BrainDen.com - Brain Teasers

# mmiguel

Members

142

1

## Posts posted by mmiguel

1. ### Golden Ticket

24,214

nice work!

there are 100000 candy bars,

5 of those have golden tickets

x of those are the ones Mr. Salt picks.

since there is no reason for Mr. Wonka to distinguish between any candy bar, when placing golden tickets, and no reason for Mr. Salt to prefer to buy any candy bar over any other, we may assume equal probabilities for all outcomes involving the ways the tickets may be distributed and the selection of candy bars by Mr. Salt.

Since all outcomes have equal probabilities, the probability of interest becomes the ratio of outcomes satisfying the event of interest (i.e. Mr. Salt buying at least one of the golden tickets), divided by the total number of outcomes.

To see this, consider the following. All outcomes are by definition mutually exclusive, so the union of any outcomes corresponds to the sum of probabilities for those outcomes. All outcomes have equal probability, and let's call that probability p.

The answer, P is the sum of all probabilities where Mr. Salt has at least one golden ticket: P = (# outcomes where he has at least one ticket)*p.

But the sum of the probabilities of all outcomes in the sample space must be 1, and this is only true if p = 1/(# of all possible outcomes).

Therefore P = (# outcomes where he has at least one ticket) / (# of all possible outcomes)

The denominator here is simply the number of ways of distributing 5 golden tickets in 100,000 candy bars times the number of ways of selecting x candy bars of 100,000.

This is (100000 choose 5) * (100000 choose x).

The numerator can be calculated in multiple ways.

Essentially it corresponds to outcomes in which the intersection of the two sets of candy bars(those with golden tickets, and those that were selected by Mr. Salt intersect) is not empty.

Way 1: Multinomial method - Identify 4 mutually exclusive groups of candy bars:

A - Those that have golden tickets and that were selected

B - Those that have golden tickets and were not selected

C - Those that don't have golden tickets and were selected

D - Those that don't have golden tickets and were not selected

It is easiest to calculate the probability for the case where Mr. Salt does not get any golden tickets, and then take one minus this result.

In that case, the size of A is zero.

The size of B is 5.

The size of C is x.

The size of D is 100000 - 5 - x

The multinomial probability formula says the number of ways of splitting up the candy bars into these 4 groups are:

100000!/(0!*5!*x!*(100000-5-x)!)

i.e. the total number of things getting split up is in the numerator (with factorial), while the sizes of each of the groups that are being filled is in the denominator (with factorials).

When this is taken over (100000 choose 5) * (100000 choose x), there is some simplification leading to:

(N-5)!*(N-x)!/((N-5-x)!*N!)

where N = 100000 (to save typing)

This turns out to also be equivalent to (N-5 choose x) / (N choose x) and also (N-x choose 5) / (N choose 5)

Intuitively these can be interpreted as (# of ways to choose x non-golden ticket bars divided by # of ways to choose bars)

and (# of ways to choose 5 non-selected bars divided by # of ways to choose 5 bars).

The actual probability is 1 minus these.

P = 1 - (N-5)!*(N-x)!/((N-5-x)!*N!)

or P = 1 - (N-x)*(N-x-1)*(N-x-2)*(N-x-3)*(N-x-4)/(N*(N-1)*(N-2)*(N-3)*(N-4))

Using some calculation aid, it can be seen that when x <= 24,213, P < 0.75 and when x>=24,124, P > 0.75

this gives prob 0.7500051562780645.

cannot be less than this since 24,213 gives prob 0.7499886619361548

2. ### Hard to Resist

A funky looking piece of electrical material has a length of 2 cm on the x-axis, which is the axis of current flow.

Assume a coordinate system where the object is sitting between x = -1 cm and x = 1 cm.

The cross-sectional shape of the object is circular, with radius varying as a function of x like so:

r(x) = 0.04 + 0.4*x^2+x^4

r(x) is in cm

A differential volume of this material has a resistance of R ohms in the the direction of current flow, where R varies with distance from the x-axis. If the distance of the differential element from the x-axis is p, then R(p) = 2*sqrt(p)

R(p) is in ohms

What is the total resistance of the whole object for current flowing along the x-axis?

3. ### Golden Ticket

Mr. Wonka produced 100,000 Wonka bars this year, 5 of which have golden tickets inside of them.

To appease his spoiled daughter, Veruca, Mr. Salt is willing to buy enough Wonka bars to ensure with probability >= 75% that he will have purchased at least one of those tickets.

How many does he need to buy?

4. ### Simple probability problem

Four cards are taken from a standard deck, one from each suit.

They are shuffled and dealt face down on a table.

Coins are then placed at random on two of the cards.

What is the probability that the two chosen cards are the same color?

4 cards from each suit.

2 are red, 2 are black.

Placed on table at random.

Coins placed on 2 of them.

There are 4 choose 2 = 6 ways of placing the coins.

Of these combinations, there are only two in which cards under coins are same color.

Each combination is equally likely (P = 1/6).

Two combinations are mutually exclusive and satisfy event of interest.

Hence probability is 2*1/6 = 1/3

5. ### Subjective Probability

There seems to be a bit of a problem with the question. You have stated the properties of the random number generator, for example, that it is unbiased. However, if we were to make an objective interpretation, using an objectivist interpretation, I don't think we can say that the generator is unbiased if the experiment (i.e. seeing what number it produces) has never been done before, thus the only sensible interpretation would be a subjectivist one.

Note that once the generator has output the 7, the experiment has not yet completed for Chris, so he cannot yet make an objective assessment.

To say "there is no such thing as repetition from a truly objective standpoint" is not true if we allow for abstraction, and we communicate for the most part in abstractions. You could, for example, say that there is objectively a repetition of things which are blue. The only time you cannot have repetition is if you define something enumeratively (i.e. an exhaustive extensional definition) to have only one instance. Confusion might arise if we are inconsistent with the scope in which we ascribe objectivity. Ultimately, whenever somebody states that something is objective, this is necessarily a subjective assessment with many contingencies. However, this does not mean that we cannot talk about something being objective, because we can narrow the scope by assuming various axioms, such as that we have commonality in our language.

Thank you for your input in this discussion.

Quote: "There seems to be a bit of a problem with the question. You have stated the properties of the random number generator, for example, that it is unbiased. However, if we were to make an objective interpretation, using an objectivist interpretation, I don't think we can say that the generator is unbiased if the experiment (i.e. seeing what number it produces) has never been done before, thus the only sensible interpretation would be a subjectivist one."

There are two "experiments" under discussion here though, one of which has been done before (to determine distribution), and one of which can only ever be done once. The first is the one for which I laid out the uniform distribution, which I agree is repeatable (though only due to abstraction ultimately coming from a subjective perspective), and for which there should be no confusion.

The second experiment is a unique measurement in the real world at a defined point in time.

There is a distinction between the abstract model (first experiment) and a real, physical measurement that coincides with some trial of this abstract model.

I believe the purpose of probability is prediction of such real, physical measurements.

My problem is that the claim of objectivity in probability fails when applied to a statement of any single real, physical measurement, unless the absolute truth of statement is known.

Relative frequency works fine if you are talking about a large number of physical measurements, but fails to say anything useful if applied only to any specific one (you can only assign it a value of 1 or 0, but only if you know which one it is). The Bayesian interpretation however easily applies to any specific measurement.

I agree with your statement that the only sensible interpretation here is subjective, but I think that in general, the only sensible use of probability in general is inherently subjective.

Quote: "Note that once the generator has output the 7, the experiment has not yet completed for Chris, so he cannot yet make an objective assessment."

What part is not complete? Which of the two experiments?

Quote: "To say "there is no such thing as repetition from a truly objective standpoint" is not true if we allow for abstraction, and we communicate for the most part in abstractions. You could, for example, say that there is objectively a repetition of things which are blue. "

I think that abstraction itself is subjective. Let us ponder the concept of abstraction. Abstraction is the opposite of specification. It is essentially the removal of details. As an example, there could be a classroom full of children, each child is different in at least some of their characteristics (e.g. height, weight, name, gender, spatial position, personality, ...etc). Instead of thinking about each of them as completely dissimilar entities, which would be impractical, we can, from observing their common characteristics, abstract out the concept of student by removing from consideration details which we designate as not important. We make spatial position not important in the definition of the concept of a student i.e. a student may have any spatial position and still be classified as a student. We leave important characteristics, i.e. such as being enrolled in the school. By removing information (i.e. unimportant characteristics) we have established an abstract group of specific entities.

Another example is numbers. Numbers are abstract concepts as well. You can communicate the concept of the number three by showing me three apples, then telling me to remove all characteristics associated with apples.

What is the point of this rambling? ---> Abstraction is the removal of unimportant information, however the judgment of what characteristics are unimportant is subjective. Hence the concept of abstraction can only exist in a subjective perspective. All hypotheticals that we might imagine or define, come from real, physical observations which we have watered down (i.e. removed details i.e. abstracted) to form abstract objects for the sake of practicality (and this is very much a good thing).

Your example: Repetition of things which are blue ---> remove all characterstics from the concept except those relating to whether the emitted/reflected light is within a certain frequency range (the blue range).

So I agree, repetition is possible with abstraction, but I argue that abstraction itself is subjective, so I disagree about repetition existing from a truly objective standpoint.

I know this is all largely philosophical, however I think it is important for understanding probability, since how one believes in this determines whether or not they would answer questions like the probability of the population mean falling within a confidence interval. Different beliefs about these philosophical details yield completely different answers to questions like that.

It seems like you agree that probability is inherently subjective, although I'm not certain that you do.

6. ### Subjective Probability

I don't really see that there is much of a difference between the subjectivist and objectivist views. We can state the problem you've provided in frequentialist terms and stil conclude that Chris was correct to say 1/10.

Given an infinite number of trials in which the only information which is guaranteed to stay the same is that which Chris is given, the relative frequency of x=5 would be 1/10.

Alice's question is specific to only the "trial/run of the experiment" mentioned above.

Let's call that trial 1 (the one mentioned above where Bob gets a 7).

Imagine that Alice noticed that something else important happened at the same time as Trial 1, which makes Trial 1 of dire importance to Alice, but any other trials throughout time are not important at all to Alice.

In Bob and Chris's perspective we could have more trials: 2, 3, ...N

and it is true that roughly 1/10 of the time the random variable could be 5. But this is a characterization of trials 1 through N and Alice only cares about trial 1.

Is it possible we are actually considering 2 random experiments (which overlap in trial 1)?

There is a random experiment related to the number that the machine spits out - and relative frequency is easily defined here.

There is another random experiment for which repetition is not defined - the one that is important to Alice.

Can relative frequency even be defined here? There is no possible repetition here, only the lack of knowledge on Alice and Chris's part.

Going by relative frequency, and assuming complete objectivity, the probability by definition must be the number of occurrences of the event divided by number of experiments. The experiment only occurs once (by my definition above), and the event x=5 never occurs for any (of the one) times the experiment is run, therefore the probability must be 0 (as Bob stated).

Generalizing this concept, although we as people, treat things as repetitions, and use that in our definition of relative frequency, there is no such thing as repetition from a truly objective standpoint (only from a subjective one). (On a side note, there is no such thing as true sameness from a truly objective standpoint).

This implies to me, that any rigorous probability theory that can claim objectiveness must have only two possible values for any event (1 or 0, depending on whether the statement of the event is true or false).

There is no probability theory in practice that restricts the values to 1 or 0 (that can be called probabilistic at least), and hence all probability theories must be subjective. All Objectivists would say the probability of heads in a coin flip is 1/2, yet this does not make sense because if they had more knowledge (wind currents, specific distribution of weight on the coin, tendency of flippers thumb to flip in a certain way, ...etc) then they would surely adjust their probability evaluation to reflect what they believe is more likely.

My ultimate argument here is that all probability theories are subjective regardless of any claims to the contrary.

7. ### A simple Monty Hall variant

Seeing as how every 8th grader with a math counts trophy knows the optimal play to his game, ol' Monty decided to shake things up.

He added two more doors, for a total of 5 instead of 3, behind which 4 held a goat, and 1 held a brand spanking new electric car (hey, gotta keep up with the times, eh?).

In the first round, the player chooses two doors instead of one. Then Monty opens up one of the other three doors and reveals a goat. Then the player may choose one of his two original doors or switch to one of the unchosen doors. Monty then releases one more goat (gotta give the first goat some company or else it'll eat Monty's brand new trousers), and the player is then given a final choice of the remaining three doors.

1) The second door Monty opens is random?

2) He is purposely playing against you, trying to keep that new car for himself?

And 3) If you weren't sure, how much would you have to suspect dear old Monty's honorable intentions to make either play?

Seeing as how every 8th grader with a math counts trophy knows the optimal play to his game, ol' Monty decided to shake things up.

He added two more doors, for a total of 5 instead of 3, behind which 4 held a goat, and 1 held a brand spanking new electric car (hey, gotta keep up with the times, eh?).

In the first round, the player chooses two doors instead of one. Then Monty opens up one of the other three doors and reveals a goat. Then the player may choose one of his two original doors or switch to one of the unchosen doors. Monty then releases one more goat (gotta give the first goat some company or else it'll eat Monty's brand new trousers), and the player is then given a final choice of the remaining three doors.

1) The second door Monty opens is random?

2) He is purposely playing against you, trying to keep that new car for himself?

And 3) If you weren't sure, how much would you have to suspect dear old Monty's honorable intentions to make either play?

Case A: Player selects 2 doors, one of which has a car; P[A] = 1 - (4 choose 2)/ (5 choose 2) = 2/5

Case B: Player's initial selection does not include car: 1 - P[A] = 3/5

Monty opens goat door.

In case A, one of the player's doors has a car, and both of Monty's remaining doors have goats.

In case B, both of player's doors have goats, and one of Monty's remaining doors have a car.

Case A.Stay: We are in case A, and player stays i.e. picks one of original 2 selected doors; With 1/2 conditional probability, will select car, conditional on being in case A. The net probability for pointing at right door here is 2/5*1/2 = 1/5

A.Stay.A = Pointing at right door (P = 1/5)

A.Stay.B = Pointing at wrong door (P = 1/5)

Case A.Switch: We are in case A, and player switches to one of Monty's doors; Player switches to goat with conditional probability 1, car with conditional probability 0. Net probability for pointing at right door here is 2/5*0 = 0;

A.Switch.A = Pointing at right door (P = 0)

A.Switch.B = Pointing at wrong door (P = 2/5)

Case B.Stay: We are in case B, and player stays; Player pointing at goat with condtional probability 1. Net probability for pointing at right door is 3/5*0 = 0.

B.Stay.A = Pointing at right door (P = 0);

B.Stay.B = Pointing at wrong door (P = 3/5)

Case B.Switch: We are in case B and player switches. Player points at right door with conditional probability 1/2; Net probability is 3/5*1/2 = 3/10

B.Switch.A = Pointing at right door (P = 3/10)

B.Switch.B = Pointing at wrong door (P = 3/10)

Monty is not allowed to open the door the player is pointing at, or the car door.

If player is pointing at the car, Monty may open any door the player is not pointing at.

If player is pointing at a goat, Monty, may not open the player's door, nor the car's door.

At this point, if player move was stay, player is pointing at right door with Prob 1/5

If player move was switch, then, player is pointing at right door with prob 3/10

If player is pointing at right door, Monty may open any of the 3 other doors. (Depending on strategy choice, we can make the probability of this case 1/5 or 3/10).

If player is pointing at wrong door, Monty may only open 2 of the other doors (since one of the other 3 is as car). (Depending on strategy choice, we can make the probability of this case 4/5 or 7/10).

Most likely to win if player is wrong at this point, and Monty gets rid of another goat.

If wrong, and Monty get's rid of another goat, then switch to one of 2 other doors, one of which is the car.

To maximize chance of being wrong at this point, should stay at first opportunity(P[wrong] in that case is 4/5), then switch at 2nd choice (P[win|was wrong] = 1/2).

With this strategy the probability of winning is 4/5*1/2 = 2/5.

To summarize, strategy is pick 2 of the 5 doors at random. After Monty opens a goat door, choose randomly of the 2 doors you initially picked.

After Monty opens another goat door, choose randomly of the other two that remain (not the one you were just pointing at). You will win 2/5 times regardless of how Monty picks doors and whatever his motives are.

8. ### Prime question

By the logic above 97 is the oddest prime number because it is the only prime that has the property that it is not not equal to 97.

9. ### You live in Killville. Can you stay alive?

everyone will die doesn't matter regardless whether you are P or K, if you meet a K, you will die. Your only hope is that all the other K's kill themselves. This will never happen since there are an even number of K's and they can only remove themselves in pairs. If there were an odd number of K's, then you would have a shot. The steady state of such a system would be one K left standing. If there are N K's and N P's and N is odd, then your chance of living would be 1/N due to assumed symmetry amongst all K's.

10. ### You live in Killville. Can you stay alive?

You live in Killville - a town populated by 10 killers and 10 pacifists.

When a pacifist meets a pacifist, nothing happens.

When a pacifist meets a killer, the pacifist is killed.

When two killers meet, both die.

Assume meetings always occur between exactly two persons

and the pairs involved are completely random.

Are your odds of survival better if you are a killer? or a pacifist?

Or does it matter?

Regardless of whether you are a pacifist or a killer,

you may disregard all events in which a pacifist other than yourself is involved

and consider only events in which you are killed

or a pair of killers other than yourself is killed.

everyone will die doesn't matter regardless whether you are P or K, if you meet a K, you will die. Your only hope is that all the other K's kill themselves. This will never happen since there are an even number of K's and they can only remove themselves in pairs. If there were an odd number of K's, then you would have a shot. The steady state of such a system would be one K left standing. If there are N K's and N P's and N is odd, then your chance of living would be 1/N due to assumed symmetry amongst all K's.

11. ### A simple sublimation problem

You have a spherical mothball which sublimes at a rate of k cm3/s per cm2 of exposed solid surface (assuming constant T and P in the room).

When will it be at half its original radius? When will it have completely sublimed?

Radius at time t in cm = r(t)

dV/dt = -k*A(t)

A(t) = 4*pi * r(t)^2

V(t) = 4/3*pi*r(t)^3

dV/dt = 4*pi*r(t)^2 *dr/dt = A(t)*dr/dt = -k*A(t) = -k*4*pi*r(t)^2

dr/dt = -k

r(t) = -kt + r(0)

Half sublimed is

r(0)/2 = r(0) - kt

t = r(0)/(2k)

Fully sublimed is

0 = r(0) - kt

t = r(0)/k

12. ### How many heats?

* Split them into five heats. Take the 3 fastest in each heat, meaning you now have the 15 fastest.

* So do 3 heats now. Take the 3 fastest in each heat. You have the 9 fastest horses.

* Do a single heat with 5 random horses of the 9, cutting out the 2 last-place horses, leaving you with 7.

* Take another random 5 horses and do it again, cuttnig out 2, leaving you with 5 horses.

* Do one final heat and take the top 3 from that race.

Those top 3 horses will only have to race 5 times each.

In my solution, the top horse races twice, the 2nd and 3rd to best horses race 3 times each (the max for any horse).

13. ### How many heats?

The annual Horse racing tournament is a week away.

You own 25 horses, and you're allowed to enter up to three.

You decide to hold a series of races to find your three fastest horses.

You have access to a track that permits five horses to race against each other.

Here's the deal:

[1] No two of your horses are equally fast - there will be no ties.

[2] You have no stopwatch - you can't compare performances from different heats.

[3] You do not want to tire the horses needlessly - the fewer heats needed, the better.

[4] You want to know which horse is fastest, which is next fastest and which is third-fastest of the 25.

How many heats are needed?

7?

Randomly divide into 5 groups of 5 and race each group.

5 races so far

Let G(X,Y) be the Yth fastest horse from group X

Race G(x,1) for all x in 1..5 i.e. fastest horse from each group

i.e.

G(1,1),

G(2,1),

G(3,1),

G(4,1),

G(5,1)

Let H(Y) be the number of the original group (of the 5 original groups) of the Yth fastest horse from the 6th race.

Race the following 5 horses:

G(H(1),2)

G(H(1),3)

G(H(2),1)

G(H(2),2)

G(H(3),1)

The 2nd and 3rd fastest are the 1st and 2nd fastest in this 7th race respectively

14. ### Pole in lake

**But what makes you assume that it is 1/3 of the entire pole? The problem states: "One half of the pole is in the ground, another one third is covered by water and eight feet is out of the water."

Another- would imply 'in addition to' or 'along with' So why would one assume that this reffering to a fraction of the entirety?

Please correct me if im wrong!

Whenever fractions are used, there is an implicit "of the" following the ratio being discussed.

"One half of the pole is in the ground, another one third of the ______ is covered by water and eight feet is out of the water."

What makes more sense to go in the blank based on common English?

A.) pole

B.) portion of the pole which is not in the ground

Let's insert each and see which one looks more correct:

A.) "One half of the pole is in the ground, another one third of the pole is covered by water and eight feet is out of the water."

B.) "One half of the pole is in the ground, another one third of the portion of the pole which is not in the ground is covered by water and eight feet is out of the water."

In this case, the word "another" implies that they are talking about the same thing i.e. the whole pole.

If they changed the object to be something else, they wouldn't use the word "another" there.

15. ### Subjective Probability

Both Bob and Chris are correct.

Probability is contingent upon given information.

If you create a hypothetical scenario in which you stipulate that some relative frequency distribution exists and that the experiment can only be performed once, then it is valid. Beyond mere stipulation, though, how are we to determine a relative frequency distribution if we have never performed the experiment before?

I was trying to see how many people would argue for a frequentist interpretation (probability is objective ratio of occurrences to infinite number of experiments) vs. a bayesian one (probability is a subjective degree of certainty).

This all originally came out of something a statistics professor said to my class about two years ago.

We were talking about confidence intervals regarding the population mean of a set of measurables.

He asked the class, for a 90% confidence interval, what is the probability that the population mean is within our calculated interval?

Some people said 0.9, but he said wrong - it is either 1 or 0, since the population mean can either be inside the interval or it cannot be.

The book agreed with him, and I disagreed with both him and the book.

I thought to myself, what if knowledge wasn't a concern - what if something knew everything, i.e. could determine whether any statement was true or false.

Wouldn't the answer to any probability question then be one or zero? Can't all probabilistic events be restated as a question of whether some statement is true or false? I concluded probability as a tool only makes sense in the context of missing knowledge, and that this is in essence the same thing as subjectivity.

I have some philosophical qualms with the frequentist perspective and was hoping to get some representation from someone who better understands it.

It seems like the responders all believe that probability is subjective though.

I don't believe the frequentist interpretation can be used to answer any question about any specific event occurring and still be consistent with the claim of objectivity. By specific event, I mean like: what is the probability that Bob got a 5 from the machine at this point in space-time?

There is no such thing as repetition for Bob drawing from the machine at this point in time (I don't think it matters if the time under discussion is past, present, or future) , so is the frequentist definition of probability even valid in this case?

I agree it may be valid for characterizing the output of the random number generator over many requests, but how can it ever be justifiably used to answer any question about any specific event?

Yet I don't think a frequentist would hesitate to answer "what is the probability that my next coin flip is heads?", which is a specific event.

To be objective, the answer must be independent from any specific subjective mind. If you took two "objective" frequentists and asked them the same question, but showed one of them that the coin had double-heads and did not show the other, then with certainty they would have different answers. What if the outcome was not so obvious as double-heads, but in general, I would expect someone with extraordinary predictive power or relevant hidden knowledge to not always have the same probability response those with duller minds and lack of knowledge.

I am convinced that probability cannot be objective, but feel there must be a good reason that so many people believe it to be.

I am interested in the insight of others.

Thanks,

16. ### Prime question

2 Since everything else is not even.

17. ### Subjective Probability

A discrete, uniform random number generator, which produces any number between 1 and 10 inclusive, all outcomes of which are equally likely, gives Bob a random number, which we shall call x.

Bob knows what x is (he knows that it is 7).

Chris does not know what x is, but knows that the random number generator produces whole numbers between 1 and 10, inclusive, equally likely.

Alice poses the following question to both Bob and Chris.

"What is the probability that x is 5?"

Chris says: 1/10

Bob says: 0

Who is right?

What answer should the person who was wrong have given?

Does your definition of probability depend on how much knowledge you have (i.e. subjective), or is it independent of knowledge (i.e. objective)?

Does the concept of relative frequency make sense for one-time, non-repeatable experiments, for which an outcome is not known?

×

• #### Activity

• Riddles
×
• Create New...