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A little math puzzle

S

SchrodingersFinch

New Member
arg-fallbackName="SchrodingersFinch"/>
Here's a little math puzzle that I came up myself. The math you need to solve it is quite simple.


The Wheel of Fortune is a perfect circle. There is one winning point at the edge of the circle.
When you spin the wheel it will spin for a while and stop at a random point. If the arrow is pointing directly at the winning point, you win (obviously).

wheeloffortune.gif


You get to spin the wheel an infinite amount of times. What is the probability that you will win?


Everyone might not agree with the answer I have for this puzzle, and I haven't ruled out the possibility that I might be wrong. But let's not get ahead of ourselves. First I want to hear your answers and then we can debate who's right. ;)
 
arg-fallbackName="Aught3"/>
Is it something to do with a circle being made up of an infinite number of points?
 
arg-fallbackName="Ciraric"/>
As a mathematician I can very boldly say that the answer is the same as 1/0.

AKA it is undefined.

Or to put it another way. You've badly defined the question (which I guess is on purpose to confuse people).

Okay.

If the spinner is a mathematical line then you must have the response in saying that there are infinite lines mapping from the centre of the circle to some point on the circle.

However, you have infinite spins and so you should be able to hit every single point right? Wrong.

Infinity is an undefined term. It is not even a number.

It is also a thought experiment with impossible terms. So I refuse to think about it any more. :lol:
 
arg-fallbackName="Ciraric"/>
Oh wait... I've thought of another one because my mind hates me.

Making a circle out of points is impossible.

Because a point is so small that if you start drawing a circle with points then the points would be at a 180 degree angle to each other and therefore no matter how many points you added you would just have a line

Pure Mathematics sucks.
 
arg-fallbackName="borrofburi"/>
Alright alright, I don't have any math, just intuitive reasoning (which has served me well in my life). [err didn't have any math, way to steal far too many minutes of my life :p]

I think the proper way to solve this problem is as two limits. You can model the circle as two arcs, each with probability 50%, then you can model the circle as four arcs, each with probability 25% (only 1 of which is your chance to win), etc. You take the limit of that as the number of arc segments approach infinity. Alternatively, if you have only two arcs, and you try once you have 50% probability, if you try twice it goes up to 75%, etc. (if there are 4 arcs, you try once and have 25% probability, if you try twice, you have 44% probability, ; s=1-(1-p)^j, where s is likelihood of success after i tries at p percentage success per try). You take the limit of that as the number of arc segments approach infinity.

I admit I had no good reason to think infinity of tries trumps the infinity of arc segments.

Thinking through it now, the expanding number of arcs will result in 50% reduction probability of success with each iteration, so p=1/(2^n) where n is the number of iterations of doubling. So, lim(j->~) lim(n->~) 1-(1-(1/(2^n)))^j, and that's just a confusing limit. So I dunno, someone who actually remembers how to do double infinite limits can handle this one.
 
arg-fallbackName="borrofburi"/>
Yah yah, double post. I thought I had left my original post there too long to merely edit it.

It occurs to me that since we're only concerned with a single infinitesimally small arc segment, we don't have to double the number of arc segments each time, and the probability of success can instead be modeled as 1/n.

The problem here is of course that you have two infinities to reconcile, that of an infinity of arc segments (or alternatively the infinitesimally small arc segment), and that of an infinity of tries. It's clearly obvious that an infinite number of tries and a finite number of arc segments then probability is 1, if you have an infinite number of arc segments and a finite number of tries then the probability is 0, but if you combine the two...

Well, which infinity grows faster? I think they're both the same infinity, which would indicate there's no answer.
 
arg-fallbackName="Squawk"/>
I'm going with 0.

Reasoning is simple. We are dealing with infinities, and these come in different sizes. I can argue that between any two spins of the wheel there are an infinite number of points, meaning that the probability of winning is 1/infity which is undefined but might as well be 0.

I cannot argue that there are an infinite number of places that a spin can stop at between each point because, by definition, that would create a new point between the two points I am in between.

Therefore the answer is 0. You can never win.
 
arg-fallbackName="Master_Ghost_Knight"/>
Som people may have acidently runed into the answer, but so far nobody has given a reasonable explenation to why.
It is in fact due to a particular proprety rather then what you arrive whit a calculation. I learned the answer to this problem on the second week of coledge math, but just not to spoil the fun I will keap quiet. (and come to think of it you may never actually arrive to the answer, it is just simply mindblowing)

And you can forget about limits, because the number of points in the circle is a constant and so is the number of available spins (i.e. infinite), they don't grow anywhere.
 
arg-fallbackName="G-Smo"/>
I think i got some kind of answer...

Earlier I read, the chance could be 1/0, resulting in infinity. That gave me my idea.

I believe we agee:
the circle exists of an infite number of point (all at the same distance of the center)
AND
We have an infinite number of tries.

We need to calculate the chance it will hit a specific point

my simple logic would result in the divsion of infinity through infinity. because we are going to hit that specific point an infinte number of times, if e keep spinning the wheel to infinity.

Another thought that crossed my mind is the uncertainty principle, even though it has little to do with mathematics.

Our spin, will hit the specific point at least once, because we define it. If we agree, on the definition that there is one point, we can't allow the circle to have an infinite number of other points. Albeit, the observer influences the experiment.
 
arg-fallbackName="borrofburi"/>
mmm I want to say Squawk wins, but I recall that all infinities of counting numbers (counting numbers squared, etc.) are equal. A larger infinity is all the numbers between zero and 1 (or zero and 100 (or any two floating point numbers, discounting repeating floating points)). It seems to me, the set of all points on a line segment can be described by counting numbers: you start at zero, you move one point over, and then one more point over, etc. This, however, is wrong because a point has no distance. However, we *can* describe every point on that line segment as a specific distance from the original point, which will require floating point numbers. And therefore the infinity of points is the infinity of floating point numbers.

However the number of spins is incremental, it grows by one each time, and therefore the infinity is infinity of counting numbers.

Therefore, the infinity of points is bigger than the infinity of spins, therefore probability 0, Squawk is right, my intuition is wrong :( Unless of course someone can convince me otherwise.

Master_Ghost_Knight said:
And you can forget about limits, because the number of points in the circle is a constant and so is the number of available spins (i.e. infinite), they don't grow anywhere.
Click to expand...
You are incorrect, there are many limits that go to infinities, and often times competing infinities. Also, common limits that go to infinities are integrals.

You are correct that "grow" is an improper term. Below is irrelevant description of what it means and why I use it
I apologize for using it, it was an archetype I created in order to understand limits like the limit as n approaches infinity of the following quantities ((n^2)/(n^3)), ((n^3)/(n^2)), ((4n^9)/(5n^9)). While those are all obvious examples, the difficulties came when you started adding in logarithms or exponentials or factorials, and the easiest way for me to answer those more insane quantities and combinations of those was to ask "which grows fastest", so does n squared factorial grow faster than n factorial squared? Well, the first goes 1, 24, 362990 while the latter goes 1, 4, 36, so the former dominates because it "grows faster".
 
arg-fallbackName="Marcus"/>
Pulsar said:
Zero.
The number of tries is countable, while the number of points on a circle is uncountable.
Unless I missed some kind of catch...
Click to expand...

That's exactly what I thought. Put it another way:

If you marked with an infinitely small dot every point on the wheel that was pointed at by the arrow in your infinite spins between now and forever, then between every pair of those points there would be an infinity of points that have never been selected - an infinity of the same order as the number of points on the entire circle.
 
arg-fallbackName="Ciraric"/>
Pulsar said:
Zero.
The number of tries is countable, while the number of points on a circle is uncountable.
Unless I missed some kind of catch...
Click to expand...
If you begin at zero spins then yes it is countable.

If you begin at the idea of drawing the circle out of an infinite number of points then it is also countable.

It would also not be a circle (as I said earlier it would be a line).

i like to confuse myself.
 
arg-fallbackName="Marcus"/>
Ciraric said:
If you begin at zero spins then yes it is countable.

If you begin at the idea of drawing the circle out of an infinite number of points then it is also countable.

It would also not be a circle (as I said earlier it would be a line).

i like to confuse myself.
Click to expand...

No, any continuous line (circular or straight) has uncountably many points.
 
arg-fallbackName="Ciraric"/>
I never argued against that claim.

in fact that is a main cornerstone of my theory.
 
arg-fallbackName="Master_Ghost_Knight"/>
borrofburi said:
You are incorrect, there are many limits that go to infinities, and often times competing infinities. Also, common limits that go to infinities are integrals.
Click to expand...
Am I?
Yes there are many limits that go to different infinite, So?
That doesn't change the fact that those "infinites" are part of a function that tend to infinite as approaching a certain value and not infinite in their vicinity, and that when you do the limit you are effectively comparing the behavior of two infinite approaching functions. If the functions are constant equaling infinite you just run out of luck.
You can create a function to account for all the tries in function of the tries, but how exactly are you going to create a function that accounts the number of points of a continuum perfect circle in function of what?
(Answer: You don't! Sense the answer as already been given I can had that "because R is uncountable and therefore it is impossible to come up with anything that would do it")
It is quite understandable that people fall in this mistake because it is not apparently obvious why this should be.
 
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