Two coin puzzle
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Well now, I'll have to wait a week for your response I see. Anyone else is welcome to sit down for the same terms.Quote: BleedingChipsSlowly[To Sonuvabish:] Are you in?
Quote: SonuvabishI read it, but I don't agree. If you are saying you pull both coins out and look only at one, which is always white, then randomly guess which one of three sides is white, then there is a 2/3 chance you are right. Regardless, your thinking is obviously superior to most of the forum!
You may not agree but the 100 trials I did do agree (at least with the multi trial format), I was an avid 1/2 believer until I did the trials and changed to the 2/3 camp. P.S don't take bleedingchips challenge he will bleed your chips at a rate of $6 for him to every $5 you win!
I do agree that issue has caused significant disagreement. However, I take issue with your suposition that the questions asked was the one you answered in the first spoiler. I belive the question asked was the one you addressed in your second spoiler.Quote: BleedingChipsSlowlyPosters are answering two questions on this thread: the one asked and one that is pretty close but not the one asked.
The post that started it all:
I believe the sentence:Quote: WizardThis is by far the most controversial problem on my mathproblems. I don't post it here because I don't know the answer, but for the edification and education of the rest of the forum.
There are two coins in a bag. One is white on one side and black on the other. The other is white on both sides.
You randomly draw one coin, and observe one side only, which is white.
What is the probability the other side of that coin is white?
Please put solutions in.Have a nice day
Is best interpreted as:Quote: WizardYou randomly draw one coin, and observe one side only, which is white.
"You randomly draw one coin from the bag. You observe nothing about the coin you did not draw. The coin you do draw, you pull out in a closed fist (so you can't observe anything about the coin visually). When you open your hand the only side visible is white." I believe it is meant to tell you that you only observe a side of coin and nothing more.
Understanding it this way there are a variety of explanations in this thread which show 2/3rds to be the correct answer to the question asked. My personal favorite is quoted below:
This is equivalently demonstrated, but perhaps more intuitively, with a four-sided top (e.g. a dreidel) with three white sides and one black side. What are the chances that a spin of such a top, if it lands on a white side, has a white side face down? The same logic as above applies: there are three ways for a white side to land face-up, and in two of those, another white side will be face down.
The equivalence between the two coins and the four-sided top is as follows: label the black side of the coin 1 and the opposite white side 3. Label the white/white coin 2/4. Then label the black side of the top 1 and the other three sides 2/3/4.
This process gves us four possible four letter sequences: WWWB, WWBW, WBWW, and BWWW (full enumeration of the sample space). In the absence of additional information (knowledge the color of a side for any pull) those sequences clearly have equal probability, 1/4th. The probability question "What is the probability the other side of that coin is white?" Is now equivalent to what is the chance W is in the second position (WWWB, WWBW, or BWWW).
When we are given that the first side observed is "W," that eliminates the possibility of BWWW from the sample space. That then leaves the sample space as WBWW, WWBW, and WWWB. Now the answer is clearly the chance of WWWB and WWBW out of WWWB, WWBW, and WWBW, given all 3 events are equally likely. Clearly the answer in this case is 2 out of 3.
This was an interesting observation. While I think any plain text reading of the problem in the OP or the link makes it clear that the observed side should be at random, this was an interesting tangent. Thank you.Quote: kubikulannTo play the lawyer: it does not say how you choose the side you observe. Everybody assumed it was at random. But it could be a "Monty Hall"-like problem, where the operator deliberately chooses the white side to show to you, in the event you drew the bicolor coin.
Then the answer is not the same.
This process gves us four possible four letter sequences: WWWB, WWBW, WBWW, and BWWW (full enumeration of the sample space). In the absence of additional information (knowledge the color of a side for any pull) those sequences clearly have equal probability, 1/4th. The probability question "What is the probability the other side of that coin is white?" Is now equivalent to what is the chance W is in the second position (WWWB, WWBW, or BWWW).
When we are given that the first side observed is "W," that eliminates the possibility of BWWW from the sample space. That then leaves the sample space as WBWW, WWBW, and WWWB. Now the answer is clearly the chance of WWWB and WWBW out of WWWB, WWBW, and WWBW, given all 3 events are equally likely. Clearly the answer in this case is 2 out of 3.
Endermike, I thought the problem states you only draw one coin for possible letter sequences of WW WW( if coin 1) and WB BW (if coin 2) since each coin is equally likely to be drawn (1/2 chance), and white was observed, the combinations are WW if coin 1 or WB if coin 2. The options are not WW, WW (coin 1)and WB (coin 2) or else you are saying coin 1 is twice as likely to be drawn as coin 2.
Note that the four combinations you refer to come into play when you perform multiple trials because you have 3 ways to see white first with two of those ways having W on the flip side. In the OP you only have two ways to see the white first, either on coin 1 or coin 2 with equal chance. Not two ways on coin 1 and one way on coin 2.
You randomly draw a coin, and observe one side. It's black. What is the probability you drew the white/white coin? If you say 1/2, I will reach through the internet and slap you.
Quote: dwheatleyFor anyone that thinks it's 1/2, try it backwards:
You randomly draw a coin, and observe one side. It's black. What is the probability you drew the white/white coin? If you say 1/2, I will reach through the internet and slap you.
Thank you. The answer is obviously zero! Once information is gained (seeing white) the chances do change that it is a particular coin. once you see white you are twice a likely to have drawn the WW coin, even though the orinal selection was 50/50.
Please disregard my above post to endermike thinking a one trial answer was different to a multiple trial approach. Firmly back on the 2/3 bandwagon. I have flipped more times than the coins in this problem!
Ah, you have done well grasshopper! The fact that one person has changed their thinking makes all my postings worthwhile.Quote: 1call2manyYou may not agree but the 100 trials I did do agree (at least with the multi trial format), I was an avid 1/2 believer until I did the trials and changed to the 2/3 camp. P.S don't take bleedingchips challenge he will bleed your chips at a rate of $6 for him to every $5 you win!
What the bloody HELL!! "Best interpreted as?" How about sticking to what was stated. Clearly the OP logical proposition is based on a coin having already been selected. If The Wizard intended otherwise his statement of the problem is on error. I sense I sharp intake of breath among the forum readers: yes, I am calling out The Wizard. To illustrate the difference, consider my bar bet proposition modified so that the bets are placed AFTER a coin has been drawn and a white side is observed. In that case I'm still willing to go head-to-head with any takers - provided I get to wager the $5 the flip side being black against a $3 wager it is white.Quote: endermikeI do agree that issue has caused significant disagreement. However, I take issue with your suposition that the questions asked was the one you answered in the first spoiler. I belive the question asked was the one you addressed in your second spoiler.
The post that started it all:I believe the sentence:Quote: WizardThis is by far the most controversial problem on my mathproblems. I don't post it here because I don't know the answer, but for the edification and education of the rest of the forum.
There are two coins in a bag. One is white on one side and black on the other. The other is white on both sides.
You randomly draw one coin, and observe one side only, which is white.
What is the probability the other side of that coin is white?
Please put solutions in.Have a nice day
Is best interpreted as:
"You randomly draw one coin from the bag. You observe nothing about the coin you did not draw. The coin you do draw, you pull out in a closed fist (so you can't observe anything about the coin visually). When you open your hand the only side visible is white." I believe it is meant to tell you that you only observe a side of coin and nothing more.
There are 4 sides on two coins:
Side 1 on Coin 1
Side 2 on Coin 1
Side 3 on Coin 2
Side 4 on Coin 2
------------------------
Side 1 is B for Black
Side 2 is W for White
Side 3 is W for White
Side 4 is W for White
--------------------------
There are 4 possibilities:
B / W
W / B
W / W
W / W
When you draw, take away the B/W possibility. That means there are 3 left, and one has the B on the other side.
2/3. Done.
Quite simply, you are wrong. The intent of the question is as a conditional probability question.Quote: BleedingChipsSlowlyI have no argument about the logic you posted. The argument I have is your logic doesn't address the OP question.
1) A plain text reading of the OP shows that:
"You randomly draw one coin, and observe one side only, which is white." This clearly is meant to say that all sides are equally probable, but you know that the black one was not seen. If you choose to ignore the color information, the sentence construction is less logical than standard English. "You randomly draw one coin." Would be the reasonable way of writing what your interpretation requires.
2) As does the wizard's thoughts on the most common solution in this thread:
3) As does his answer on the mathproblems site.Quote: WizardI forgot that I posted this before. Evidently this is a bright group, as the majority are right.
Maybe you do not agree. In that case, best of luck with your interpretation, I will agree to disagree.
Quote: BleedingChipsSlowlyWhat the bloody HELL!! "Best interpreted as?" How about sticking to what was stated. Clearly the OP logical proposition is based on a coin having already been selected. If The Wizard intended otherwise his statement of the problem is on error. I sense I sharp intake of breath among the forum readers: yes, I am calling out The Wizard. To illustrate the difference, consider my bar bet proposition modified so that the bets are placed AFTER a coin has been drawn and a white side is observed. In that case I'm still willing to go head-to-head with any takers - provided I get to wager the $5 the flip side being black against a $3 wager it is white.Quote: endermikeI do agree that issue has caused significant disagreement. However, I take issue with your suposition that the questions asked was the one you answered in the first spoiler. I belive the question asked was the one you addressed in your second spoiler.
The post that started it all:I believe the sentence:Quote: WizardThis is by far the most controversial problem on my mathproblems. I don't post it here because I don't know the answer, but for the edification and education of the rest of the forum.
There are two coins in a bag. One is white on one side and black on the other. The other is white on both sides.
You randomly draw one coin, and observe one side only, which is white.
What is the probability the other side of that coin is white?
Please put solutions in.Have a nice day
Is best interpreted as:
"You randomly draw one coin from the bag. You observe nothing about the coin you did not draw. The coin you do draw, you pull out in a closed fist (so you can't observe anything about the coin visually). When you open your hand the only side visible is white." I believe it is meant to tell you that you only observe a side of coin and nothing more.
I am willing to take that bet for anything over 20 trials. Trials to be independently verified.
We agree: The intention is clear but the problem statement is flawed. Perhaps this problem would not have been so controversial if stated properly. This is The Wizard's world so I can understand why you would pitch your tent in his camp. I have no compunction stating "The Wizard has no clothes" for this one.Quote: endermikeQuite simply, you are wrong. The intent of the question is as a conditional probability question.
1) A plain text reading of the OP shows that:
"You randomly draw one coin, and observe one side only, which is white." This clearly is meant to say that all sides are equally probable, but you know that the black one was not seen. If you choose to ignore the color information, the sentence construction is less logical than standard English. "You randomly draw one coin." Would be the reasonable way of writing what your interpretation requires.
2) As does the wizard's thoughts on the most common solution in this thread:
3) As does his answer on the mathproblems site.
Maybe you do not agree. In that case, best of luck with your interpretation, I will agree to disagree.
Quote: BleedingChipsSlowlyPosters are answering two questions on this thread: the one asked and one that is pretty close but not the one asked.
The question asked:
You randomly draw one coin and observe one side only which is white. What is the probabilty that the other side of that coin is white? [My emphasis.]
The question many posters thnk was asked but was not:
If you randomly draw a coin and observe one side only as white, what is the probability the other side will be white?
I'm not following. What do you think is the substantive difference in information between these two questions, such that the answer would be different? What's the difference between "one coin" and "a coin"?
Quote: BleedingChipsSlowlyTo illustrate the difference, consider my bar bet proposition modified so that the bets are placed AFTER a coin has been drawn and a white side is observed. In that case I'm still willing to go head-to-head with any takers - provided I get to wager the $5 the flip side being black against a $3 wager it is white.
I'll take that bet.
And you would win, oh Great and Mighty Wizard! We are not worthy! Mea culpa, I love the taste of crow. There is a 1/2 chance that the flip side of a coin will be black, but the double-white coin is in play twice as often.Quote: WizardQuote: BleedingChipsSlowlyTo illustrate the difference, consider my bar bet proposition modified so that the bets are placed AFTER a coin has been drawn and a white side is observed. In that case I'm still willing to go head-to-head with any takers - provided I get to wager the $5 the flip side being black against a $3 wager it is white.
I'll take that bet.
maybeQuote: sodawaterI haven't read this thread at all, but are there really seven pages of debate over the most well known conditional probability problem of all time?
seems like the basic problem for many is defining the sample space
I see to start
3 sides = white
1 side = black
the condition is
now I see 1 white side
that leaves 2 white sides I have not yet seen
and
1 black side I have not yet seen
this = 3 total sides not seen
1 side = black
so the probability of the other side being black = PB =
1 side / 3 total sides not seen
probability of the other side being white = 1 - PB
Sally
It has to be 1/2. There are two coins. The color configurations are just the identifying information for each coin, and the question is asking for the probability that you chose one coin over the other.
It might as well be two otherwise identical coins of two different years (year A and year B for example). If you select one at random and happen to see its tails side first, there's still a 50% chance that the heads side will show that the coin is year A.
Quote: pmpk
It has to be 1/2. There are two coins. The color configurations are just the identifying information for each coin, and the question is asking for the probability that you chose one coin over the other.
It might as well be two otherwise identical coins of two different years (year A and year B for example). If you select one at random and happen to see its tails side first, there's still a 50% chance that the heads side will show that the coin is year A.
Sorry, this is totally wrong. Looking at one of the coins provides information you can't ignore. The answer is 2/3.
Here's how. There are 4 ways you could pull a coin out and look at one side.
1. Observe top is white, bottom is white
2. Observe top is white, bottom is white
3. Observe top is white, bottom is black.
4. Observe top is black, bottom is white.
You pull out a coin and look at the top and see it is white. This leaves 3 options:
1. Observe top is white, bottom is white
2. Observe top is white, bottom is white
3. Observe top is white, bottom is black.
Two of these three are the all-white coin.
The answer is 2/3
Think of it this way. If you reach in and pull out a coin without looking at it, it's 1/2. But now that you look at it and see that the top is not black, that hurts the chances you have the black and white coin. Sometimes when you have the black and white coin, the top will be black. Since it is not, you have eliminated those times.
If you're still not getting it, think of this:
Say it's the same set up, but you pull out one coin and the top is black. Now the chances of picking the all-white coin have dropped from 50 percent (unseen) all the way down to 0 percent (seen.) Seeing extra information changes probability.
One final example. You buy a Mega Millions ticket and wait till the drawing to look at your ticket. Before you look at your ticket, the chance you have hit the jackpot is 1 in 259 million. Now you pick up the ticket with your thumb covering the Mega ball and you see with excitement that you've matched the first 5 numbers. With your thumb over the Mega Ball, your probability has increased from 1 in 259 million to 1 in 15 for the jackpot. Note that you have either won the jackpot or you haven't, but it's not 50-50.
Quote: sodawaterQuote: pmpk
It has to be 1/2. There are two coins. The color configurations are just the identifying information for each coin, and the question is asking for the probability that you chose one coin over the other.
It might as well be two otherwise identical coins of two different years (year A and year B for example). If you select one at random and happen to see its tails side first, there's still a 50% chance that the heads side will show that the coin is year A.
Sorry, this is totally wrong. Looking at one of the coins provides information you can't ignore. The answer is 2/3.
Thanks for the detailed explanation! I stand corrected.
Quote: sodawaterI haven't read this thread at all, but are there really seven pages of debate over the most well known conditional probability problem of all time?
Yes
Quote: BleedingChipsSlowly... In that case I'm still willing to go head-to-head with any takers - provided I get to wager the $5 the flip side being black against a $3 wager it is white.
Hence your name?
Edit : What about $4000 winner on black , $3000 winner on white (me)? 100 random draws?
Quote: sodawaterQuote: pmpk
It has to be 1/2. There are two coins. The color configurations are just the identifying information for each coin, and the question is asking for the probability that you chose one coin over the other.
It might as well be two otherwise identical coins of two different years (year A and year B for example). If you select one at random and happen to see its tails side first, there's still a 50% chance that the heads side will show that the coin is year A.
Sorry, this is totally wrong. Looking at one of the coins provides information you can't ignore. The answer is 2/3.
Here's how. There are 4 ways you could pull a coin out and look at one side.
1. Observe top is white, bottom is white
2. Observe top is white, bottom is white
3. Observe top is white, bottom is black.
4. Observe top is black, bottom is white.
You pull out a coin and look at the top and see it is white. This leaves 3 options:
1. Observe top is white, bottom is white
2. Observe top is white, bottom is white
3. Observe top is white, bottom is black.
Two of these three are the all-white coin.
The answer is 2/3
Think of it this way. If you reach in and pull out a coin without looking at it, it's 1/2. But now that you look at it and see that the top is not black, that hurts the chances you have the black and white coin. Sometimes when you have the black and white coin, the top will be black. Since it is not, you have eliminated those times.
If you're still not getting it, think of this:
Say it's the same set up, but you pull out one coin and the top is black. Now the chances of picking the all-white coin have dropped from 50 percent (unseen) all the way down to 0 percent (seen.) Seeing extra information changes probability.
One final example. You buy a Mega Millions ticket and wait till the drawing to look at your ticket. Before you look at your ticket, the chance you have hit the jackpot is 1 in 259 million. Now you pick up the ticket with your thumb covering the Mega ball and you see with excitement that you've matched the first 5 numbers. With your thumb over the Mega Ball, your probability has increased from 1 in 259 million to 1 in 15 for the jackpot. Note that you have either won the jackpot or you haven't, but it's not 50-50.
The coin is always white when you pull it out, 100% of the time. It is immaterial which white side comes out on the all-white coin because you are not observing 2 coins. That means there is a 50% the other side is white. Re-read the question. 2/3 is the probability that one of the remaining 3 faces is white; you either have the all-white coin, or you don't.
2/3 is simply the ratio of white faces to total unseen faces. This is the initial thought that everyone should have; everyone knows the chance of randomly selecting a white face, excluding the given, is 2/3--that was learned in elementary. If this is actually the correct answer, it is a trick question with poor wording. I would not call that controversial.
Quote: SonuvabishThe coin is always white when you pull it out, 100% of the time. It is immaterial which white side comes out on the all-white coin because you are not observing 2 coins. That means there is a 50% the other side is white. Re-read the question. 2/3 is the probability that one of the remaining 3 faces is white; you either have the all-white coin, or you don't.
2/3 is simply the ratio of white faces to total unseen faces. This is the initial thought that everyone should have; everyone knows the chance of randomly selecting a white face, excluding the given, is 2/3--that was learned in elementary. If this is actually the correct answer, it is a trick question with poor wording. I would not call that controversial.
False. Don't blame the question.
Quote: WizardYou randomly draw one coin, and observe one side only, which is white.
What is the probability the other side of that coin is white?
I just couldn't let the last word be incorrect.
If black side can be observed, then it's 2/3.
If only white side can be observed, then it's 1/2.
Quote: sonuvabish
The coin is always white when you pull it out, 100% of the time. It is immaterial which white side comes out on the all-white coin because you are not observing 2 coins. That means there is a 50% the other side is white. Re-read the question. 2/3 is the probability that one of the remaining 3 faces is white; you either have the all-white coin, or you don't.
2/3 is simply the ratio of white faces to total unseen faces. This is the initial thought that everyone should have; everyone knows the chance of randomly selecting a white face, excluding the given, is 2/3--that was learned in elementary. If this is actually the correct answer, it is a trick question with poor wording. I would not call that controversial.
False. This is very similar to the Monty Hall problem (check wikipedia). Three closed doors - one has a car, the other two have nothing behind them. You point at one of the three doors. The host walks to the doors and opens one of the other two doors, and reveals that the car isn't behind that one. You can now choose to either keep your door or switch doors. What do you do? Open your door or open the other one? Straight from the movie 21! You can't ignore the information you're given.
Quote: RSThe question is worded poorly. I understand why one would say 2/3, but the question suggests the white side must be observed (and the black side cannot be observed).
Hahaha, what question are you reading?
Quote: Wizard
There are two coins in a bag. One is white on one side and black on the other. The other is white on both sides.
You randomly draw one coin, and observe one side only, which is white.
What is the probability the other side of that coin is white?
That sounds like you randomly draw one coin, and observe one side only, which happens to be white.
