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.