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Forum Index » » English (General) » » Maths R Fun
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 Author Maths R Fun
Julian Delphiki
Admiral

Joined: June 19, 2002
Posts: 170
Posted: 2010-11-14 07:38   
I agree with Geja, and the others that said 50%.. I first would have thought 2/3, but after re-reading it... Just because its on a math website, doesn't make it absolutely true..

If you selected the coin after flipping both and getting heads on both, it'd hold at 2/3.



But look at it this way. Taking the extra information out(that in the problem isn't really necessary, as it isn't worded correctly to be relevant to the question itself)

Quote:
Suppose you have two coins. One of these coins, let's call it Coin A, has heads on both sides. The other coin, Coin B, is a regular coin of which one side is heads, and the other is tails.

You randomly grab one of these coins. You flip this coin, and the result is heads.

What is the probability of you having grabbed Coin A?



Becomes

Quote:
Suppose you have two coins. Let's call them Coin A and Coin B.

You randomly grab one of these coins.

What is the probability of you having grabbed Coin A?



Without the information that is worded in a way that makes it irrelevant to the final question, it is 50%.

The probability of grabbing one coin out of two before there are any unique circumstances such as both appearing the same on one side, is 50%




The only ways it would be 2/3 is if it was worded:

Quote:
Suppose you have two coins. One of these coins, let's call it Coin A, has heads on both sides. The other coin, Coin B, is a regular coin of which one side is heads, and the other is tails.

Both coins are on a flat opaque surface. Both coins are showing "heads" on the top. You slide one towards you.

What is the probability of you having chosen Coin A?



In this question, the fact that the coin is "heads-up" is relevant to the question, because you have to choose between two coins that are displaying heads.

Another way would be:

Quote:
Suppose you have two coins. One of these coins, let's call it Coin A, has heads on both sides. The other coin, Coin B, is a regular coin of which one side is heads, and the other is tails.

You randomly choose a coin and flip it.

What is the probability of the result being "heads?"



This is a completely different question after the information, but it still gets the 2/3 result from the information given. In the original question, the probability of the coin flip resulting in heads isn't the probability that you chose Coin A, just the probability that you could have ended up with heads after the flip.


In the original question, the fact that the coins are different is irrelevant. You choose the coin before the "heads" appearance takes place. Therefore when you choose the coin, the probability is still at 50%.

[ This Message was edited by: Julian Delphiki on 2010-11-14 07:40 ]
_________________


SpaceAdmiral
Grand Admiral

Joined: May 05, 2010
Posts: 1005
Posted: 2010-11-14 11:52   
Quote:

On 2010-11-14 07:38, Julian Delphiki wrote:
I agree with Geja, and the others that said 50%.. I first would have thought 2/3, but after re-reading it... Just because its on a math website, doesn't make it absolutely true..

If you selected the coin after flipping both and getting heads on both, it'd hold at 2/3.



But look at it this way. Taking the extra information out(that in the problem isn't really necessary, as it isn't worded correctly to be relevant to the question itself)

Quote:
Suppose you have two coins. One of these coins, let's call it Coin A, has heads on both sides. The other coin, Coin B, is a regular coin of which one side is heads, and the other is tails.

You randomly grab one of these coins. You flip this coin, and the result is heads.

What is the probability of you having grabbed Coin A?



Becomes

Quote:
Suppose you have two coins. Let's call them Coin A and Coin B.

You randomly grab one of these coins.

What is the probability of you having grabbed Coin A?



Without the information that is worded in a way that makes it irrelevant to the final question, it is 50%.

The probability of grabbing one coin out of two before there are any unique circumstances such as both appearing the same on one side, is 50%




The only ways it would be 2/3 is if it was worded:

Quote:
Suppose you have two coins. One of these coins, let's call it Coin A, has heads on both sides. The other coin, Coin B, is a regular coin of which one side is heads, and the other is tails.

Both coins are on a flat opaque surface. Both coins are showing "heads" on the top. You slide one towards you.

What is the probability of you having chosen Coin A?



In this question, the fact that the coin is "heads-up" is relevant to the question, because you have to choose between two coins that are displaying heads.

Another way would be:

Quote:
Suppose you have two coins. One of these coins, let's call it Coin A, has heads on both sides. The other coin, Coin B, is a regular coin of which one side is heads, and the other is tails.

You randomly choose a coin and flip it.

What is the probability of the result being "heads?"



This is a completely different question after the information, but it still gets the 2/3 result from the information given. In the original question, the probability of the coin flip resulting in heads isn't the probability that you chose Coin A, just the probability that you could have ended up with heads after the flip.


In the original question, the fact that the coins are different is irrelevant. You choose the coin before the "heads" appearance takes place. Therefore when you choose the coin, the probability is still at 50%.

[ This Message was edited by: Julian Delphiki on 2010-11-14 07:40 ]



if both coins you flipped are heads, and you take one, is that not the same as taking one head?
you flip 2 coins, both are heads. you take one, so there are 3 heads and coin A has 2 heads, so the probability is 66.66% Notice that flipping both coins did not change my logic.
the only time where both heads will matter is when one is tails.

PS: Apparently some earlier poster posted a link that can probably explain better than i can, so i wonder why some still say 50% :/
[ This Message was edited by: SpaceAdmiral on 2010-11-14 11:55 ]
_________________


Bardiche
Chief Marshal

Joined: November 16, 2006
Posts: 1247
Posted: 2010-11-14 12:21   
The maths problem does continue to raise controversy, and I agree that a professor in maths claiming one answer is right doesn't necessarily mean the answer is right.

Actually, no, I disagree.

Julian, you misunderstand probabilities. Probabilities assumes a set of possible outcomes and then judges how many of those outcomes conform to a number of conditions.

Our condition is that a set of outcomes must belong to Coin A.

There are two possible outcomes, out of three, that fit that condition.

There are only three possibilities because we have eliminated the only possible fourth.

Therefore, probabilities demand that the chance is 2/3rds.

It's not about the odds of grabbing one of two coins. It's about the probability. And the probability that, after observing at least one side is heads that the other may be heads as well is 2/3rds.



If you viewed two heads on two different coins, the probability would not actually become 2/3rds, but 1/2th. Observe.

There are two coins on the table. Both coins display heads. You know that there are four possibilities on the onset, had you viewed neither coin, assuming H1a, H2a, Hb and Tb.

H1a and Hb
H2a and Hb
H1a and Tb
H2a and Tb

It's impossible for any other result to be seen together.

Therefore, the moment you observe two coins with both showing heads, the only possibilities remaining are:

H1a and Hb
H2a and Hb

As there is no way there can be a "Tails" viewing.

Let's be creative and expand on ORDER as well.

H1a and Hb
H2a and Hb
Hb and H1a
Hb and H2a
H1a and Tb
H2a and Tb
Tb and H1a
Tb and H2a

Only four out of eight possibilities are possible after the observation, therefore reducing the probability of either of two coins being the two-headed coin to... you guessed it, 50%.

Therefore, there is a 50% probability that either of the given coins is the two-headed coin.

As demonstrated.
_________________


Xydes
Grand Admiral

Joined: August 07, 2009
Posts: 276
From: England
Posted: 2010-11-14 16:33   
It is 50/50

Reasons Why:

The Fact that you end up with Heads is irrelevant to the question. You RANDOMLY grabbed a coin. You flipped it.

You ended up with heads. What is the probability you picked Coin A?

50/50. 2 coins, the out-come is unknown until you have flipped ONE of the coins.

Your question in lay-man terms is: I have two coins, One is called A, One is called B, I flip One, Which coin did I flip.

But you have stated it in a way where: You would have to base the probability on the Unknown outcomes.

So in the end, its still 50/50.

And the guy does admit it himself that he is wrong, and that it is 50/50

Should read what before you post.
_________________


Aradrox
Grand Admiral

Joined: March 12, 2007
Posts: 133
From: Tennessee
Posted: 2010-11-14 17:15   
Quote:

On 2010-11-14 02:48, SpaceAdmiral wrote:
Quote:

On 2010-11-14 00:34, Grand Admiral Thrawn. wrote:
50% duh for the question what is on the coins has absulotely nothing to do with it. Same question could of been there are 2 pillows Pillow a and pillow b. One has a star on both sides the other has a star on one side. You randomly grab a pillow what is the chance you grabbed pillow A?
50%



except it says you grabbed a pillow with a star on top.
too many people miss that tiny detail that matters.



its a trick question. The question was. WHAT IS THE CHANCE YOU GRABED COIN A. has nothing to do with the result of the flip.
_________________
[

Aradrox
Grand Admiral

Joined: March 12, 2007
Posts: 133
From: Tennessee
Posted: 2010-11-14 17:19   
Quote:

On 2010-11-14 12:21, Germane Riposte wrote:
The maths problem does continue to raise controversy, and I agree that a professor in maths claiming one answer is right doesn't necessarily mean the answer is right.

Actually, no, I disagree.

Julian, you misunderstand probabilities. Probabilities assumes a set of possible outcomes and then judges how many of those outcomes conform to a number of conditions.

Our condition is that a set of outcomes must belong to Coin A.

There are two possible outcomes, out of three, that fit that condition.

There are only three possibilities because we have eliminated the only possible fourth.

Therefore, probabilities demand that the chance is 2/3rds.

It's not about the odds of grabbing one of two coins. It's about the probability. And the probability that, after observing at least one side is heads that the other may be heads as well is 2/3rds.



If you viewed two heads on two different coins, the probability would not actually become 2/3rds, but 1/2th. Observe.

There are two coins on the table. Both coins display heads. You know that there are four possibilities on the onset, had you viewed neither coin, assuming H1a, H2a, Hb and Tb.

H1a and Hb
H2a and Hb
H1a and Tb
H2a and Tb

It's impossible for any other result to be seen together.

Therefore, the moment you observe two coins with both showing heads, the only possibilities remaining are:

H1a and Hb
H2a and Hb

As there is no way there can be a "Tails" viewing.

Let's be creative and expand on ORDER as well.

H1a and Hb
H2a and Hb
Hb and H1a
Hb and H2a
H1a and Tb
H2a and Tb
Tb and H1a
Tb and H2a

Only four out of eight possibilities are possible after the observation, therefore reducing the probability of either of two coins being the two-headed coin to... you guessed it, 50%.

Therefore, there is a 50% probability that either of the given coins is the two-headed coin.

As demonstrated.


all that really has nothing to do with the question you asked.
You can take out everything but the following and get the same answer to the question. What is the probality you grabbed Coin A.

There are 2 coins. Coin A and Coin B.
You randomly grab one coin.
What is the probality you grabed coin A.
Answer 50%
_________________
[

Aradrox
Grand Admiral

Joined: March 12, 2007
Posts: 133
From: Tennessee
Posted: 2010-11-14 17:21   
the people that are getting other then 50% are trying to solve for what is the chance of getting heads. not the chance of you grabing coin A
_________________
[

Bardiche
Chief Marshal

Joined: November 16, 2006
Posts: 1247
Posted: 2010-11-14 18:24   
Quote:
And the guy does admit it himself that he is wrong, and that it is 50/50

Should read what before you post.




Quote:
Yes, the answer is 2/3.



Quote:
To illustrate why the 1/2 answer is wrong



... are the first few words of that page. It ends with a principal somewhere admitting that after various mathematics experts said it was 2/3rds, that it was 'probably' 2/3rd. He never "admitted" it was wrong, he only carried a few examples on why it is right, involving mathematical formulae, simplified explanations and exaggerated explanations, and illustrated with two emails received from a Biology teacher on how it is a prevalent problem with people who don't get along with Probability maths well.

There is no trick question, and there is no trick. The moment you know the result of "heads" is the moment probabilities are changed. It's not about chance, or odds, it's about probabilities.

It's asking what the probability is you grabbed the coin with two heads. You already know your coin has at least one head. At that point, probability changes due to getting more information.

--

I understand probabilities boggle the mind, but 2/3rds is correct. It's not:

I grabbed Coin A,
OR I grabbed Coin B.

It's I grabbed a Coin, result was Heads, what is the probability this is Coin A? If it was Tails you would all agree that it must've been Coin B he picked up, because three out of four possibilities are eliminated. But when one out of four is eliminated you guys stubbornly insist that two possibilities must have been eliminated, or that there were only three possibilities to begin with.

That is simply not true if you approach it the mathematical way, and all those who argue that it must be 1/2 and that everyone saying 2/3rds is wrong really should re-examine the problem. It's asking about Probabilities.

So let me repeat.

It's [u]NOT a question of "What is the chance you grabbed a coin", it's asking, "Based on the information that one of the sides of Heads, what is the probability that the coin you picked is the two-headed coin?", and that first part is not necessary to be said because the information is given and nothing implied it was a trick question.

Probabilities are [u]not the "chance I randomly picked one of two choices", it's saying how probable it is you picked one coin over the other given observed data collected from experiment or other means of observation.


If the coin came up Tails, none of you would claim it was a 50% chance that it's Coin A. Why do you take that observed data into account, but throw away the othe?
[ This Message was edited by: Germane Riposte on 2010-11-14 18:26 ]
_________________


SpaceAdmiral
Grand Admiral

Joined: May 05, 2010
Posts: 1005
Posted: 2010-11-14 19:30   
Quote:

On 2010-11-14 05:31, Germane Riposte wrote:
For more detailed information on this problem, please consult the following page and be marveled at the mysteries of mathematics:

http://mathproblems.info/prob16s.htm



Will anyone click this link...
AND SEE THE OFFICIAL EXPLANATION/ANSWER?
_________________


$yTHe {C?}
Grand Admiral
Sundered Weimeriners


Joined: September 29, 2002
Posts: 1292
From: Arlington, VA
Posted: 2010-11-14 22:00   
Sure is poor understanding of probability ITT
_________________


Shigernafy
Admiral

Joined: May 29, 2001
Posts: 5726
From: The Land of Taxation without Representation
Posted: 2010-11-15 06:28   
Think of it this way: Two coins, sure. And the question asks "You randomly grab one of these coins. You flip this coin, and the result is heads."

At the point where you randomly grab one of these coins, the chances that you have coin A are indeed 50% - because you're just picking one of two coins and have no further information. However, as soon as you flip it and see a result of heads, you have to change your thinking. After all, a heads is twice as likely from coin A as it would be from coin B - because A is going to get you a heads 100% of the time. Coin B will only do so 50% of the time. So given that you have a heads, and coin A will give you a heads twice as often as coin B, you have basically 3 options: heads (from coin A), heads (from coin B) or tails (from coin B). Hence, 2/3 chance of it being from coin A.

I'm not sure that's actually much clearer, but it is in fact true.
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* [S.W]AdmBito @55321 Sent \"I dunno; the French had a few missteps. But they're on the right track, one headbutt at a time.\"

  Email Shigernafy
jamesbob
Grand Admiral

Joined: August 22, 2009
Posts: 410
Posted: 2010-11-15 07:04   
Quote:

On 2010-11-15 06:28, Shigernafy wrote:
Think of it this way: Two coins, sure. And the question asks "You randomly grab one of these coins. You flip this coin, and the result is heads."

At the point where you randomly grab one of these coins, the chances that you have coin A are indeed 50% - because you're just picking one of two coins and have no further information. However, as soon as you flip it and see a result of heads, you have to change your thinking. After all, a heads is twice as likely from coin A as it would be from coin B - because A is going to get you a heads 100% of the time. Coin B will only do so 50% of the time. So given that you have a heads, and coin A will give you a heads twice as often as coin B, you have basically 3 options: heads (from coin A), heads (from coin B) or tails (from coin B). Hence, 2/3 chance of it being from coin A.

I'm not sure that's actually much clearer, but it is in fact true.





unfortunately this topic is entirely debateable till you are allowed to flip it a certian amount of times (then it becomes clear)


we could sit here for 10 days doing this and all we get is one big post thing that will to be fair will more then likely get locked


as this thing depends on your point of view.


as he gave us bare info he says one at ramdom and flips only says flip once


now depending on how you do your maths you either reach

50 50 chance

or 75 25 percent chance.

this question is very simlar as whether or not 1 is a prime number

because it can be techinely devided by one and its self.


however because it is the number one it is also NOT a prime number

it is entirely dependent on how you do your maths one path will lead to the right answer and to be honest maths this late is never a good idea.

as such let us end this blasted topic



ps: if my post confused you VICTORY
[ This Message was edited by: jamesbob on 2010-11-15 07:18 ]
_________________


Bardiche
Chief Marshal

Joined: November 16, 2006
Posts: 1247
Posted: 2010-11-15 07:35   
Maths isn't open to interpretation, nor is it debatable. The answer is 2/3rd. Anyone saying otherwise doesn't understand Probability. A single observed fact is enough to influence Probabilities.
_________________


Shigernafy
Admiral

Joined: May 29, 2001
Posts: 5726
From: The Land of Taxation without Representation
Posted: 2010-11-15 09:11   
Yeah, the number of flips only matters between 0 and 1. If you don't flip it at all, you just have a coin, and thus you have a 50% chance of it being one of the two coins. As soon as you flip and get heads, and you know that one coin is double headed, the information shifts it to being a 2/3 chance of being coin A.

Flipping a thousand times will prove the point, but once is enough to make it true.
_________________
* [S.W]AdmBito @55321 Sent \"I dunno; the French had a few missteps. But they're on the right track, one headbutt at a time.\"

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Bardiche
Chief Marshal

Joined: November 16, 2006
Posts: 1247
Posted: 2010-11-15 09:56   
It's like...

There are 200 ships in Tau Ceti and 200 ships in Kaus Borealis.

In Tau Ceti, 199 of the ships are Dreadnoughts, and one is a Cruiser. In Kaus Borealis, 199 of the ships are Cruisers, and one is a Dreadnought.

If you enter a wormhole that randomly exits in either of the two systems, you have a 50% probability of being in either system.

The moment you observe a Dreadnought, you think, "Oh, I'm probably in Tau Ceti", because you have a 0.5% chance of seeing a Dreadnought in Kaus Borealis.

99.5% probability of being in Tau Ceti. You CAN still be in Borealis, the Probability's just low.

You can't say, "Well, the probability is still 50/50 because I entered that wormhole and bugger all with observations".

It's more likely that you will see a Dread in Tau, so you're more likely to be in Tau.

Just the same. It's more likely you'll see a Heads on Coin A, so it's more likely you grabbed Coin A than Coin B.
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