Two Coins: Probability

DeletedUser9470

Ive generated 100 random flips on excel:
smartass2.png

out of 100 possible outcomes 20 are TT
so that leaves us with 80 possible outcomes where at least 1 coin is heads.
what percentage of this total has the outcome HH?
If I'm not mistaken I count 24.
24/80
thats 30%
nowhere close to 50%

like I said, do it yourself...
 

DeletedUser

100 flips is kinda small......and those flips are merely pseudo-random.......
 

DeletedUser9470

100 flips is kinda small......and those flips are merely pseudo-random.......

show us a million flips then.
the result will be even more accurate tending towards 33.33%.
 
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DeletedUser28032

*Sighs* someones going to spend all of tomorrow flipping coins aren't they
 

DeletedUser

If I had the money to buy this fancy little thing, I would. And squander my days getting ever closer to an infinite number of flips.
 

DeletedUser

I think that my problem comes with determining when you're calculating the probability. If you say before flipping them that you want to know the probability of both being heads if one of them is, then the answer would be 1:3. If you flip them and are told that one of them is definitely heads and to calculate the probability that the other one is also heads, the answer is 1:2.

It's been a lot of years since I took a logics/statistics class, but that's what I co e up with using my somewhat foggy memory.
 

DeletedUser9470

I think that my problem comes with determining when you're calculating the probability. If you say before flipping them that you want to know the probability of both being heads if one of them is, then the answer would be 1:3. If you flip them and are told that one of them is definitely heads and to calculate the probability that the other one is also heads, the answer is 1:2.

It's been a lot of years since I took a logics/statistics class, but that's what I co e up with using my somewhat foggy memory.

the flip has already occurred
thus the only options with only one head are
HT
TH
and HH
in all 3 of these cases, at least one of them is definitely heads.
 

DeletedUser

"The answer is 50%"
...
"the question is ambiguously presented"
...
"So, as you can see, it is NEVER 33% [sic]"
...
"the probability that both are heads. That probability is 25%"
...
"There is no correct answer"


All the above quotes from Hellstrom in this thread. Looks like squirming to me.
The question is not ambiguous - it's a simple and easily-understood puzzle. It requires appreciation of a simple principle that is not usually immediately intuitively obvious. That's not to say that everyone will 'get it' or that people won't hide behind some fancied imprecision in the formulation. That's why this is such a refreshing thread - there's a right answer and a wrong one and there really isn't any wriggle-room in between, unlike most of the subjects in D & D.

Any quibbles about edge-landing, fair coins, angular momentum or whatever are just flim-flam. It's like disputing a chess problem that delivers mate in three moves by saying a cat may jump on the board and the players forget the position.:p

One last time - toss 2 coins repeatedly and 1/4 of the time you will get double-tails, a quarter of the time you will get double-heads, and half the time you will get a head-and-tail combination (order immaterial). By specifying that there is at least one head we eliminate the double-tail possibility, so the chance of having a double head rises from one in four to one in three.

Imagine drawing 2 cards from an 'infinite' and randomised pack (i.e. drawing a card does not prevent the likelihood of drawing the same card again, unlike with a finite pack of 52 cards). You are told that one of the cards is an ace. What is the likelihood of having a pair of aces?
Intuition would likely tell you that the chances are 1 in 13 (at least, that's what mine did). You have one ace and the other card surely has a 1 in 13 chance of being an ace, no?
No.
If you actually work it out, the chance is 1 in 25.
If you can't figure that, think of it this way>
If the first card you draw is an ace then you really do have a 1 in 13 chance of making an ace-pair, as you would expect.
But if only the second card is an ace then you have a 0 in 13 chance of making an ace-pair, but the condition that one of the cards is an ace is still met.
So if you take all the 2-card combos that contain an ace clearly a lot less than 1 in 13 are ace-pairs, and it works out to 1 in 25.

Exactly analagous to the coin problem.
 

DeletedUser9470

lol, nice to see Neo is consistently missing the boat. --- http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

paha, it is you my friend who has missed the boat, and it looks like you need a lifejacket too!

"Gardner argued that a "failure to specify the randomizing procedure" could lead readers to interpret the question in two distinct ways"

in this brainteaser the randomising procedure is very specific.

1-> YOU FLIP TWO COINS.
possibilities:
HH
TT
HT
TH

2->YOU ARE TOLD: "at least one is heads"
possibilities:
HH
HT
TH

3->WHAT ARE THE CHANCES OF HH?
1/3

its very simple to understand this randomising procedure.
 

DeletedUser

Ok, I finally figured out where my problem really is. (At least my problem with this question!)

If you are asking what the chances are that both coins are heads if one is, the answer is 1:3. I think a lot of us have been reading it as asking what the chances are of the other coin also being heads if one of them is; the answer to that would be 1:2. :huh: In that case, I have to change my response, since it is asking the former rather than the latter.
 

nashy19

Nashy (as himself)
By selectively removing TT which shouldn't actually be TT but should have at least one H, you are artificially increasing the probability of HH in your results.

Of course heads has more of a chance if you ignore a boatload of T flips.

Normal TT = 25%
25% / 3 = 8.333...
25 + 8.333... = 33.333...

Using Neo's results:
TH = 76%
HH = 24%

Very close to 25%.

How do you think a condition that requires "at least one heads" would impact something that already has at least one heads anyway?
 
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DeletedUser

"The answer is 50%"
...
"the question is ambiguously presented"

etc. As I said in one of my earlier posts (my third post, page 2), I continued the debate, for craps and giggles, because people wanted to continue it, despite evidence presented that indicated the question was ambiguous, and thus there is no correct answer. Or, if you wish to argue cup full vs cup empty, there are two or more correct answers depending on your interpretation of the question, which is in itself ambiguous:

"The Boy or Girl paradox surrounds a well-known set of questions in probability theory which are also known as The Two Child Problem,[1] Mr. Smith's Children[2] and the Mrs. Smith Problem. The initial formulation of the question dates back to at least 1959, when Martin Gardner published one of the earliest variants of the paradox in Scientific American. Titled The Two Children Problem, he phrased the paradox as follows:
  1. Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
  2. Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?
"Gardner initially gave the answers 1/2 and 1/3, respectively; but later acknowledged that the second question was ambiguous.[1]Its answer could be 1/2, depending on how you found out that one child was a boy. The ambiguity, depending on the exact wording and possible assumptions, was confirmed by Bar-Hillel and Falk,[3]and Nickerson.[4]

"Other variants of this question, with varying degrees of ambiguity, have been recently popularized by Ask Marilyn in Parade Magazine,[5] John Tierney of The New York Times,[6] and Leonard Mlodinow in Drunkard's Walk.[7] One scientific study[2] showed that when identical information was conveyed, but with different partially-ambiguous wordings that emphasized different points, that the percentage of MBA students who answered 1/2 changed from 85% to 39%.

"The paradox has frequently stimulated a great deal of controversy.[4] Many people argued strongly for both sides with a great deal of confidence, sometimes showing disdain for those who took the opposing view. The paradox stems from whether the problem setup is similar for the two questions.[2][7] The intuitive answer is 1/2.[2] This answer is intuitive if the question leads the reader to believe that there are two equally likely possibilities for the sex of the second child (i.e., boy and girl),[2][8] and that the probability of these outcomes is absolute, not conditional.[9]" ~ http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

The example posed in the initial post is a duplicate of the second question, which is that of, "Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?"

As such, the answer is, "it's ambiguous."
 
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DeletedUser9470

etc. As I said in one of my earlier posts (my third post, page 2), I continued the debate, for craps and giggles, because people wanted to continue it, despite evidence presented that indicated the question was ambiguous, and thus there is no correct answer. Or, if you wish to argue cup full vs cup empty, there are two or more correct answers depending on your interpretation of the question, which is in itself ambiguous:

"The Boy or Girl paradox surrounds a well-known set of questions in probability theory which are also known as The Two Child Problem,[1] Mr. Smith's Children[2] and the Mrs. Smith Problem. The initial formulation of the question dates back to at least 1959, when Martin Gardner published one of the earliest variants of the paradox in Scientific American. Titled The Two Children Problem, he phrased the paradox as follows:
  1. Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
  2. Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?
"Gardner initially gave the answers 1/2 and 1/3, respectively; but later acknowledged that the second question was ambiguous.[1]Its answer could be 1/2, depending on how you found out that one child was a boy. The ambiguity, depending on the exact wording and possible assumptions, was confirmed by Bar-Hillel and Falk,[3]and Nickerson.[4]

"Other variants of this question, with varying degrees of ambiguity, have been recently popularized by Ask Marilyn in Parade Magazine,[5] John Tierney of The New York Times,[6] and Leonard Mlodinow in Drunkard's Walk.[7] One scientific study[2] showed that when identical information was conveyed, but with different partially-ambiguous wordings that emphasized different points, that the percentage of MBA students who answered 1/2 changed from 85% to 39%.

"The paradox has frequently stimulated a great deal of controversy.[4] Many people argued strongly for both sides with a great deal of confidence, sometimes showing disdain for those who took the opposing view. The paradox stems from whether the problem setup is similar for the two questions.[2][7] The intuitive answer is 1/2.[2] This answer is intuitive if the question leads the reader to believe that there are two equally likely possibilities for the sex of the second child (i.e., boy and girl),[2][8] and that the probability of these outcomes is absolute, not conditional.[9]" ~ http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

The example posed in the initial post is a duplicate of the second question, which is that of, "Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?"

As such, the answer is, "it's ambiguous."

Second question

Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?
This question is identical to question one, except that instead of specifying that the older child is a boy, it is specified that at least one of them is a boy. In response to reader criticism of the question posed in 1959, Gardner agreed that a precise formulation of the question is critical to getting different answers for question 1 and 2. Specifically, Gardner argued that a "failure to specify the randomizing procedure" could lead readers to interpret the question in two distinct ways:
From all families with two children, at least one of whom is a boy, a family is chosen at random. This would yield the answer of 1/3.
From all families with two children, one child is selected at random, and the sex of that child is specified. This would yield an answer of 1/2.[3][4]
Grinstead and Snell argue that the question is ambiguous in much the same way Gardner did.[11]
While it is certainly true that every possible Mr. Smith has at least one boy - i.e., the condition is necessary - it is not clear that every Mr. Smith with at least one boy is intended. That is, the problem statement does not say that having a boy is a sufficient condition for Mr. Smith to be identified as having a boy this way.
Commenting on Gardner's version of the problem, Bar-Hillel and Falk [3] note that "Mr. Smith, unlike the reader, is presumably aware of the sex of both of his children when making this statement", i.e. that 'I have two children and at least one of them is a boy.' If it is further assumed that Mr Smith would report this fact if it were true then the correct answer is 1/3 as Gardner intended.


from your own wiki justification to your erroneous answer to this brainteaser mr HS.
here Both coins are selected at random right at the start of the problem.

and thus there is NO ambiguity.
the answer is 1/3
 
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DeletedUser

lol, did you actually bother to read what was presented? Wow, you're being stubborn to the point of blinders.

Here, take the time to read about probabilities and how "interpretation" dictates: http://www.math.dartmouth.edu/~prob/prob/prob.pdf

If that's not sufficient, how about reading all these?

http://psycnet.apa.org/?&fa=main.doiLanding&doi=10.1037/0096-3445.133.4.626

http://www.ncbi.nlm.nih.gov/pubmed/15584810

http://www.sciencedirect.com/science/article/pii/001002778290021X

Essentially, they all demonstrate that you are making assumptions and that those assumptions are allowing you to feel confident in your answer. That doesn't make you right, it just makes you certain.

Because of this certainty, you are most definitely incorrect. I.e., you are incorrect in being certain that you are correct. Why? Because ambiguity indicates there is more than one interpretation and, as such, there is more than one possible answer, depending upon your interpretation, your assumptions.
 

DeletedUser

Nice try HS, but no cigar.

In RL we have to acquire information in a certain way and this affects our calculations.
But what we have here is clearly a mathematical problem - an abstract formulation. In such problems is it assumed that we are given ALL the information necessary to arrive at a solution. That is why we can ignore the possibility of the coin falling on its edge - if we were supposed to allow for that we would have been given the dimensions of the coin. I go back to the cat and chessboard analogy I gave earlier. Likewise, we can ignore the process by which it is established that at least one coin falls heads-up.

Neo is right (that wasn't easy to say!). The answer is one-third. The way we are to approach the problem is implicit in its formulation. This is not about psychological certainty - it's a mathematical problem with a definite and unique answer and it's odd that you can't (or don't want to) get your head round that. (But thank you for giving me the opportunity, which would never otherwise have arisen, of pwning you! )
 
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DeletedUser

Lo, np. And you're wrong, I'm right. :p



(btw, tried to send u a pm earlier, but it seems you have them turned off, or your mailbox is full)
 

DeletedUser

We can also prove Neo's answer using his rephrasing that the question is "The chance of HH given it's not TT", which is statistically written as P(HH|TT')

Using the formula P(A|B) = P(AnB)/P(B), we can say that:
P(HH|TT') = P(HH n TT') / P(TT')

As P(HH) and P(TT) are mutually exclusive, we can say that P(HH n TT') = P(HH), which a simple tree diagram tells us is 0.25

As P(TT) is also 0.25, P(TT') is 0.75

Therefore, P(HH | TT') = 0.25 / 0.75, which simplifies to 1/3

Hoorah!

Source: Stats revision :p
 
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