Two-coin problem
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53 members have voted
Quote: MathExtremistI'll gladly offer anyone a wager of $100 to my $130 that *tails* will show when we play the game. If the odds of heads or tails are truly 1/2, that yields a healthy +15% EV. I'll play for as many trials as you want.
To be clear, what event(s) are you offering to bet upon?
- Pulling a coin from the box and seeing heads on the top side.
- Flipping over a coin in your hand that shows heads and seeing tails on the other side
- The parlay of events one and two
Quote: kpTo be clear, what event(s) are you offering to bet upon?
- Pulling a coin from the box and seeing heads on the top side.
- Flipping over a coin in your hand that shows heads and seeing tails on the other side
- The parlay of events one and two
The third. The whole point of the problem statement was to eliminate as a starting condition that you saw tails first. In this wager, if you saw tails first, you would have zero chance of seeing tails on the flip side. The premise is to evaluate the conviction of those who believe that if you see *heads* first, that tails will occur on the flip side exactly 50% of the time. If true, the wager at 130-to-100 has a large player advantage.
1. Setup box from problem.
2. Pick random coin, then random face.
3. If face is tails, GOTO 1
Otherwise,
4. Reveal opposite face of selected coin.
If it's tails, you win ($130)! if it's heads, MathExtremist wins ($100)
Spoiler alert:
He will take your money.
*edited to prevent unnecessary sex changes*
Quote: dwheatleyMathExtremist's game would work like this:
1. Setup box from problem.
2. Pick random coin, then random face.
3. If face is tails, GOTO 1
Otherwise,
4. Reveal opposite face of selected coin.
If it's tails, you win ($130)! if it's heads, MathExtremist wins ($100)
Spoiler alert:
She will take your money.
Spoiler alert:
He. :)
Quote: KelmoIt's 1/2.
If the coin was shown to be tails, we would know the other coin with 100% certainty. The fact that it's heads tells us nothing. Heads does not affect the measure of probability, whereas we gain perfect knowledge with a tails.
The mistake is thinking that "heads gives you no information". In reality it tells you that the side is not "tails". That eliminates 1 out of 4 possibilities. So out of the remaining 3 possibilities 2 of them have a head on the other side of that coin.
Those two possibilities are that you have side A or side B of the two headed coin.
Sheeesh! Put away the numbers already.
Quote: pacomartinThe mistake is thinking that "heads gives you no information". In reality it tells you that the side is not "tails". That eliminates 1 out of 4 possibilities. So out of the remaining 3 possibilities 2 of them have a head on the other side of that coin.
Those two possibilities are that you have side A or side B of the two headed coin.
If you want to look at 3 possibilities, there is a 50% chance that tails is on the other coin (25% that it is on the upside, 25% that it is on the down side). The other 50% is that it is face down on the coin you are looking at.
Quote: KelmoMy god. Why is everyone overthinking this problem. There are only 2 objects not 4. Heads tells you nothing. No information. Nada! Probability is a measurement based on information. With no additional info, it is still a 50/50 chance.
Sheeesh! Put away the numbers already.
Two objects, but we're not dealing with objects, we're dealing with faces. And there are four faces.
With four faces, a probability of an event would be expressed as X/4. However, since heads is the qualifying condition, and there are only 3 heads, the probability MUST be expressed as X/3.
To give the coins and faces identity, you have Coin 1 with 1 face labeled H1 and the other face as H2. Coin 2 has a face labeled as H3 and the other face as T4.
With the condition that heads is drawn and observed, an example beginning with T4 cannot count. Therefore, the only possibilities left are...
Draw Coin 1, see H1, H2 is on back.
Draw Coin 1, see H2, H1 is on back.
Draw Coin 2, see H3, T4 is on back.
In two out of three (2/3) possibilities, you will see heads on back.
Quote: KelmoMy god. Why is everyone overthinking this problem. There are only 2 objects not 4. Heads tells you nothing. No information. Nada! Probability is a measurement based on information. With no additional info, it is still a 50/50 chance.
Sheeesh! Put away the numbers already.
I suggest you make an experiment. Please tell us your results after 100 tries. Then you may "overthink" it again.
Quote: FaceTwo objects, but we're not dealing with objects, we're dealing with faces. And there are four faces.
With four faces, a probability of an event would be expressed as X/4. However, since heads is the qualifying condition, and there are only 3 heads, the probability MUST be expressed as X/3.
To give the coins and faces identity, you have Coin 1 with 1 face labeled H1 and the other face as H2. Coin 2 has a face labeled as H3 and the other face as T4.
With the condition that heads is drawn and observed, an example beginning with T4 cannot count. Therefore, the only possibilities left are...
Draw Coin 1, see H1, H2 is on back.
Draw Coin 1, see H2, H1 is on back.
Draw Coin 2, see H3, T4 is on back.
In two out of three (2/3) possibilities, you will see heads on back.
Its not 2/3 POSSIBILITIES that you would see heads on the other side of the coin
its 1/2 POSSIBILITIES that you will heads on the other side of the coin
The PROBABILITY that you will see heads on the other side is 2/3
This is were i messed up saying the answer was 1/2, because the question clearly states what is the PROBABILITY that you will see heads on the other side of the coin.
I was mixing PROBABILITY theory with POSSIBILITY theory and The Math Extremist pointed it out to me.
The answer to the question as it was worded is for sure 2/3. But after you pull the coin from the box and "observe" heads there are only 2 POSSIBILITIES left to be on the other side 1 heads and 1 tails as at least 1 heads will always still be left in the box.
EDIT: its like when you are watching poker on TV and a guy has 4 outs going into the river. His PROBABILITY of winning the hand is 9%. But you might have observed all 4 of his outs folded by other players, so his POSSIBILITY of winning the hand is 0%. Personally I would think it would be cool if they showed both %'s especially on flush draws.
Quote: vert1276Its not 2/3 POSSIBILITIES that you would see heads on the other side of the coin
its 1/2 POSSIBILITIES that you will heads on the other side of the coin.
'Possibilities' refered to the possible ways the sequence could occur, not possible faces that could be on the other side of the coin.
Let's look at it differently.Quote: KelmoMy god. Why is everyone overthinking this problem. There are only 2 objects not 4. Heads tells you nothing. No information. Nada! Probability is a measurement based on information. With no additional info, it is still a 50/50 chance.
Sheeesh! Put away the numbers already.
Let's say you pulled out a coin and saw tails.
What does that say about the odds of the other side being heads? Still 50/50?
I am shocked at the number of people who answered 1/2 (I mean seriously lol). All of you should make the real experiment MangoJ proposed before commenting any further.
Quote: MangoJI suggest you make an experiment. Please tell us your results after 100 tries. Then you may "overthink" it again.
I sugguest that, if you don't beleive me, ask Kevin Spacey!
Quote: KelmoMy god. Why is everyone overthinking this problem. There are only 2 objects not 4. Heads tells you nothing. No information. Nada! Probability is a measurement based on information. With no additional info, it is still a 50/50 chance.
Sheeesh! Put away the numbers already.
Perhaps it would help if I posed an alternative question. Given that I flip the coin and get a head, what is the probability that the coin has double sided heads? With regard to that question, you still have no information that helps you narrow down the answer. The probability is still 50/50 to the alternative question.
But with regard to the original question you have eliminated 1 of 4 possible faces. There are 3 left (2 are heads, and 1 is tails).
Now advance to the next level. How does having a given number of cut cards change probability in black jack?
Quote: KelmoMy god. Why is everyone overthinking this problem. There are only 2 objects not 4. Heads tells you nothing. No information. Nada! Probability is a measurement based on information. With no additional info, it is still a 50/50 chance.
Sheeesh! Put away the numbers already.
Are you sure?
Put three red cards and one black card into a deck and randomly shuffle them. Draw a red card. What's the chance of the next card being red?
Take the same four cards and tape them back-to-back so you have two double-faced cards. Now randomly flip and shuffle them. Draw a red card. What's the chance of the back-side also being red?
Aren't they the same?
Quote: MathExtremistAre you sure?
Put three red cards and one black card into a deck and randomly shuffle them. Draw a red card. What's the chance of the next card being red?
Take the same four cards and tape them back-to-back so you have two double-faced cards. Now randomly flip and shuffle them. Draw a red card. What's the chance of the back-side also being red?
Aren't they the same?
That is one of the best explanations of the problem I have heard. Even I can understand it.
The original question did not state that a random coin face would be chosen from the three faces showing heads. It simply stated that a coin was chosen and shown to be heads (this is still a choice between two coins) and whichever coin was chosen, there is a heads on it that can be shown.
Again, there are still only two coins that can be chosen and there is still a 50/50 chance that the coin chosen will have the tails on one of its faces.
You pull a coin and could be looking at any of four sides. You see a face, so you're reduced to three sides. I.E.:
There is NO CHANCE that you're looking at the tails side of the regular coin.
There is one chance that you're looking at the face side of the regular coin.
There is one chance that you're looking at the front face side of the double-headed coin.
There is one chance that you're looking at the back face side of the double-headed coin.
Add them up and there's two chances out of three that you're holding the double-headed coin.
Quote: DJTeddyBearOK. Let me try a different approach.
You pull a coin and could be looking at any of four sides. You see a face, so you're reduced to three sides. I.E.:
There is NO CHANCE that you're looking at the tails side of the regular coin.
There is one chance that you're looking at the face side of the regular coin.
There is one chance that you're looking at the front face side of the double-headed coin.
There is one chance that you're looking at the back face side of the double-headed coin.
Add them up and there's two chances out of three that you're holding the double-headed coin.
I'm still missing the part of the question that states one of the 3 faces (heads) of the coin are selected at random. I'm only seeing that one of the two coins are selected at random. How do you turn that into a selection of 3 objects?
Quote: KelmoIt simply stated that a coin was chosen and shown to be heads (this is still a choice between two coins) and whichever coin was chosen, there is a heads on it that can be shown.
Ok, let's do it more simple. You have 2 dices, each with 6 sides. One dice is a normal dice with numbers 1-6 on them. The other dice is a special dice with 6 on both sides. The room is completely dark, but there is one person with night-vision, so he can see the dices.
You grab one of the dices and roll it. The night-vision guy says you rolled a 6.
What is the probability that you had chosen the normal dice. What is the probability you had chosen the special dice ?
What is on the other side of the dice ?
Still not convinced ?
Lets think you have a dice with 1 million sides. One dice has numbers 1 to 1 million on it. The other dice has "1 million" on each of it's million sides.
You grab a dice, and roll it. The night-vision guy says you rolled a million.
What is on the other side of the dice ?
Quote: KelmoAgain, there are still only two coins that can be chosen and there is still a 50/50 chance that the coin chosen will have the tails on one of its faces.
That's true, but that's not the question that was asked.
One of two coins is fair, the other has two heads. What is the probability of picking the fair coin?
That's 1/2.
One of two coins is fair, the other has two heads. If you randomly pick a coin and observe heads on the side you look at, what is the probability that it is fair?
That is *not* 1/2.
See the difference? You're answering the first question, but the second question has been asked (actually, the opposite - the probability that it is unfair).
True, you are selecting only one coin. But, by not looking at the other side, you are also selecting one side of the coin. You see that it is heads, but that in itself doesn't tell you which coin you have. It just changes the odds of which coin it was. If you see tails, then would you know which coin it was.Quote: KelmoI'm still missing the part of the question that states one of the 3 faces (heads) of the coin are selected at random. I'm only seeing that one of the two coins are selected at random. How do you turn that into a selection of 3 objects?
So, out of the four possible sides you could have picked, you're looking at one of the three heads. Two of those three heads are on the double-headed coin.
So it's a 2/3 chance you got the double header.
Quote: MathExtremistThat's true, but that's not the question that was asked.
One of two coins is fair, the other has two heads. If you randomly pick a coin and observe heads on the side you look at, what is the probability that it is fair?
That is *not* 1/2.
The question does not state that after randomly selecting the coin, you randomly select the side of the face to look at. It simply states "the face of one side is observed" (this not stated as a random act of choosing between two faces, which means that the face chosen to be observed can always be the side of heads, regardless of the coin chosen). "If heads is observed" may appear to imply a random choice from the subset of heads, but it does not clarify that it is a random choice between three faces. If the side of heads is always chosen to be shown (which the first statement allows for), heads will always be observed.
This is the Monty Hall is disguise.
Quote: KelmoThe question does not state that after randomly selecting the coin, you randomly select the side of the face to look at. It simply states "the face of one side is observed" (this not stated as a random act of choosing between two faces, which means that the face chosen to be observed can always be the side of heads, regardless of the coin chosen). "If heads is observed" may appear to imply a random choice from the subset of heads, but it does not clarify that it is a random choice between three faces. If the side of heads is always chosen to be shown (which the first statement allows for), heads will always be observed.
You have to assume randomness or the concept of probability doesn't make sense as we've been discussing it. For example, if I showed you one of the coins and it had heads on it, but my selection criteria is "always show you the fair coin", then the "probability" of the other side being heads is zero. It's as if the other coin didn't exist in the first place.
Given that the randomness -- or an equal selection of all possible outcomes -- must be part of the problem for it to make sense, all you need is to determine what those possible outcomes are. The mistake is in thinking that there is only one possibility to see heads on the double-headed coin. There are two. Add in the third possibility that the heads is shown from the fair coin, and there are three ways the flip side of a coin can be heads. Only one of those is on the fair coin, so the other two must be on the double-headed one. Thus, under the assumption of equal likelihood, 2/3 of the time you observe heads, heads is also on the reverse.
Quote: KelmoI wish there was a Wizard who could talk some sense into some of these 1/3'ers..
Quote: Wizard of Odds (from his site)
Here is my favorite e-mail I received on this problem. I x'd out some private information.
Dear Michael,
I am a biology teacher at Xxxxxxx High School in Xxxxxxx, MA. I recently was searching for an extra credit question to put on a test for tenth graders. I used the three coins in a bag problem (#16). Unfortunately, none of my students got the correct answer. They all answered 1/2. I have done my best to explain the problem and some of the students now understand why the correct answer is 2/3. The debate has spread throughout the whole school. Most teachers and students beleive the answer to be 1/2. So I started an organization called Team 2/3 - The Probability Masters. Students take an oath to become official members of the team. The mission of the team is to recruit as many new members as possible. Team members get a membership card that has your original question on the back which helps them recruit new team members. Many people think I have gone off the deep end, but I find it amazing that so many people including most teachers and the principal still do not see the truth. If fact, the day after I started the Team 2/3, the principal got on the morning announcements and stated the the answer is 1/2 and the group that thinks otherwise should rethink the problem.
One student created a computer program that simulates 10,000 coin picks in 1 second. The data clearly show the 2/3 answer. But that same student insists that the probability is 2/3 only if many trials are conducted. He doesn't understand that the probability is the same in the first and last trial. People are stuck on the idea that you must have either the H-H or H-T coin and say it is 1 out of two. I have tried to simplify your solution without the T-T coin to help people understand.
If two coins: H-H, H-T
Conditional Prob = 1/2 divided by 3/4 = 2/3
Team 2/3 will recruit for two more days and then reveal the truth. We will try to get the principal to announce that Team 2/3 was correct all along. All team members will hold their heads high if that happens. I only have one math teacher on my side at this time. One of my student team members had a hat made for me that says "Team 2/3 Captain". I wore it on Friday and will wear it this week. We have almost everyone in the school discussing your coin problem. Many of the students are giving it a great deal of thought.
We are very confident that the answer is 2/3, however most people still disagree.
It would be most helpful if you would place a call to the school and talk to Mr. Xxxxxxx Xxxxxxxxx, the principal. This would be greatly appreciated by the team, your team, Team 2/3.
Xxxxxxx High School 1-(xxx)-xxx-xxxx
Sincerely,
Troy X. Xxxxxxx
Biology Teacher P.S. Please feel free to call me at home: 1-(xxx)-xxx-xxxx
I did call the principal and tried to explain why the answer was 2/3. He seemed pretty entertained that somebody would call all the way from Baltimore about this. After going over the problem he seemed to partially understand but added that "he didn't want to be confused with facts." Later the Biology teacher wrote me back, here is his second e-mail:
I really appreciate you talking to the principal. He did talk to me afterward and he seemed very unsure of the correct answer. Team 2/3 has 42 oath-taking official members including an English teacher and the guidance secretary. We allowed people to join unofficially by signing the unofficial membership list. That way they still get a membership card without taking the oath; it's our "chicken list". The Latin teacher, the US history techer, and the alternative education teacher are on that list. Our membership card states-"TEAM 2/3 the Probability Masters" and there is a picture of a man holding two coins. On the back of the card is your original problem. Another team formed (not as organized as our team) called Team 1/2. They also have membership cards. Just this past Thursday, a good week after your call, I reminded the principal that we needed closure to the debate. I just assumed he believe the answer was 2/3. And I told him it would be nice to recognize the courage of team 2/3 and congratulate them. He got on the announcements and stated "After consulting with several mathematical experts" (and other people heard about the controversy and contacted him with explanations for 2/3) "the answer to the coin probability problem is 'probably' 2/3, so congratulations to Team 2/3, you are probably correct. While we would have liked a stronger statement(without the probablies), we have accepted the principal's announcement as victory and feel that we have been successful in our mission.
Thanks for your help,
Troy
Before you go down that road, read this thread, as well as the link to the interview with Monty Hall himself:Quote: KelmoThis is the Monty Hall is disguise.
https://wizardofvegas.com/forum/off-topic/general/4396-the-monty-hall-paradox-interview-with-monty-hall/
Paco -
Thanks for finding that.
Wiz -
GREAT story / letter.
The who are stuck on 1/2 are missing a very important element in the problem. You keep saying that once you've pulled one of the coins, if you're looking at a head then what is on the other side MUST be either a head or a tail and therefore it is a 50/50. It seems obvious. BUT, look at the question again.
"A box contains two coins. One coin is heads on both sides and the other is heads on one side and tails on the other. One coin is selected from the box at random and the face of one side is observed. If the face is heads what is the probability that the other side is heads?"
The point is that you do not begin the puzzle with a coin in your hand. You begin with 2 coins in a box and you select one at random not knowing which one it is.
The question could be asked, "There are 2 coins in a box. One coin is heads on both sides and the other is heads on one side and tails on the other. You select one at random and look at the face. What are the odds that you are looking at a head?" The answer to this question is obviously 3/4. Now, let's call the sides of the 2 coins:
>HEAD "A"
>HEAD "B"
(We'll call the 2 headed coin Coin #1 and this coin has HEAD "A" and HEAD "B".)
>HEAD "C"
>TAIL
(The second coin, COIN #2)
The original question says One coin is selected from the box at random and the face of one side is observed. At that point there are 4 different possibilities when you select the coin, not 2.
Although there are only 2 coins to choose from the parameters of the problem are clear - the coin you've selected MUST show a HEAD. If you select a coin and you see a TAIL, you cannot continue. Because of this critical detail in the wording of the problem, here are 4 possible outcomes when selecting the coin:
COIN #1 - HEAD "A"
COIN #1 - HEAD "B"
COIN #2 - HEAD "C"
COIN #2 - TAIL
You'll see that 2 times out of 4 selections you've chosen COIN #1. The other 2 times you've chosen COIN #2. So in effect, you're not selecting 1 of 2 coins, you're selecting 1 of 4 faces. According to the parameters of the problem, the 1 time out of 4 when you select the TAIL, you must start over - think of it as a no-roll in craps; it never happened.
This leaves you with 3 possible outcomes from your selection (as the TAIL outcome cannot be counted).
COIN #1 - HEAD "A" (With HEAD "B" on the other side)
COIN #1 - HEAD "B" (With HEAD "A" on the other side)
COIN #2 - HEAD "C" (With the TAIL on the other side)
Now it's easy to see that as 2 out of the 3 acceptable selections will be COIN #1, 2 out of the 3 acceptable selections will result in seeing a HEAD when the coin is flipped over. Therefore if one coin is selected from the box at random and the face of one side is observed and that face is heads, the probability that the other side is heads = 2/3.
If the question had been worded, "You are given one of two coins. One coin is heads on both sides and the other is heads on one side and tails on the other. You look at the coin in your hand and see a head. What is the probability that the other side is heads?" The correct answer would be 1/2... but that wasn't the question.
Which possibility is that? That of drawing side 2 of the two-headed coin?Quote: Asswhoopermcdaddy.... The act of drawing heads removes another possibility.
Quote: TheNightflyIf the question had been worded, "You are given one of two coins. One coin is heads on both sides and the other is heads on one side and tails on the other. You look at the coin in your hand and see a head. What is the probability that the other side is heads?" The correct answer would be 1/2... but that wasn't the question.
It wasn't the question, but 1/2 wouldn't be the answer if it had been. The answer in that case is "I don't have sufficient information to properly answer the question."
Without knowledge of the selection criteria ("You are given one of two coins"), it is *impossible* to know the likelihood of the other side being heads. For example, if I was the one giving the coin to you, and I always gave you the fair coin, the probability of the other side being heads would be zero and not 1/2. But you wouldn't know that, because you wouldn't know that I'm never going to give you the unfair coin. If you assume that I select the from the coins randomly (with 50/50 probability), then the question reverts to the prior statement.
Bonus question:
Instead of selecting coin HH or HT equally (50% each), suppose I show you coin HH with probability P and HT with probability 1-P (and you know what P is). What does P need to be in order to make the probability of heads being on the other side equal to 1/2?
Quote: MathExtremist
Bonus question:
Instead of selecting coin HH or HT equally (50% each), suppose I show you coin HH with probability P and HT with probability 1-P (and you know what P is). What does P need to be in order to make the probability of heads being on the other side equal to 1/2?
p=1/3.
Quote: FaceWonderful. An innocent posing of this question to fellow co-workers devolved into my boss repeatedly screaming at me 'IT'S 1/2! THERES ONLY TWO SIDES!' etc and so forth.
I learned not to do that after I "solved"the two guards "problem" at a seminar a few years ago. I spent the next two weeks explaining why the truthful guard had to tell a lie...
I am going to now say, using the EXACT wording of the question, that the answer cannot be ascertained. Although the question states that the coin is chosen at random, it DOES NOT say that which side is looked at is random. It just says 'the face of one side is observed'. There is nothing in the original question which states that a 'randomly chosen face of one side is observed'. It would not be inconsistent with the exact wording of the original question if someone (Monty Hall?) preferentially selected one side. If the question is corrected to state "One coin is selected from the box at random and a randomly chosen face is observed" then the explanations of 2/3 all make sense.
Quote: FaceIn a way I wished you would have told me this sooner,
Wouldn't have helped. I did it long after I cut a Moebius strip in half :)
Quote:but as it stands I'm told I only have to worry for another 20 hours or so.
Dind't you get the memo? It's God's day off tomorrow. No way He'll end the world then :P
Quote: MathExtremistRandom face selection isn't a requirement. Any tails on the fair coin will be discarded, and the two heads on the other coin are indistinguishable. All that's important is the condition that the face will be heads.
Incorrect, MathExtremist. If when the coin with a head and a tail is picked, but I chose to look at the tail EVERY time, then if a head is looked at the odds of the other side being a head is 100%. You MUST state that a random side is looked at, or else the question cannot be answered.
It clearly states that a coin is selected and one face is observed.
The original post NEVER states that a head is observed. It asks what the chances are that the back is a head, if the observation shows a head.
By that phrasing, if a tails is observed, the question becomes moot. Therefore, the answer is 2/3.
Quote: DJTeddyBearSorry Soopoo. Read the original post again.
It clearly states that a coin is selected and one face is observed.
The original post NEVER states that a head is observed. It asks what the chances are that the back is a head, if the observation shows a head.
By that phrasing, if a tails is observed, the question becomes moot. Therefore, the answer is 2/3.
Incorrect. It says a face is observed. It does NOT say that a face is randomly observed. If EVERY time I selected the coin with a tail and a head, I CHOOSE to observe the tail, then every time a head is observed the other side will be a head. It is really that simple. Unless you state that the side which is observed is a random selection, the person CAN choose which side to observe. So in the original question, I can ALWAYS opt to look at the tail if it is the two sided coin. Thus EVERY time a see a head the other side must be a head. I can't see how you can argue this.
Read it again. And this time READ it.
Only ONE side is observed. It NEVER states what the result of that observation is.
It asks, IF the observation is heads, what are the odds that the other side is also heads?
Quote: DJTeddyBearSoopoo -
Read it again. And this time READ it.
Only ONE side is observed. It NEVER states what the result of that observation is.
It asks, IF the observation is heads, what are the odds that the other side is also heads?
I read it. Since I am CHOOSING to always look at the tail side when I pick the coin with a head and a tail, whenever the side i look at is a head, the other side will always be a head. UNLESS YOU FORCE ME TO RANDOMLY CHOOSE A SIDE, then EVERY TIME A HEAD IS SEEN THE OTHER SIDE WILL BE A HEAD. Not complicated. If you argue with me one more time I will choose someone else to DJ my next wedding!
Edit- I will wait for the Wiz to weigh in, hopefully. And if the world ends then this is moot, anyway.
Quote: SOOPOOI am going to now say, using the EXACT wording of the question, that the answer cannot be ascertained. Although the question states that the coin is chosen at random, it DOES NOT say that which side is looked at is random. It just says 'the face of one side is observed'. There is nothing in the original question which states that a 'randomly chosen face of one side is observed'.
I would say that it is easily implied that a random side of the random coin is observed. You seem to be saying that a host is deliberately manipulating the game to always show a heads. If he were, then he would be observing both sides if the first side was tails. Since no such information is provided about such a mischievous host so it can be assumed there isn't one. There is no host at all, by the way, YOU choose the coin.
If we allow unstated information into the problem then I could say that the probability is 0%, because a magician chooses the coin and changes the it into a dove.
Quote: WizardI would say that it is easily implied that a random side of the random coin is observed. You seem to be saying that a host is deliberately manipulating the game to always show a heads. If he were, then he would be observing both sides if the first side was tails. Since no such information is provided about such a mischievous host so it can be assumed there isn't one. There is no host at all, by the way, YOU choose the coin.
If we allow unstated information into the problem then I could say that the probability is 0%, because a magician chooses the coin and changes the it into a dove.
I will accept your interpretation about the side chosen to be viewed as 'implied to be random'. I would have preferred it to be stated affirmatively. The simple coin toss question has been used to show why the exact conditions of a problem should be stated. If i ask you, 'if I flip a coin twice and it is heads both times, what is the chance of it being heads next time?' Most reading the question naturally think of a fair 2 sided coin and answer 50%. If I ask 'if I flip a coin 1 million times and it is heads each time, what is the chance of it being heads the next time?" I am sure before answering that question you would be more inquisitive as to whether the coin had two heads, rather than just assume it is a fair coin.
Quote: SOOPOOIncorrect, MathExtremist. If when the coin with a head and a tail is picked, but I chose to look at the tail EVERY time, then if a head is looked at the odds of the other side being a head is 100%. You MUST state that a random side is looked at, or else the question cannot be answered.
I see the distinction you're making. I would argue that in the case of an entirely non-random process where you choose what the answer is before asking the question, it's not just that the question cannot be answered, but it doesn't make sense. If you're predetermining the answer, the "probability" is either 0 or 1, I just don't know which. That's a different class of question altogether.
You're absolutely right.Quote: SOOPOOI read it. Since I am CHOOSING to always look at the tail side when I pick the coin with a head and a tail, whenever the side i look at is a head, the other side will always be a head. UNLESS YOU FORCE ME TO RANDOMLY CHOOSE A SIDE, then EVERY TIME A HEAD IS SEEN THE OTHER SIDE WILL BE A HEAD. Not complicated. If you argue with me one more time I will choose someone else to DJ my next wedding!
Edit- I will wait for the Wiz to weigh in, hopefully. And if the world ends then this is moot, anyway.
If you CHOOSE to look for the tails side, and fail to find one, then you've got the double-headed coin.
But who gave you permission to look at both sides?
