Poll: A little math problem

[Alex_P]

"Each coin toss is independent, but the distribution of results within the set is a known function. If you knew an honest, unbiased coin was flipped 10 times and the first 9 times came up heads, you'd guess the 10th time would be tails even though as an honest, unbiased coin the chances of that toss were 50-50 just like all the other tosses." [http://www.escapistmagazine.com/forums/jump/18.73797.809901]

This is an incorrect statement. If you keep flipping fair coins in order, each is 50-50. If you flip a set together, however, and count up the outcomes, it is much more likely to contain (9 heads, 1 tails) than (10 heads, 0 tails).

-- Alex

 

[Alex_P]

Another example to maybe break the deadlock.

First scenario: same as the question, assume you guys are right, answer is 33%

Second scenario: Puppy Washing Man doesn't just say "Yes!", he says "yes, the dog I have in my hand is male." What's the answer to the question of what are the chances they are both male? Is it 50%

If so, why does the man answering the question "yes" because of what he knows about the dog in his hand make it more likely that the other dog is male than if he just answers on the basis of what he knows about the set as a whole? That's what I'm really trying to find out. Werepossum mentioned sequential probability vs. set probability a while back, but I couldn't find out anything about that. Does anyone know about that?

In other words, I'm looking for why if our knowledge is about the binary, on/off, yes/no characteristic of one member or the other of a set of two, and not about the characteristic of a specific member of a set of two, that makes a difference.

Consider a set of a hundred "fair" puppies.

If you pick a single puppy out of the set without knowing its sex, what is the chance you'll successfully find a male?

If you go through the set and pick out a male puppy instead, what is the chance you'll successfully find a male?

-- Alex

 

[Saskwach]

Well then what is it? Because other people are telling me it's a permutation problem.

This:

A few pages back I brought up permutations and combinations myself, and explained why this question was a permutation problem where the first dog (the one that's a male) is uncertain, so we have to factor that in.

It is a permutation problem, but one in which the order is uncertain, so M/F and F/M still have to be considered separate. In other words, which dog are we starting with - Jesse or Osan? Which one is the male who we will, after the fact, have counted first (which one is the male)? No one knows - and since we don't we have to keep in mind that there are two ways that I might end up with 1 male and 1 female. Jesse could be the male, and thus the dog we have already checked, or maybe it's Osan. Who knows? There are thus twice the configurations in which we'd be expected to check a female after a male than there are configurations in which we'd find a male after the first male (M/M).

 

[Alex_P]

If you pick a single puppy out of the set without knowing its sex, what is the chance you'll successfully find a male?

If you go through the set and pick out a male puppy instead, what is the chance you'll successfully find a male?

A. 50%

B. the wording of your question is confusing, but from what I think it says, 100%

I think you've got the gist of it.

"A" is, as you said, exactly 50%.

"B" is pretty close to 100%. The actual fraction is 1 - (1/2)^100, to be exact. There is one situation in which case you won't get a male puppy at all, which is if all 100 puppies are female.

Does that make sense?

-- Alex

EDIT: Oh, shit, you also said the wording was confusing. Let me be specific: imagine I said "Okay, find me a male puppy. You can keep looking for as long as you want (until you totally run out of puppies)."

 

[Saskwach]

It is a permutation problem, but one in which the order is uncertain, so M/F and F/M still have to be considered separate.

Okay, and why doesn't considering them in the way I did work:

http://www.escapistmagazine.com/forums/jump/18.73797.816359

This is why:

It is a permutation problem, but one in which the order is uncertain, so M/F and F/M still have to be considered separate. In other words, which dog are we starting with - Jesse or Osan? Which one is the male who we will, after the fact, have counted first (which one is the male)? No one knows - and since we don't we have to keep in mind that there are two ways that I might end up with 1 male and 1 female. Jesse could be the male, and thus the dog we have already checked, or maybe it's Osan. Who knows? There are thus twice the configurations in which we'd be expected to check a female after a male than there are configurations in which we'd find a male after the first male (M/M).

The way you did ignores that we don't know which dog is the male: you've assumed that, because we know one of them is, we must know which one is. You might not think that yourself, but your working at least has discounted that possibility.
Example 1: Jesse is male and Osan is female. If I checked Jesse and then Osan I woulf get M then F.
Example 2: Jesse is female and Osan is male. If I checked Jesse first again I would get F and then M.
So, knowing only that Jesse and/or Osan is male, but not which one, I can't say "let's check Osan now, because I know Jesse is the male". That is essentially what your working does - ignore the possibility that we might have to check Jesse because Osan is the male. You've declared that the first dog you'll check is Jesse and Jesse must be the male. Again, this is a permutation problem with no sure order of dog checking, so we have to include both of the two orders of dog checking that woudl come about if one or the other dog was male.

Both male = .25 (.5 x .5)

One of each = .50 (.5 x .5 + .5 x .5 -XOR- .5 x .5 + .5 x .5)

Both female = .25 (.5 x .5)

This is absolutely correct - 1 male/1 female is twice as likely to occur then either of the two other outcomes. Those probability distributions arise from the following possible outcomes:
J O
M M
M F
F M
F F
As you can see, 50% of those outcomes are the combination M/F but they are different permutations: Jesse being a male and Osan a female is quite different to Osan a male and Jesse a female. Your own reasoning accepts this.

Now, what actually happens when we get that information from the Puppy Washing Man? We know at least one dog is male. Now, what happens to the probability of one dog being male when we know one dog is male: it becomes 1, and therefore the probability of one dog being female becomes 0. Well, here's what I think happens: I've put the changes that I think are correct in bold.

Both male = .5 (1 x .5)

One of each = .5 (1 x .5 + 0 x .5 -XOR- 0 x .5 + 1 x .5)

Both female = 0 (0 x .5)

That's...about as well as I think I'm going to be able to explain myself.

You reached your conclusion from the "at least one male" information. But what does this tell us, really? It tells us only that the F/F possibility is impossible. Both the outcome is which Jesse is the one male and the one in which Osan is the one male - both of them satisfy the "at least one male" condition. So we can't take either of the outcomes out, both of which contribute to the 0.5 probability.
To return to Jesse and Osan, looking at which of the above outcomes would be possible with the new information:
J O
M M - still possible
M F - still possible
F M - still possible
F F - not possible

Before "at least one is male":
P(two males)=0.25
P(1 male, 1 female)=0.5
P(two females)=0.25

After "at least one is male":
P(two females)=0
P(two males)=0.25
P(1 male, 1 female)=0.5
but
P(total)=0.75=/=1

The thing is that we have taken out 0.25, leaving us with 0.75 total probability, instead of the necessary 1. But we've only taken out the information that's no longer possible and are left with the only possible outcomes adding up to 0.75. Since the only possible outcomes sum to 0.75, we have to do a simple operation - one any probability student would know to do - that makes 0.75 now equal 1, and apply that same operation to the probabiliites of all possible outcomes.

P(total|two females is impossible)=0.75*(4/3)=1
Applied to all probabilities:
P(two males|two females is impossible)=0.25*(4/3)=0.333...
P(1 male, 1 female|two females is impossible)=0.5*(4/3)=0.666...

 

[fedpayne]

I've pretty much convinced myself it's a third by thinking of it in terms of they need to both be male. and the options are M,M, F,M, M,F. But I kind of wonder if we need another M,M. The only thing that makes me think that it could be 50% is if you consider what the other puppies could be in relation to the one picked out.
Could be an older sister
Could be a younger sister
Could be an older brother
Could be a younger brother

If we talk about F,M and M,F as being different based on which one has been born first, surely it needs considering? Why should it be ruled out? I mean, I know it should be, but why?

Or is it just that it is a possibility that the other dog could be older/younger male, it is just less likely. Yeah, I think I understand now. It does need to be thought about with maths. My beloved words have failed me. I'm going to go change courses :'(

 

[werepossum]

Damn. I had abandoned this thread and then I just HAD to look at it one more time.

Samirat had the clearest explanations, I think. But I'll start with the ten coin tosses.
Scenario #1: Toss a fair coin ten times in a row. If the first nine tosses come up heads, the tenth toss still has a 50-50 chain of being heads. Each coin flip is independent. This is sequential probability.

Scenario #2: Toss a fair coin ten times in a row without looking at the results. Now begin looking at the results for the first nine. They are all heads. What is the chance of the tenth being heads? About one in a thousand. This is set probability.

Why the difference? In the first scenario, each coin toss is independent. In the first scenario, by grouping ten tosses together AFTER you know the results of the first nine, you've already eliminated the vast majority of possible permutations for a set of ten coin tosses in forming the set. In the second scenario, you've grouped ten coin tosses together without knowledge of any set member's value. Thus you preserve all possible permutations for a set of ten coin tosses.

Now shift to the pup problem. For clarity, let's assume one pup is brown and one black. There is a 50% chance each is male. Therefore our possible permutations are 25% brown male/black female, 25% brown male/black male, 25% brown female/black male, and 25% brown female/black female. Our possible combinations are thus 25% all male, 50% pair, and 25% all female. That's pretty simple, and everyone should be able to agree up to this point. At this point, we can work the problem as either a combination problem or a permutation problem. Here's a matrix:
Brown Male/Black Female 25%
Brown Male/Black Male 25%
Brown Female/Black Male 25%
Brown Female/Black Female 25%

Now comes the difficult part. We are told that one pup is male, but not which one. Let's look first at the permutation problem. This eliminates the "Female/Female" permutation, obviously. Can we eliminate another permutation? No, because Brown Male/Black Female, Brown Male/Black Male, and Brown Female/Black Male all satisfy the requirement that there be at least one male. The question then asked is what is the chance the other dog is also male? Only one permutation in the remaining three can possibly satisfy that question - thus the chance is 33%. Our probability of having two males in our set has increased from 25% to 33% because we have eliminated one possible permutation prior to asking the question.

Let's look next at the combination problem. Our possible combinations are thus 25% all male, 50% pair, and 25% all female. We can first eliminate the all female combination because we have been told there is at least one male. Can we eliminate another combination? No, because "all male" and "pair" both satisfy the requirement that there be at least one male. Our possible remaining combinations are 25% all male, 50% pair. The question then asked is what is the chance the other dog is also male, which would require the all-male combination. Thus the chance of having two males is one in three or 33%, up from 25% because prior to being asked the question we were given information allowing us to remove the "all female" combination.
MATH:
Permutation: 25%/(25% + 25% + 25%) = 33%
Combination: 25%/(25% + 50%) = 33%

I can't possibly be any more clear.

 

[BigKingBob]

Wow, I never thought I'd see a mathematical flame war...

All this debate about permutations and combinations is of no use!

You know there are two dogs and that one (Dog A) is male. The gender of Dog B is completely independent and unrelated to the gender of Dog A (i.e the chance of the Dog B being born a male is in no way effected by the fact that Dog A was born a male.) And that the probability for both dogs of being born male is 0.5

Once you get your head around that then you will see that all the question is really asking is what the chance of Dog B being male is. As stated before, it has a 0.5 chance of being born a male, therefore the answer is 50%.

There, its been solved.

(I will also explain the monty hall problem in less than 100 words on request, the length of some of these answers is staggering!)

 

[Samirat]

Wow, I never thought I'd see a mathematical flame war...

All this debate about permutations and combinations is of no use!

You know there are two dogs and that one (Dog A) is male. The gender of Dog B is completely independent and unrelated to the gender of Dog A (i.e the chance of the Dog B being born a male is in no way effected by the fact that Dog A was born a male.) And that the probability for both dogs of being born male is 0.5

Once you get your head around that then you will see that all the question is really asking is what the chance of Dog B being male is. As stated before, it has a 0.5 chance of being born a male, therefore the answer is 50%.

There, its been solved.

(I will also explain the monty hall problem in less than 100 words on request, the length of some of these answers is staggering!)

How do you know that Dog A is male? What if Dog A is female, and Dog B is male? There are three possibilities:

FM
MF
MM

Saying that one is male only cuts out the option of FF. So the chances that the other one if male are 33 percent.

 

[Alex_P]

Wow, I never thought I'd see a mathematical flame war...

All this debate about permutations and combinations is of no use!

You know there are two dogs and that one (Dog A) is male. The gender of Dog B is completely independent and unrelated to the gender of Dog A (i.e the chance of the Dog B being born a male is in no way effected by the fact that Dog A was born a male.) And that the probability for both dogs of being born male is 0.5

Once you get your head around that then you will see that all the question is really asking is what the chance of Dog B being male is. As stated before, it has a 0.5 chance of being born a male, therefore the answer is 50%.

There, its been solved.

(I will also explain the monty hall problem in less than 100 words on request, the length of some of these answers is staggering!)

Fail. (*)

You know there is a set of two dogs and that at least one of the dogs in the set must be male. You cannot say anything definitive about DogA or DogB, only about the set (DogA and DogB).

-- Alex

* - The snark is only there because you're really sure of yourself, but really wrong.

 

[Lukeje]

Wow, I never thought I'd see a mathematical flame war...

All this debate about permutations and combinations is of no use!

You know there are two dogs and that one (Dog A) is male. The gender of Dog B is completely independent and unrelated to the gender of Dog A (i.e the chance of the Dog B being born a male is in no way effected by the fact that Dog A was born a male.) And that the probability for both dogs of being born male is 0.5

Once you get your head around that then you will see that all the question is really asking is what the chance of Dog B being male is. As stated before, it has a 0.5 chance of being born a male, therefore the answer is 50%.

There, its been solved.

(I will also explain the monty hall problem in less than 100 words on request, the length of some of these answers is staggering!)

Fail. (*)

You know there is a set of two dogs and that at least one of the dogs in the set must be male. You cannot say anything definitive about DogA or DogB, only about the set (DogA and DogB).

-- Alex

* - The snark is only there because you're really sure of yourself, but really wrong.

Read the question again; it states that 'at least one is male', of course, but it also refers to a dog as the 'Other'. This is a source of ambiguity, and leads to the fact that you can solve it two different ways, to get two different answers (3, if you count my solution of zero, which no-one seems to be, despite the infallable logic).

 

[Aurora219]

This is a very simple logic puzzle. You have to realise these are separate issues, the fact that the first one is male doesn't change the probability of the other one being male or female. So say 63% of the population of all beagles was female, then this'd have a 37% chance of being male.

 

[Samirat]

This is a very simple logic puzzle. You have to realise these are separate issues, the fact that the first one is male doesn't change the probability of the other one being male or female. So say 63% of the population of all beagles was female, then this'd have a 37% chance of being male.

Who says the first one is male? What if the second one is male? You, like most others here, are making the mistake of assuming that the male dog must be the first one, so out of this set:

MM
MF
FM
FF

You are ruling out both FF and FM, when the question only prohibits FF. Since either dog could be male, there are two situations where the other dog will be female and only one where it will be male.

 

[Samirat]

Wow, I never thought I'd see a mathematical flame war...

All this debate about permutations and combinations is of no use!

You know there are two dogs and that one (Dog A) is male. The gender of Dog B is completely independent and unrelated to the gender of Dog A (i.e the chance of the Dog B being born a male is in no way effected by the fact that Dog A was born a male.) And that the probability for both dogs of being born male is 0.5

Once you get your head around that then you will see that all the question is really asking is what the chance of Dog B being male is. As stated before, it has a 0.5 chance of being born a male, therefore the answer is 50%.

There, its been solved.

(I will also explain the monty hall problem in less than 100 words on request, the length of some of these answers is staggering!)

Fail. (*)

You know there is a set of two dogs and that at least one of the dogs in the set must be male. You cannot say anything definitive about DogA or DogB, only about the set (DogA and DogB).

-- Alex

* - The snark is only there because you're really sure of yourself, but really wrong.

Read the question again; it states that 'at least one is male', of course, but it also refers to a dog as the 'Other'. This is a source of ambiguity, and leads to the fact that you can solve it two different ways, to get two different answers (3, if you count my solution of zero, which no-one seems to be, despite the infallable logic).

All right, let's name the dogs.

There are three situations.

Sparky could be male, Othello could be male.
Sparky could be female, Othello could be male.
Sparky could be male, Othello could be female.

The fact that they refer to "the other dog," doesn't mean that they've defined which dog that is. For instance, in situation 1 and 3, Sparky is the first male dog she found, and hence, the other dog is Othello. In situations 2, the second dog she checked, Othello, is male, and so the other dog is Sparky.

What was this infallible logic again? Refresh me, and I will point out the fallacies. Are you really going to try to maintain that if you actually performed this problem, the chances that the other dog would be male are 0?

 

[Alex_P]

Scenario #1: Toss a fair coin ten times in a row. If the first nine tosses come up heads, the tenth toss still has a 50-50 chain of being heads. Each coin flip is independent. This is sequential probability.

Scenario #2: Toss a fair coin ten times in a row without looking at the results. Now begin looking at the results for the first nine. They are all heads. What is the chance of the tenth being heads? About one in a thousand. This is set probability.

Your description of #2 sound fishy.

These scenarios are equivalent:
A. I flip a coin and look at it, then I flip another coin and look at it, &c., up to 10 coins.
B. I flip ten coins and then look at them in some specific order (it can be different from the order in which I flipped them).

The probability only changes when you start cherry-picking: searching until you find a male.

-- Alex