interesting coin flip problem

Here is a problem from a decision theory class that I am taking:
If you are told that two coins have been flipped, and that one of them landed heads, what is the probability that the other landed tails?

It seems to me that it would be 1/2, because the two events are independent. However, my teacher (a very intelligent man) INSISTS that the probability is 2/3. Here is his explanation: There are four possibile outcomes for flipping two coins:
HH
HT
TH
TT
Each are equally likely to occur before the trials. He says that if we know that one coin has landed heads, we have eliminated TT from the possibility set. This leaves HH,TH, and HT. In two of the three pairs, the other coin would have to be tails, meaning (according to him) P (tails) = 2/3.
I disagree. Once we know that one coin has landed heads, then obviously TT is eliminated, leaving HH,HT, and TH. However, just because these three pairs were equally likely to occur (probability of each = 1/4) BEFORE the trials, does not mean they are equally likely GIVEN THAT WE KNOW ONE COIN HAS LANDED HEADS. That is to say, in the set HH, HT, TH, there are 4 Hs and 2 Ts. 2 of the Hs are found in HH, one is found in HT, and one is found in TH. Therefore, GIVEN that one coin landed heads, the (posterior) probability that it came from HH is 1/2 (because HH contains 1/2 of the Hs). Similarly, the probability that it came from HT is 1/4 and the probability that it came from TH is also 1/4. Therefore there is a 1/2 chance that the other coin landed Heads and a 1/4 + 1/4 = 1/2 chance that it landed tails.
Who is correct?

HE GOT GAME
ZZzzZZzz

[PaultheS]

Think of it this way:

H1 = first coin was heads
T1 = first coin was tails
H2 = second coin was heads
T2 = second coin was tails

Your question is: What is the probability of (T1 AND H2) OR (H1 AND T2) given (H1 OR H2)?

P(T1H2 OR H1T2 | H1 OR H2) = P((T1H2 OR H1T2) AND (H1 OR H2)) / P(H1 or H2)
= (0.5^2 + 0.5^2) / (0.5 + 0.5 - 0.5^2)
= 0.5 / 0.75
= 2/3

So, as you see, your teacher is correct. The key is that you aren't told which coin is heads, just that at least one of them is.

Of course, if the question were different and you were told that the first coin was heads then the probability that the second is tails would be 1/2, and vice-versa.

[PrayingMantis]

The question is somewhat vague and could be interpreted in several ways, but as a default (it that's even possible) I'd say that you are correct and your teacher is wrong.

from the 4 equally likely possibilities of HH,HT,TH,TT, we are sure that the outcome was HH,HT or TH. While those 3 options are equally likely, it's twice as probable that "we are told" about the H result in the first outcome (HH) than in one of the other two (HT or TH), since there are twice as many heads in that specific results. Thus the probability for tails for the second flip is 1/2.

But there are other ways to interpret the information.

[AaronBrown]

I agree with PrayingMantis, the question is not fully specified. We could interpret it as either:

(1) Two coins are flipped. If they are both tails, we reflip both. We continue until we get an outcome that contains at least one head. What is the probability that we end up with two heads? The answer is 1/3, and your teacher is correct.

(2) A dime and a quarter are flipped. I check the dime and tell you whether it's heads or tails. What is the probability that the quarter is the same? Here the answer is 1/2, you are correct.

A practical application of this effect, which proves that it's real and not just a verbal game, is the law of restricted choice in Bridge. Basically, if there are two honors missing in a suit in certain situations, the odds are nearly 50/50 that they are both in one hand or split among both opponents' hands. However, once one opponent plays one of them, whichever one it is, the odds go up to 2/3 that the other honor is in the other hand. It's possible for the opponents to change this, for example, by always playing the higher of two honors. In that case, half the time you know for certain where the other honor is, the other half of the time you get no information from the play. But whatever they do, as long as you know the rule, they have to give you information.

[Tom1975]

You're talking to a man you've never met before. He mentions that he has only one natural sibling. What the chances he has a brother? What the chances he has a sister?

[PaultheS]

[ QUOTE ]
(1) Two coins are flipped. If they are both tails, we reflip both. We continue until we get an outcome that contains at least one head. What is the probability that we end up with two heads? The answer is 1/3, and your teacher is correct.

(2) A dime and a quarter are flipped. I check the dime and tell you whether it's heads or tails. What is the probability that the quarter is the same? Here the answer is 1/2, you are correct.

[/ QUOTE ]

This is true. Re-reading the question, it does sound like it could really mean either of these scenarios. However, I assume that when the teacher gave the question (if he really is a very intelligent guy) he probably outlined it more like the first scenario than the original poster did.

[PaultheS]

[ QUOTE ]
You're talking to a man you've never met before. He mentions that he has only one natural sibling. What the chances he has a brother? What the chances he has a sister?

[/ QUOTE ]

Well, if we're assuming that a parent is 50/50 to have a son or daughter, then the probability he has a sister is 1/2.

This is the same as scenario 2 in AaronBrown's post where you are told one specific result and have to find the probability of the other.

On the other hand, if a parent told you "I have two kids and at least one of them is a boy" then the probability that the other is a daughter is 2/3, like in the coin flipping examples given earlier in this thread.

[Abbaddabba]

When you're given information that reveals that "one coin was heads", you aren't supposed to assume that this information was arrived at via random sampling. It's just a given piece of information.

If you DO assume it was arrived at via random sampling, you can tell that it is twice as likely that the H that we see is a member of the set that contains two H's than it is for each individual set that contains one H and one T. Consequently, it would be 1/2. But again, you aren't assuming that the initial piece of information is arrived at via sampling; you just take it at face value.

All coin flips are completely independent. Regardless of how many you intend to flip and and how many you have flipped already and there results, each coin flip will have a 1/2 chance of being heads or tails. This deals with the law of large numbers. We can't apply the law of large numbers to a scenario in which there are only 2 trials. The greater the amount of trials preformed, the lower and lower the standard deviation becomes. When tossing coins, no matter what, chances are 50-50.
Knowing that you're tossing two coins does not change that.

[KenProspero]

Your teacher is right, and exactly for the reasons he stated.

There are two coins -- let's call them coin 1 and coin 2. You've been told that one of them comes up heads, but don't know which one.

When we started, there were four possibilities. The information given eliminates one and only one of the possibilities, so there's a 2/3 chance that the coin is heads.

This problem is kind of like the Monte Hall -- Let's Make a Deal problem.

In that show, contestents were given the opportunity to go for the "Big Deal" which was behind one of three Doors.

A contestant chooses Door Number 1. Monte (who knows where the loot really is) shows the contestant that the Big Deal was not behind Door Number 2, and then offers the contestent the chance to change his choice of doors (i.e., swap door 1 for door 3). What should the contestant do.

The answer -- Assuming that Monty would have shown a door and offered the contestant the swap in all circumstances, is that the contestant should swap. There's a 1/3 chance the Big Deal is behind door number 1, and a 2/3 chance it's behind door number 3.

[PaultheS]

[ QUOTE ]
All coin flips are completely independent. Regardless of how many you intend to flip and and how many you have flipped already and there results, each coin flip will have a 1/2 chance of being heads or tails. This deals with the law of large numbers. We can't apply the law of large numbers to a scenario in which there are only 2 trials. The greater the amount of trials preformed, the lower and lower the standard deviation becomes. When tossing coins, no matter what, chances are 50-50.
Knowing that you're tossing two coins does not change that.

[/ QUOTE ]

Yes, all coin flips are completely independent, but flipping two coins in a row is not the situation that is being dealt with here.

The question is: I flip two coins. Given that at least one landed heads, what is the probability that one coin was tails?

This is a very different question and has nothing to do with the independence of repeated coin flipping trials.

[AaronBrown]

[ QUOTE ]
When you're given information that reveals that "one coin was heads", you aren't supposed to assume that this information was arrived at via random sampling. It's just a given piece of information.

[/ QUOTE ]
The trouble with this view is the interpretation of the information is affected by how it is given. To interpret a signal you need to know not just the signal, but what the signal might have been in different circumstances. Remember Sherlock Holmes and the dog that didn't bark.

When a person tells you "one coin was heads," we need to know what she would have said if both were heads and if both were tails. So the information is subject to interpretation.

[PaultheS]

[ QUOTE ]
[ QUOTE ]
When you're given information that reveals that "one coin was heads", you aren't supposed to assume that this information was arrived at via random sampling. It's just a given piece of information.

[/ QUOTE ]
The trouble with this view is the interpretation of the information is affected by how it is given. To interpret a signal you need to know not just the signal, but what the signal might have been in different circumstances. Remember Sherlock Holmes and the dog that didn't bark.

When a person tells you "one coin was heads," we need to know what she would have said if both were heads and if both were tails. So the information is subject to interpretation.

[/ QUOTE ]

Good post. This is probably the best answer to the original question.

[PrayingMantis]

In your post you say this:

[ QUOTE ]
Your teacher is right, and exactly for the reasons he stated.

[/ QUOTE ]

And then, after claiming it is like the Monte-Hall problem, you say this:

[ QUOTE ]
The answer -- Assuming that Monty would have shown a door and offered the contestant the swap in all circumstances, is that the contestant should swap. There's a 1/3 chance the Big Deal is behind door number 1, and a 2/3 chance it's behind door number 3.

[/ QUOTE ]

However, the thing with the original and unspecified Monte-Hall problem is that you don't know if the assumption above about Monte and his "duties" is correct at all. Without more specified information (like you find in more detailed verstions of the problem), you don't know whether Monte indeed shows you a door and lets you swap in all circumstances.

That's the reason why you can't simply say that his teacher is right. His teacher might be right or wrong, according to different interpretations of the information he gives. And without any particular assumptions, the answer looks more like 1/2 than 2/3, as you need fewer assumptions in order to get to the 1/2 answer.

[kelvin474]

Three things are equally likely to be the state of the two coins, HH, TH, HT. We agree on that.

100% of the time any of these 3 events occurs, it is correct that "one of the coins landed heads".

So, 1/3 of the time, he has flipped HH. 1/3 of the time he flipped HT, and 1/3 of the time he flipped TH.

That means that 2/3 of the time "at least one head occurred", there is also an occurrence of tails.

The answer is 2/3.

If you don't believe this, send me a PM with where we can bet on this. Preferably for $1000/trial.

[BoogerFace]

[ QUOTE ]
This problem is kind of like the Monte Hall -- Let's Make a Deal problem.

In that show, contestents were given the opportunity to go for the "Big Deal" which was behind one of three Doors.

A contestant chooses Door Number 1. Monte (who knows where the loot really is) shows the contestant that the Big Deal was not behind Door Number 2, and then offers the contestent the chance to change his choice of doors (i.e., swap door 1 for door 3). What should the contestant do.

The answer -- Assuming that Monty would have shown a door and offered the contestant the swap in all circumstances, is that the contestant should swap. There's a 1/3 chance the Big Deal is behind door number 1, and a 2/3 chance it's behind door number 3.

[/ QUOTE ]

And that's why the Monte Hall problem is stupid. The people who switch are only right 1/2 of the time.

[PrayingMantis]

[ QUOTE ]
The answer is 2/3.

If you don't believe this, send me a PM with where we can bet on this. Preferably for $1000/trial.

[/ QUOTE ]

I'm quite amazed by the inability of people who are posting on a gambling forum to see that there could be few interepretations to the information given in the original post. Betting on "2/3" like you seem be willing to do is really a poor decision IMO for the simple reason that the conditions of the bet are not clear, and you are at the mercy of the person who decides how to manipulate the information he gives you. Why is that so difficult to understand?

/images/graemlins/confused.gif

[KenProspero]

I think you miss the point, both the coin flip problem and the monte hall problem are probablility problems, not psychology problems.

The problem is designed to determine if you understand how probability works, and the assumptions which lead to the 2/3 answer in the coin flip, and "who knows what monte is thinking" in Monte Hall are not part of the probablistic analysis.

Further, in the OP, it was stated that the answer to the coin flip problem was 50-50 because they are independent events. At the time he posted, OP apparantly didn't understand why 2/3 was a valid answer.

Now, after reading this thread, if OP wants to go back and argue semantics with his/her prof, go right ahead. However, my academic experience is that with many profs, this would be a -ev exercise. Better in this case to understand the principle that the problem is trying to get at than to argue the nits.

Now, would I make a bet on this if someone offered me. Nope, if I did, I'm sure I'd end up with an earful of cider (Reference to Damon Runyon for anyone who didn't pick it up).

Some very excellent posts. Consider the 2 sibling boy/girl problem. Say there are 24 mothers who have two children. 6 of them have two boys, 12 have one boy and one girl, and 6 have two girls. They all come up to you and tell you the gender of one of their kids. Obivously, the 6 mothers who have two boys tell you "One of my children is a boy." The six mothers who have two girls say "One of my children is a girl." If you want it to be true that P(G|there is one boy) = 2/3, then you must assume that the twelve mothers who have both a boy and a girl tell you about their boy. This means that 18 mothers tell you "One of my children is a boy." In 12 of those cases, the other child is a girl. However, we also want it to be true that P(B|G) = 2/3. For that to be true, all 12 mothers must chose to tell us about their girl. Clearly, all 12 mothers can't chose to tell us about their girls and their boy. Thoughts?

[PrayingMantis]

[ QUOTE ]
I think you miss the point, both the coin flip problem and the monte hall problem are probablility problems, not psychology problems.

[/ QUOTE ]

I never claimed that these are "psychology problems". But if you think these are "probability problems", in the purest sense, it seems to me that you don't fully understand both of these problems.

[ QUOTE ]
The problem is designed to determine if you understand how probability works, and the assumptions which lead to the 2/3 answer in the coin flip, and "who knows what monte is thinking" in Monte Hall are not part of the probablistic analysis.

[/ QUOTE ]

No. You assume that "the problem is designed to determine if you understand how probability works", but you have little ground to base this on. With regard to the Monte-Hall problem, originally it was clearly NOT designed to determine if you understand how probability works, but rather to see if there's any advantage in switching doors. And I repeat - in the original unspecified Monte-Hall problem, there's no one simple solution, because you HAVE to know what Monte is thinking and doing in order to get to any reasonable solution, as opposed to what you suggest above. The probabilites on such a problem cannot be worked out in a vacuum.

There's a lot of work done with regard to the Monte-Hall problen, and I suggest you read some of it. The version of Monte-Hall problem you are thinking about (the one where the solution is 2/3) is not the interesting one, as that one is very easy to understand and solve.

[ QUOTE ]
Further, in the OP, it was stated that the answer to the coin flip problem was 50-50 because they are independent events. At the time he posted, OP apparantly didn't understand why 2/3 was a valid answer.

[/ QUOTE ]

Reading the OP, It is clear to me that the poster understood why 2/3 is a valid answer. It is still a valid answer, BTW, but not the only valid answer.

[ QUOTE ]
Now, after reading this thread, if OP wants to go back and argue semantics with his/her prof, go right ahead. However, my academic experience is that with many profs, this would be a -ev exercise.

[/ QUOTE ]

This is not "semantics", in the sense that understanding the nature of the information you have is crucial to solving the problem. In certain types of probability problems, the exact wording is very important, as opposed to most other kinds of problems. There are many examples for this.

[ QUOTE ]
Now, would I make a bet on this if someone offered me. Nope, if I did, I'm sure I'd end up with an earful of cider (Reference to Damon Runyon for anyone who didn't pick it up).


[/ QUOTE ]

Examine this (note: this is not the original problem, but a version of it):

I have just flipped 2 unbiased coins, wrote the results and put in an envelope. I gave the envelope to a very respected poster here, that we both trust.

Now I'm telling you that one of the two results is heads. this result has a little star right next to it, in the envelope. I'm offering you 5:1 on a $1000 bet that the other result is tails. That is, if it's tails I pay you $5000, if it's heads you pay me $1000. That's the information you have, that's the bet I'm offering you. No further questions.

Would you (or anyone who thinks the answer to the first problem is definitely "2/3") take it, yes or no? And if not, why?

[KenProspero]

[ QUOTE ]
There's a lot of work done with regard to the Monte-Hall problen, and I suggest you read some of it. The version of Monte-Hall problem you are thinking about (the one where the solution is 2/3) is not the interesting one, as that one is very easy to understand and solve.

[/ QUOTE ]

Of course it's easy to solve -- yet many people don't believe the answer. Other versions (which I believe were spawned to some extent by an interview with Monte, in which he said no one knows what he was thinking or something like this) which rely on psychology and game theory. Also they need to take into account the fact that Monte (and his producers) really want to give away the big prize as much as necessary to maximize ratings).

To my knowledge, when the problem started out, it was meant as the easily calculable exercise, but because it was phrased colloqually, people were able to turn it into a different kind of problem.

[ QUOTE ]
This is not "semantics", in the sense that understanding the nature of the information you have is crucial to solving the problem. In certain types of probability problems, the exact wording is very important, as opposed to most other kinds of problems. There are many examples for this.

[/ QUOTE ]

Not having been in the class, I can't say for sure what the Prof was getting at. However, I have a strong suspicion that he was trying for the easily calculable mathmatical solution, but phrased the problem colloqually. In cases like these, you can absolutely bust the Prof's chops and explain the ambiguity, or you can recognize what he was trying to do. Either approach is 'correct'. With many professors, only one has a +ev.

[ QUOTE ]
Would you (or anyone who thinks the answer to the first problem is definitely "2/3") take it, yes or no? And if not, why?

[/ QUOTE ]

Of course not -- in order to take the bet, you have to assume that the game isn't rigged (i.e., a 'random series of two coin flips was chosen). In a problem that is phrased colloqually, Like Monte, this condition is assumed.

Now, now if you were to come to me before the coins were flipped, and guarantee that you'd offer me 5-1 if at least one of the coins comes up heads -- well that would be a good bet. But we all knew that, right?

[PrayingMantis]

[ QUOTE ]
In cases like these, you can absolutely bust the Prof's chops and explain the ambiguity, or you can recognize what he was trying to do. Either approach is 'correct'. With many professors, only one has a +ev.

[/ QUOTE ]

I think that overall we are agreeing. There's ambiguity in the problem, and it can be answered differetnly according to different assumptions. I'm really not interested in that prof and what should the OP tell him. For all I care, always agreeing with your prof, no matter how smart or stupid he is, is the best play in most circumstances, and obviously that has pretty much nothing to do with solving or thinking serioulsy about a problem.

And about the bet I've offered you, I can't see why you think it's rigged. Let's change it a bit: I will FILM myself flipping 2 unbiased coins, and give the film to a witness you trust. There will be no editing done on the film. I'm telling you that the outcome for one of the coins on the film was heads. Now, I offer you the same bet on the second flip filmed, 5:1 if it's tails, for $1000. What's rigged about it? If you are able to put a number on the probability for the 2nd coin, why not take such a bet? You know what, I'll offer you 10:1. Even 100:1 if you insist.

I can't see why anyone who really thinks that the answer for the original problem is definitely 2/3 (and no room for different interpretations) will not take such a bet, as a rational gambler. This is essentially the same problem with exactly the same details, as I'm giving you the same information about something that had _already happened_.