Bit of a maths puzzle

[raynerd]

I was talking to some students a few weeks ago and they showed me this puzzle, I know the answer and have been amazed with it for the last few days. I thought with you guys being use to dimensions and sizes you might like it.

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Imagine a perfect sphere, about the size of the Earth and tie a string around it, pull it tight...really tight, so tight that you can`t even slide a razor blade under it at any position. Now you cut the string and add a meter. So now if you like, you have the original length that fit perfectly around the Earth plus this meter slack. Now if you got millions of people/robots to line up around the string and pull upwards perfectly evenly to move the string off the Earth, would that meter now give you enough slack to now slide a razor blade under it all the way around?

No tricks...just maths.
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Chris

[z3t4]

Yep.

Circumference of sphere, of radius r, = 2πr.
Increase circumference by 1, so new circumference = C+1 = 2πr + 1 =2π(r+1/2π).

The string is then 1/2π metres (about 6") above the surface of the sphere.

Very counterintuitive.

HTH

John

[cidrontmg]

It is indeed. Itīs the same amount (15.9 cm) no matter what the original sphere diameter. If the "sphere" is a point, it is the radius of a circle with a circumference of one meter.
Another counter-intuitive fact: Most people have more legs than the average.      Huh???   
Most people have 2 legs. But not all. Even if thereīs just one one-legged man, the average is below 2. I.e. most people have more than the average number of legs. 
 

[andyf]

And another: if 23 people are in a room, there is more than 50% probability that two of them will share the same birthday (day and month, not the year of birth). With 50 people, the probability rises to 97%.

Andy

[mklotz]

Suppose you're on a game show and you're given the choice of three doors [and will win what is behind the chosen door]. Behind one door is a car; behind the others, goats [unwanted booby prizes]. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one [uniformly] at random. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice?

I'll publish the answer after you've had a chance to confuse yourselves.

[andyf]

 
I'd switch to door 2.

To avoid confusion with the existing numbers, there's one door with a car behind it and another two hiding goats - call those goat doors X and Y.

There are 3 possible scenarios:
1. My first choice was the door which hides the car. When offered my second choice, I switch to another door (either X or Y) so I lose.
2. My first choice is door X which hides a goat. If when offered my second choice, I switch to door Y, I win  (the host having exposed the other goat before offering me that second choice).
3. My first choice is door Y which hides a goat. The host is forced to open door Y and show me the other goat. If I switch when offered my second choice, I win the car.

In the first of those scenarios, I lose, but in either of the other two I win, so the odds are 2 to 3.

If I don't switch when offered my second choice, then:
1. I win - I choose the right door first time, and stick with it.
2. I lose - I choose door X first time round and stick with it, but it hides a goat.
3. I lose - I choose door Y first time round and stick with it, but it hides a goat.

In the first of those scenarios I win, but in either of the other two I lose, so the odds are only 1 in 3 when I stick with my original choice.

QED (I think!)

Andy
 

[cidrontmg]

Yes, like Andy said, changing ones mind in this situation gives better odds, if you want the car. I already have a car, though, so Iīd  be happier with the goat....   

[z3t4]

Just noticed that Wiki seems to be in a state of some disquiet re the Monty Hall problem. As with most Maths stuff, it does seem to cover it very well, IMHO.

Goats are cute.

John

[kvom]

The goat problem is similar to a principle in bridge called "restricted choice".  Assume your holdings in a suit is Axxxx vs K10xx.  As declarer you play the A and RHO plays the Q or J.  It is 2-1 to finesse when playing towards the K10.

[jim]

the earth/string one is one of my favorites.

the version i've heard was "how longer would the string be if it was lifted up an inch?" i've seen very clever people resorting to calculators to prove its more than a few inches!!

[mklotz]

Andyf nailed it right off the bat.

The most concise description of the solution I've found goes like this...

Switching loses if and only if the player initially picks the car, which happens with probability 1/3, so switching must win with probability 2/3

[raynerd]

Yes, they are both good but I was really blown away with the string around the Earth version. And when I calculated that you could pretty much fit a Bunny Rabbit under it at any point never mind a razor blade I couldn`t get it in my head. Of course I realise now but it certainly doesn`t feel right!

Chris

[ksor]

Here is another one - a little like the "string/earth" problem.

1) You have a ball
2) You drill a hole through the center of that ball (= the center line of the hole goes through the center of the ball - not easy in real life but ... )
3) This hole turns out to be 6 cm long

How much is left of the ball ?   

No tricks - just pure math !

[John Hill]

Rather good for children but some adults get confused too!

"Three guys walk into a cheap hotel and ask to stay the night, the clerk says they will have to share and the room is $30, so they each pay $10 and up they.  The manager was listening from his office and came out to chew a strip off the clerk.  The room is only $25 and the clerk was to take the change to the room and not do that again.  On the way upstairs the clerk decides it is too hard to divide $5 by 3 so he just gives each man one dollar and keeps the other two dollars for himself.

Now then, each man paid ten dollars and got one dollar back, so nine dollars each,  three nines is twenty seven and the two dollars in the clerk's pocket makes twenty nine so what happened to the missing dollar?"

[ksor]

Yeah - it's confusing !

I think the solution is something like:
When you leave out the 3X1$ of the change money to get the 9$ each payed, you can't "add in" the 2$ - you have to leave them out too.

[foozer]

Rather good for children but some adults get confused too!

"Three guys walk into a cheap hotel and ask to stay the night, the clerk says they will have to share and the room is $30, so they each pay $10 and up they.  The manager was listening from his office and came out to chew a strip off the clerk.  The room is only $25 and the clerk was to take the change to the room and not do that again.  On the way upstairs the clerk decides it is too hard to divide $5 by 3 so he just gives each man one dollar and keeps the other two dollars for himself.

Now then, each man paid ten dollars and got one dollar back, so nine dollars each,  three nines is twenty seven and the two dollars in the clerk's pocket makes twenty nine so what happened to the missing dollar?"

3 9's make 27 and "and" is the trick. Gets ya to think its a positive addition when its really a minus 'and the two dollars in the clerk's pocket makes twenty nine' rather the clerks 2 bucks is a -2 as was the 3 bucks given back, so it  makes 25