Math oddities

[Kurzyk]

Someone pointed something interesting out to me today.

If you take how old you'll be on your birthday this year and add it to the last two numbers of your birthyear, it will equal 111.

So for me, I was born 5/4/1972.

Ill be 39 this year + 72 = 111.

Try it. Everyone so far equals 111 and it seems to just be for this year.

[Jude]

"If you take how old you'll be on your birthday this year and add it to the last two numbers of your birthyear, it will equal 111."

If you are 10 or younger this actually doesn't hold up.  Only for people born in the last century does it pan out.  However, in all circumstances you will get 11 or 111, and that's just because you're adding when you were born + how long you've been alive to get the current year, 2011, basically.  The extra 100 comes in because we're truncating the first two digits (19) from your birth year, so that 100 doesn't add to it to make it (20).

[Kurzyk]

Cool thank you. I was wondering how that worked.  :-)

[Wolfy]

I was born in 90.

This year I turned 21, which does add up to 111.

However, next year I'll be 22, and that adds up to 112.

:/

[Oniya]

But next year will be 2012.

[Y]ou're adding when you were born + how long you've been alive to get the current year, 2011, basically.  The extra 100 comes in because we're truncating the first two digits (19) from your birth year, so that 100 doesn't add to it to make it (20).

[Kurzyk]

Ok here's another one. I found it on a math oddities website:

Three guys walk into an hotel and ask for a room with three beds. They just want to stay for one night. The man at the hotel tells them that the cost will be 30 dollars. They pay 10 dollars each and walk upstairs to their room. After some minutes the man knocks at their door, apologizing for a mistake. He made them pay the high season price and not the low season one, and since now it's the low season, he must give them 5 dollars back.

They are happy and appreciate the honesty of the man, so they tell him: "from these 5 dollars we will just take back 1 dollar each and we will give you the remaining 2 dollars as a tip for your honesty". The man thanks the guys and walks out.

Ok, they paid 10 dollars each (30 in total). They take back 1 dollar each (3 dollars in total). In this way they have paid 10-1=9 dollar each. So in total they have now paid 9*3=27 dollars. Plus the 2 dollars tip given to the man we get 27+2=29 dollars. Where is the missing dollar?

In reading this, my impression is that the dollar isn't really missing. Is it? I mean they gave 30, it was supposed to be 25, so they received 5 back. They kept 3 and gave 2.

Yet looking at that math it does look like 1 is missing.

[Oniya]

Actually, they paid 27 to the hotel, 25 of which went to the room, and 2 of which went to the man.  The 2 is subtracted from the 27 to get the price of the room.

[Kurzyk]

Oh.. so they initially gave him 30, got 3 back and gave him 2, leaving 1 missing.

But if the price of the room is 25 off season, and he gave them 5 back, does that mean that the innkeeper kept the extra dollar?

[Oniya]

No.  The way it goes is $30 (what they paid initially) - $3 (what they got back) = $25 (what the room cost) + $2 (what the manager got).

[Kurzyk]

Right. That's what I was initially thinking. So then there is no extra dollar? They paid 30, got back 3 and gave him 2. It still totals 30.

[Oniya]

No extra dollar.  If you think of the $5 as five $1 bills, two of them never leave the possession of the hotel manager.

[NotoriusBEN]

yep, what oniya said.

plus, that money swap with the hotel manager is one of the "oldest" scams in the book.  When I worked in retail years ago, I fell for it a couple times.

If you work in retail, do like they do in Vegas. Keep all money on the table and only transition it when the final amounts are known. Beware the person talking a lot of different numbers and asking or trying to change money mid transition. :P

[DreamlandDenizen]

"Ok, they paid 10 dollars each (30 in total). They take back 1 dollar each (3 dollars in total). In this way they have paid 10-1=9 dollar each. So in total they have now paid 9*3=27 dollars. Plus the 2 dollars tip given to the man we get 27+2=29 dollars. Where is the missing dollar?"

The problem with the math is you should not add the two dollars to the 27 near the end. You say they take 3 dollars back (not five) so the 2 dollar difference stays with the manager the whole time and is thus already accounted for. So leaving out that error you're left with 27, the amount they actually paid in the end, and to get back to the 30 you just add the $3 they took back.

[ReflectManSmile]

This last oddity reminds me the monty hall problem, in where you have a result that seems absurd, but which is demonstrable.

[Kurzyk]

Interesting example Reflect. Pulling it out from the article:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which he knows has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Wouldn't the chances be equal between 1 and 2? So knowing 3 wouldn't be an advantage regarding your choices between 1 and 2.

[Remiel]

Ok here's another one. I found it on a math oddities website:

Three guys walk into an hotel and ask for a room with three beds. They just want to stay for one night. The man at the hotel tells them that the cost will be 30 dollars. They pay 10 dollars each and walk upstairs to their room. After some minutes the man knocks at their door, apologizing for a mistake. He made them pay the high season price and not the low season one, and since now it's the low season, he must give them 5 dollars back.

They are happy and appreciate the honesty of the man, so they tell him: "from these 5 dollars we will just take back 1 dollar each and we will give you the remaining 2 dollars as a tip for your honesty". The man thanks the guys and walks out.

Ok, they paid 10 dollars each (30 in total). They take back 1 dollar each (3 dollars in total). In this way they have paid 10-1=9 dollar each. So in total they have now paid 9*3=27 dollars. Plus the 2 dollars tip given to the man we get 27+2=29 dollars. Where is the missing dollar?

In reading this, my impression is that the dollar isn't really missing. Is it? I mean they gave 30, it was supposed to be 25, so they received 5 back. They kept 3 and gave 2.

Yet looking at that math it does look like 1 is missing.

Whew!  This one threw me for a while.  I knew, obviously, that the dollar really wasn't "missing", but I couldn't for the life of me explain why.  I think I've figured it out, however.  Here goes:

The amount that the hotel offers to refund is a red herring--that is, it's a completely misleading number.  As Oniya referred to, the amount that actually matters is the total amount actually paid to the hotel (including the tip).

To illustrate, let's say our three fellows book rooms at the same hotel the next season.  Once again, they pay the hotel $10 each, for a grand total of $30.  This time, however, let's say that our hotel manager is so grateful for their business (and perhaps a little drunk) that he offers to refund $20.  Once again, our generous travelers refuse, accepting only $1 back each and giving him the other $17 as a tip.   So now our travelers have paid a grand total of  $44 ($10 - 1 = 9 dollars each x 3 = $27 + the $17 tip = $44).  Right?

Obviously that is not the case.  In both examples, the travelers have paid $27 (10 initial - 1 back = $9 each x 3 = $27).  The only difference is that, in the first example, the manager made a $2 (or $5 - 3) tip, while in the second he made $17 (or $20 - 3).   The amount of the refund only factors into how much tip the manager gets, and has nothing to do with what the guys pay.

[Oniya]

Interesting example Reflect. Pulling it out from the article:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which he knows has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Wouldn't the chances be equal between 1 and 2? So knowing 3 wouldn't be an advantage regarding your choices between 1 and 2.

Except that the chances weren't equal when you initially picked.  Also pulled from the article:

Code Select Expand

Door 1 Door 2 Door 3 result if switching result if staying
Car    Goat   Goat    Goat                    Car
Goat    Car    Goat    Car                     Goat
Goat    Goat   Car     Car                     Goat

The results columns are assuming that you picked Door 1, and Monty revealed one of the goats.

[Kurzyk]

Its true that the chances were different when the initial choice of door 1 was chosen, but the question was whether it was an advantage in changing my choice. Its still 50/50 as to whether its 1 or 2.

[Remiel]

Also, the movie 21 has a pretty good explanation of the problem and the solution.

"21" explains the Monty Hall problem

[Jude]

Its true that the chances were different when the initial choice of door 1 was chosen, but the question was whether it was an advantage in changing my choice. Its still 50/50 as to whether its 1 or 2.

The problem with that logic is that Monty Hall knows which door the car is behind and cannot pick it to be the odd one out by the rules of the game.  Look at it this way:

Lets say you choose door 1 then break it down into cases from there.  There are 3 possibly situations, and each is defined by the door the car is behind.

Door 1 Car:  his choice of which door to exclude is arbitrary so no additional information is given by it.
Door 2 Car:  he actually has no choice on which door to exclude, it has to be number 3 because 2 has the car.  Thus he is exposing door 2 as the car.
Door 3 Car:  he actually has no choice on which door to exclude, it has to be number 2 because 3 has the car.  Thus he is exposing door 3 as the car.

In two out of three situations his actions give away what door the car is behind, so changing your choice based on his actions will more often than not lead to a better outcome.

[samantha_cs]

I find it easier to think about the probability shift if you take it to a ridiculous extreme.

Let's say there are 100 doors, with 1 prize behind one of them. You pick one door. Monty Hall opens 98 doors you didn't pick and shows there is nothing behind any of them.

You now have the option of changing doors.

Isn't it pretty obvious that you should switch doors in this example?

[Shjade]

I can't really see that the 100 to 1 example is more/less obvious than the 3 to 1 case. In either circumstance the rules end up the same: either you picked the door with the car behind it and the opened doors are arbitrary or you didn't pick the car and the door remaining has the car behind it. Works the same whether there are 3 doors or 3000. Right?

[WhiteyChan]

Changing slightly (read: a lot), another mathematical oddity for you, a lot more abstract but nonetheless amusing if you can understand it.

It is completely possibly for 1 + 1 = 0 to be true.

...

If you are in the Set F₂, that is.

[samantha_cs]

I can't really see that the 100 to 1 example is more/less obvious than the 3 to 1 case. In either circumstance the rules end up the same: either you picked the door with the car behind it and the opened doors are arbitrary or you didn't pick the car and the door remaining has the car behind it. Works the same whether there are 3 doors or 3000. Right?

Yes, but the odds you picked the right door on your initial try are much, much smaller. Ignoring the math, which doesn't care if it's 3 or 3000, your 'common sense' should tell you to switch doors in the 100 (or 1000 or 10,000) to 1 case.

[Shjade]

Did you just say "ignoring the math" in a thread about math? ;p

Still, fair point.

[ReanimateMagnus]

In Euclidean geometry, the Pythagorean theorem no longer always works. For the instance if you stand on the north pole and draw a line down the prime meridian and then one down the 90th meridian either east or west it doesn't matter. Then go to the equator and draw a line that connects the two. You just created a triangle with all 90 degree angles.

[Oniya]

Except that, by definition, is non-Euclidean geometry. 

[ReanimateMagnus]

yeah I wanted to edit that T_T

[Oniya]

Well, the 'problem' is that, in order to prove the statement 'the sum of the angles in a triangle = 180 degrees' to be true (which isn't the Pythagorean theorem, by the way - that's distance), you have to use the Parallel Postulate.  Once you toss that and start going into non-Euclidean spaces, triangles can have more than 180 degrees (elliptic geometry) or less than 180 degrees (hyperbolic geometry, which is useful when dealing with things like gravity wells).

[ReanimateMagnus]

I was just pointing something out...I don't need a "lesson" on Lobachevskian geometry.

[Oniya]

Well, then, consider the explanation a courtesy for those others reading the thread.  Perhaps if you get approved, we can have a conversation about the analytic and algebraic topology of locally Euclidean parameterization of infinitely differentiable Riemannian manifolds.

[ReanimateMagnus]

Well the Riemannian manifolds were only really applicable for advanced physics and general relativity and would be no where if it wasn't for Gauss.

[Trieste]

I have no idea how I followed that conversation, but I did, and I can tell you with absolute certainty that you have moved quite a bit beyond the scope of the topic of this thread. <3

At the very least, take your Gauss, your Reimann, and your Lobachevski to a thread dedicated to WHOA-ARE-YOU-EVEN-SPEAKING-ENGLISH-ANYMORE mathematics, please.

(But it's mean to make Euclid sit outside and cry, y'all. ;) )

[ReanimateMagnus]

I have no idea how I followed that conversation, but I did, and I can tell you with absolute certainty that you have moved quite a bit beyond the scope of the topic of this thread. <3

At the very least, take your Gauss, your Reimann, and your Lobachevski to a thread dedicated to WHOA-ARE-YOU-EVEN-SPEAKING-ENGLISH-ANYMORE mathematics, please.

(But it's mean to make Euclid sit outside and cry, y'all. ;) )

I'm sorry, but this made me smile. If my wife wasn't sleeping, I might have laughed out loud. ^__^ Sorry, I wont post here anymore.

[Trieste]

No worries! Just make a new thread for the new topic. :)

[Oreo]

I came across these recently and found them fun oddities:

9x9+7=88
98x9+6=888
987x9+5=8888
9876x9+4=88888
98765x9+3=888888
987654x9+2=8888888
9876543x9+1=88888888
98765432x9+0=888888888


1x1=1
11x11=121
111x111=12321
1111x1111=1234321
11111x11111=123454321
111111x111111=12345654321
1111111x1111111=1234567654321
11111111x11111111=123456787654321
111111111x111111111=123456787654321


1x9+2=11
12x9+3=111
123x9+4=1111
1234x9+5=11111
12345x9+6=111111
123456x9+7=1111111
1234567x9+8=11111111
12345678x9+9=111111111
123456789x9+10=1111111111


1x8+1=9
12x8+2=98
123x8+3=987
1234x8+4=9876
12345x8+5=98765
123456x8+6=987654
1234567x8+7=9876543
12345678x8+8=98765432
123456789x8+9=987654321

[Oniya]

No worries! Just make a new thread for the new topic. :)

It's not something I'd be likely to discuss in a thread anyways, mostly because I wouldn't want to risk any head-splodeys from an unwary passer-through.

(For what it's worth, I was quoting a line from Tom Lehrer's 'Lobachevsky' - a bit of a shout-out to any math person with a sense of humor.)

[ReanimateMagnus]

(For what it's worth, I was quoting a line from Tom Lehrer's 'Lobachevsky' - a bit of a shout-out to any math person with a sense of humor.)

I guess I don't have a sense of humor then ^_^ I'll have to look up Tom Lehrer then.

I show this to my students

1/9=.1111(these are repeating of course)
2/9=.2222
3/9=.3333
4/9=.4444

1/11=.0909
2/11=.1818
3/11=.2727
4/11=.3636
When I was a kid this was how I always remembered those fractions.

[Oniya]

I guess I don't have a sense of humor then ^_^ I'll have to look up Tom Lehrer then.

I suspect you'll like it.  He is a (now retired) Harvard math professor who had a brief run as a satirical pianist.  A number of his songs are actually educational in nature, and a few songs that made it to the Electric Company.

[Oreo]

I was always able to remember my 9's times with this. The answer always began with one less number than the multiplier and the two added up to 9.

2x9=18 (1+8=9)
3x9=27 (2+7=9)
4x9=36 (3+6=9)
5x9=45 (4+5=9)
6x9=54 (5+4=9)
7x9=63 (6+3=9)
8x9=72 (7+2=9)
9x9=81 (8+1=9)

[Falanor]

I figured someone here might find this amusing...

[ReanimateMagnus]

I love that picture btw.

[kckolbe]

Here's one I've always liked:

Proving that .999 (repeating) is equal to 1

.333 = 1/3
.666 = 2/3
.999 = 3/3
.999 = 1

[ReanimateMagnus]

Didn't I already post that? Oh well maybe not.

[frogman]

Did you know that... 7 x 13 = 28

There's another version with vacuum sales.  Ingenius... ;D

[frogman]

For a serious math oddity, though, how about the St. Petersburg Paradox:

You pay a flat fee to enter a lottery where a fair coin is flipped repeatedly until a tails appears, thus ending the game.  The pot starts at $1, and the pot doubles every time a heads appears.  You win whatever is in the pot once the game ends.  For example, if the first flip is tails, then you win $1.  If you gets heads and then tails, you win $2.  Heads, heads, tails = $4.  Heads, heads, heads, tails = $8.  And so on.  So the question is how much should you be willing to pay to enter this lottery?

To determine this, you need to find the expected payout.  The expected value is found by taking the probability of each outcome, multiplying that probability by what you would win, and adding all the probability payoffs together.  In other words, there's a 1/2 chance that you will get tails on the first flip, thus getting $1.  There's a 1/4 chance that you will get heads on the first flip and tails on the second, thus getting $2.  The chance decreases as you continue the string of getting heads in a row, but that's compensated by doubling the pot.  So mathematically, that's expressed as the following:

E(x) = (1/2 x 1) + (1/4 x 2) + (1/8 x 4) + (1/16 x 8) + (1/32 x 16) + ...

Once you get far out there, the chances of getting that payoff are very miniscule, but it's compensated for by the enormous payoff.  As you can see from simplifying the math above, there's an interesting pattern:

E(x) = 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + ...

This mathematical expression converges to infinity, meaning that if you keep adding 1/2 forever, so the total will keep growing without ever stopping.  So we see that the expected earnings of the St. Petersburg Paradox is an infinite amount.  Thus, you should be willing to spend everything you have on this game, because your probable earnings are infinite.  Would you be willing to risk everything?  Mathematically, it seems that you should!

Note: the math here is correct (unless I mistyped something), but obviously something doesn't seem right here because who do you know would actually be willing to risk everything?  There are answers to explain this paradox (some more mathematical than others), but it's nonetheless a cool mathematical oddity that I thought people would enjoy. 

[Oreo]

Did you know that... 7 x 13 = 28

There's another version with vacuum sales.  Ingenius... ;D

Gotta love Abbott and Costello.

[ReanimateMagnus]

[Kuroneko]

I think my brain just asploded ...

[frogman]

I think my brain just asploded ...

In a base 4 number system, 2 + 2 = 10.  In base 3, it's 11. :-)

[Pumpkin Seeds]

My head hurts.

[Oniya]

This might help.   >:)

Tom Lehrer- New Math

[Kuroneko]

Math bad.

[ReanimateMagnus]

undecimal

5+6=A

[MHaji]

For a serious math oddity, though, how about the St. Petersburg Paradox:

You pay a flat fee to enter a lottery where a fair coin is flipped repeatedly until a tails appears, thus ending the game.  The pot starts at $1, and the pot doubles every time a heads appears.  You win whatever is in the pot once the game ends.  For example, if the first flip is tails, then you win $1.  If you gets heads and then tails, you win $2.  Heads, heads, tails = $4.  Heads, heads, heads, tails = $8.  And so on.  So the question is how much should you be willing to pay to enter this lottery?

To determine this, you need to find the expected payout.  The expected value is found by taking the probability of each outcome, multiplying that probability by what you would win, and adding all the probability payoffs together.  In other words, there's a 1/2 chance that you will get tails on the first flip, thus getting $1.  There's a 1/4 chance that you will get heads on the first flip and tails on the second, thus getting $2.  The chance decreases as you continue the string of getting heads in a row, but that's compensated by doubling the pot.  So mathematically, that's expressed as the following:

E(x) = (1/2 x 1) + (1/4 x 2) + (1/8 x 4) + (1/16 x 8) + (1/32 x 16) + ...

Once you get far out there, the chances of getting that payoff are very miniscule, but it's compensated for by the enormous payoff.  As you can see from simplifying the math above, there's an interesting pattern:

E(x) = 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + ...

This mathematical expression converges to infinity, meaning that if you keep adding 1/2 forever, so the total will keep growing without ever stopping.  So we see that the expected earnings of the St. Petersburg Paradox is an infinite amount.  Thus, you should be willing to spend everything you have on this game, because your probable earnings are infinite.  Would you be willing to risk everything?  Mathematically, it seems that you should!

Note: the math here is correct (unless I mistyped something), but obviously something doesn't seem right here because who do you know would actually be willing to risk everything?  There are answers to explain this paradox (some more mathematical than others), but it's nonetheless a cool mathematical oddity that I thought people would enjoy.

The problem isn't just that you risk everything - it's that expected payoff isn't the same thing as the expected usefulness of the money. To put it more simply, I would rather take $50 for sure than a 1/1,000,000,000,000,000 chance of $50,000,000,000,000,000. Even if that much money existed (and it pretty much has to exist for the infinite sum to genuinely be infinite), you'd be pretty crazy to lose $50 on such a tiny chance. Risk aversion exists for a reason - it keeps people alive.

The paradox also breaks if we assume a GIGANTIC, but finite, bankroll. Let's try it with "more money than exists in the entire world, even counting derivatives."

Suppose the game goes only up to a meager $10,000,000,000,000,000 maximum, or ten quadrillion dollars. That would be about 50 rounds, giving us an expected payoff of.... $25.

So if you pay more than $25 to play this, you'd better hope that you have a use for "more money than exists in any form."

[AndyZ]

I saw Mythbusters where they did the Monty Hall problem and did an E search to see if anyone else had talked about it.  Saw this thread, decided to jump in.  Hope nobody minds.

I read the thing about the Monty Hall problem and thought that they did a poor job of actually explaining it.  It helps to break things down, I think, case by case:

Spoiler: Click to Show/Hide

Let's say that there's 3 doors, doors 1 and 2 contain nothing but door 3 has a car.  The host knows this but the contestant does not.

If you pick door 1, the host will open up door 2 to reveal it as empty.  You'd have to switch to win.

If you pick door 2, the host will open up door 1 to reveal it as empty.  You'd have to switch to win.

If you pick door 3, the host will open up either door 1 or 2, but you should not switch.

Thus, not switching only lets you win 1/3 of the time, while switching lets you win 2/3 of the time.  It seems as though it would be a 50-50 shot, but this just isn't true.

With the St. Petersburg paradox, I'm reading it over and I think the problem is recursion.  You see, out of 100% possible solutions, you have a 50% chance to earn nothing at all, 25% to earn $1, 12.5% to earn $2, 6.25% to earn $4, and you can continue from there.

Were it truly E(x) = (1/2 x 1) + (1/4 x 2) + (1/8 x 4) + (1/16 x 8) + (1/32 x 16) + ..., then you'd need to have a 50% chance of earning a dollar, plus a 25% chance of earning another $2.  This means that the jump for two successive heads would be $3.  The next would be 1+2+4 to become 7, then 15, and so on.  I may be wrong on this, but that's how I see it.