _know_ it will stop on the next toss of the coin! But both outcomes are equally likely for the next toss, so how can one outcome occur more often? Did I misunderstand the procedure?

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Vlad Dolezal June 16, 2013 at 17:52

Hi Dan,

yes, you misunderstood the procedure slightly. We’re considering 2 cases:

1. You keep flipping until HTH appears.

2. You keep flipping until HTT appears.

So in the first case, if you get HT, you have a 50% chance of stopping on the next turn, and a 50% chance that you’ll get the wrong side of the coin and have to keep flipping.

Same thing in the second case. You don’t mix the two cases, you simply pick one case, and keep flipping until the specific pattern you chose appears.

Does that make sense?

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Dan June 28, 2013 at 20:35

Vlad – the mixing is just done artificially if you will.
You claimed that on average in one of the cases you stop sooner than in the other. This I understand to mean that on average, one of HTT or HTH occurs sooner in the coin tossing process than the other.

So let me illustrate my “mixing of the two cases” by an example. Suppose that I claim that HTH occurs slightly more often and you claim that HTT occurs more often. So we agree on a game: we sit down together and we get an impartial judge to repeatedly flip one coin. Whenever HTH occurs before HTT, I get one dollar from you – but if HTT occurs first, you pay me one dollar. Then if I’m correct, I will win money from you (on average) and lose if I’m wrong.

So here we have one coin and record the outcome like in my previous post, so that my reasoning should be correct…

By the way, I agree that some outcomes tend to occur earlier than others, for example THH occurs before HHH on average. This is since if the first toss is a T, there is no way that HHH occurs before THH.

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Vlad Dolezal July 1, 2013 at 07:41

Your new example again modifies the conditions slightly, Dan.

Because under your conditions, we are now saying “count the HTH until either HTH or HTT occurs. Also count the HTT.”

A way of sticking with the original assignment would be to keep flipping until both HTH and HTT have occured, and just count how soon each one occured during each run.

Believe it or not, it makes a difference. In the first case, you would find both of them occurring equally early on average, while in the second case, you would find one of them occurring earlier on average than the other. (On average meaning the average of how quickly each of them occurred. So if it occurs once after 5 flips, and once after 11 flips, that would be 8 flips on average.)

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