Brain Teaser No 1

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Originally posted by Matthew T:
ignore ant effect of ... friction.)

Matthew

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All three will arrive at the same time.

Since there is no friction, the objects will not be forced to rotate, and they are then simply objects falling under gravity with a sideways impetus.

cheers, Martin

When on the slope, again they both experience a force along the slope: mgsin(A) (where A is the angle of inclination of the ramp). If they were sliding their acceleration would be equal at gsin(A). But their rotational acceleration is not the same because their rotational inertia is different. Thus the lighter ball, with less rotational inertia, will have a higher rotational acceleration than the heavy ball.

BAM

Paul

I have been busy these past few days.
I am therefore only now,about to give your roller-coaster question a bit of thought. At the risk of appearing somewhat dense, I am clarifying as my starting point that :-

I don't know whether the ball (sphere) is hollow or solid. I assume that 'perfectly formed' means it is absolutly spherical AND has a uniform distribution of mass and density. Likewise the cylinder, I don't know whether it is hollow (empty soup tin, with or without lid and base), or solid. The ring I assume, is infinitly thin and has no lateral dimension, otherwise it would be called a hollow cylinder!

I am therefore assuming that I will be able to demonstrate that none of these uncertainties actually matter.

I am also assuming that the comment that we can assume there is no friction was a slight slip of the pen (or keyboard), as drawn to our attention by Martin P and that you intended to say there was no slipping or sliding along the incline, only pure (and hypothetical) rolling motion, as implied by '..start rolling down the plane...'.

Perhaps others would like to comment before I start on two balls, three cylinders and a ring (is this the sixth day of Xmas??) Actually the ring and the 'hollow cylinder without ends' look kind of similar, so perhaps there are just five days.

Cheers

Don

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The current mini-question is whether objects of the SAME geometry but with different material-density will behave alike or not.

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If the moments of inertia and masses are the same then they will behave alike, regardless.

It's 'trivial'...

Paul

Re your 'two numbers in an envelope' teaser.

I presume that person A could equally place two pieces of paper, each containing one number, into a bag. Person B then picks one piece of paper out of the bag at random and then has to decide whether this number is smaller or larger than the number still in the bag??

The problem remains the same whether the numbers are selected from

a. the integers (ie whole numbers from minus infinity to plus infinity)

b. the natural numbers (ie whole numbers from 1 to plus infinity) (or possibly from 0 to plus infinity)

c. the natural numbers from 0 to 10 (ie the whole numbers 0??, 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10)

Confirmation/clarification sought whilst acknowledging this seems pedantic and appreciating that clarification about Zero being a natural number is even more pedantic.

Reminds me of the astronomer, physisist and mathematician enjoying a train journey in Scotland!!

Cheers

Don

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Being a bit pedantic, the moments are the same, the masses aren't (but are distributed the same).

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I think the masses have to match for equivalence.

Consider the yo-yo....

Paul

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What about a Yoyo ? remember, that you need a theoretical yoyo with a string mass of 0 and zero friction.

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String mass is negligible, friction exists to serve the purpose and doesn't otherwise interfere. Two yo-yos, same Moment of Inertia, different masses, which one reaches the end of its string first?
(I think maybe they both arrive together, but spinning at different rates?)

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Do you agree with my "proof" above that shows that the final velocities are equal and independent of the material density of the object ?

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Yes. But that's 'trivial' in the same way that ideal masses dropped off towers in a vacuum arrive together, or simple pendulums have a period independent of the bob's mass.

IIRC the original problem involved geometrical rather than density variations in the moment.

Paul

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This sounds unlikely, they travel the distance at the same time, but have different end velocities

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It is unlikely. The one with the lower mass will arrive later.

Paul

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RichardN,

Correct, very nice.

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But I'm buggered if I understood a word of it. Can someone explain in words of one syllable please?

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Originally posted by Duncan Fullerton:

But I'm buggered if I understood a word of it. Can someone explain in words of one syllable please?

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I didn't understand it either!!!!

Steve D

Two ways to get the answer.

One look at energy
Potential gravitational energy going to kinetic (rotational and linear) energy

Second look at forces
Gravitional acceleration puts force on object and therefore it moves(this is a complicated route becasue you have resolve forces that cause between the surface and object (with no friction!)....)

So correct answer is they get to the bottom at the same time but it you assume they roll then the sphere goes fastest!

Matthew

Matthew

Using Omer's method (which I thought is what I said in my earlier note) I am expected to guess your second choice is "lower" since there are more numbers <83 than >83. But I don't think this strategy has any real basis because I don't know what process you used to select your two numbers.

My intuition is that you have chosen a number higher than 83 in order to make your point.

Anyhow, refering to Omer's earlier post. If your first pick is a border number you are certain about the other number. If not I think you have only a 50% chance of guessing right (since you have no idea what the opponents process was for choosing his numbers). But I think this is all that can be said about the problem as stated.

Take the earlier case. Range is 0-10 and the opponent puts 1 & 2 in the bag. The suggested strategy says you will guess the second number is higher whichever you draw first. You can only be correct 50% of the time. Not 55%. This is true of any pair of numbers the opponent chooses other than if either number is 0 or 10.

0, !10 => 0.5*100%+0.5*50% = 75%
10, !0 => 75%
0, 10 => 100%
any other => 50%

These are your odds of getting it right for specific opponents numbers. The average performance of your strategy depends entirely on what the odds are of the opponent selecting a particular pair of numbers. We don't know this. So we cannot say for sure whether a particular strategy has odds better than 50-50.

Or am I still missing something?
BAM

Higher or lower than 83. Lets include 0 but exclude 83.

There are 100 random choices available to your son. 83 of them are lower than 83. 17 of them are higher than 83.

The chance of him randomly picking one lower than 83 is 0.83 and the chance of him picking one higher than 83 is 0.17.

You'd have to be daft to opt for any outcome other than 'lower'

I haven't got a headache yet, but is this relevant to Omer's original challenge?

Cheers

Don

But I think your analysis depends upon the opponent making a random selection of at least the second number. This is how you can say that if you pick 83 there is a 83/100 chance of it being the higher number. So your chance of guessing correctly is >50%. Yes. But this is obvious.

If the opponent hasn't chosen randomly the distribution is not related to the number of numbers. Knowing the value of one number tells you nothing new about the value of the other (ignoring border numbers).

Your original teaser is :-
Suppose someone chooses two different integer numbers and writes them in an envelope. You pick one of them (by chance) and open it.
Now you have to guess whether this was the smaller or the larger number.
guessing "blindly" will yield 50%. Is there a better strategy (even by a fraction ?)

After clarification we learned that 'suppose someone chooses' does NOT mean 'at random' but 'chooses numbers using a strategy and what's more he knows YOUR strategy as well!'

For the moment we have simplified the problem by confining ourselves to (say) the natural numbers 0 to 100.

I have written a simple Excel spreadsheet to simulate the problem for 0 to 100.

In the first version I have deliberately selected numbers 3 and 4. These numbers can be changed easily to any other two numbers. The spreadsheet runs the simulation 1,000 times and I have recorded the averages of 10 runs thus representing 10,000 trials. Using Bam'stechnique of 'Guessing' gives me a probability of 0.506 of getting the right answer, whilst using Richard N's technique of selecting a random number from a bag of 0 to 100 gives a probability of 0.513. Richard's technique seems to work!!!

OTOH, in my second version, I have simulated the first two numbers being selected at RANDOM. Again based on 10,000 trials, the first number chosen had a probability of 0.499 of being the smallest (so the random number generating technique is 'fair'). Using Bam's 'if its bigger than 50' technique, the probability of the right answer is 0.749 (suggesting 0.75) whilst using Richard N's technique the probability is only 0.658 (suggesting 0.667).
Now unfortunately, Excel hasn't got a "minus infinity to plus infinity" button, so I'll have to rely on Richard's technique……………except that Richard will also need a 2*infinitely large bag of numbered balls………

Interesting!!!!!

Cheers

Don

e-mail me if you would like a copy of the spreadsheet to examine

Meanwhile, my hangover has gone and I'm applying more thought. I now understand RN's To make it simple I've considered the case of the range 0-4 and consider the opponent choosing 1,2 or 1,3 or 2,3.

1,2 RN prob is 5/8, BAM prob is 6/8
1,3 RN prob is 6/8, BAM prob is 8/8
2,3 RN prob is 5/8, BAM prob is 6/8

For this range my method is always better (can you confirm this Don?).

Then, repeating for a range of 0-5:

1,2 RN is 6/10, BAM is 5/10
1,3 RN is 7/10, BAM is 10/10
1,4 RN is 8/10, BAM is 10/10
2,3 RN is 6/10, BAM is 10/10
2,4 RN is 7/10, BAM is 10/10
3,4 RN is 6/10, BAM is 5/10

This time RN's method is better in two cases whereas mine is (much) better in 4 cases. What I suspect is that if the opponent chooses randomly my method will, on average, be better. And Don's trial indicates this. If the opponent ALWAYS deliberately chooses the worst scenario for either method then I suspect RN's method will, on average, be better provided the range is wider than 0-4. But I'm not sure what scenario is the worst for each method vs range.

BAM

Now I know you are all bursting to hear my story about the astronomer, the physicist and the mathematician, but I also know you are simply too busy trying to solve Omer's 'two numbers' or Mathew's 'ringing, cylindrical balls' or Richard N's 'seven pirates' problems to inundate me with requests. So I shall take the hint and oblige!!

An astronomer, a physicist and a mathematician found themselves sharing a train journey through Scotland (highly improbable, Bam and Omer will tell us the actual figures!!). On their second day, they saw a black sheep (just as well I'm not posting this on the Mana Forum!!) in a field.

"Ah ha!" cried the astronomer! "I see that Scottish sheep are black!"
"Don't be absurd!" said the physicist " SOME Scottish sheep are black"

With a deep sigh, the mathematician turned his head towards heaven and in that weary voice of the tired lecturer said,

" Gentlemen, let us be careful, all that we know of this subject is that In Scotland there exists at least one field containing at least one sheep, at least one half of which is black"

Am I glad I haven't spent too much time travelling on trains in Scotland!!

Cheers

Don

Nonsense!!! Anyhow, please wait until I have eaten my humble pie for suggesting two balls of same size and different mass will roll down a ramp in different times!

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The side you see is red, do you bet one the other side being red or black?

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It's 2/3 red, 1/3 black.

Every card appears with equal probability, we always retry with the black/black card, the red/red card counts on every appearance, the red/black card counts on half its appearances. 2/3 of the time when we actually guess what's on the back the card will be the red/red.

Paul