   Painfully easy I may be late on this, but the solution's wrong. Four combinations - TT, TH, HT, HH. If the first coin is H then TT and TH can be excluded, leaving a 50% chance of another head. Andy West Monday, December 10, 2001 He didn't say "the first one was heads"; he said "one of them was heads". Paul Brinkley Monday, December 10, 2001 I am damn sure the solution is wrong.  My main argument against the posted solution is called "The Monte Carlo Fallacy."  Go to    http://gncurtis.home.texas.net/gamblers.html  this will fully explain the fallacy The end result is, for ANY coin toss, the odds of heads or tails will always be 50/50, no matter the previous results of other flips for any other coin.  Since two coins flipped at the same time are two independant events, the outcome of the one flip cannot have an effect on the other.  After the fact, the odds of two heads in a row will be .25, the odds of three heads in a row will be .125, and so on, but the odds of the next one being heads will ALWAYS be .5. jj Monday, December 10, 2001 The real question, which is disguised within the scenario, is, "What is the probability getting two heads?"  The statement, "one of the coins came up heads," means ONLY that TT did not happen. This leaves three possibilities: HT, TH, HH. HH has a probability of .3333... Chad Hulbert Monday, December 10, 2001 Or if you prefer, consider this a conditional probability: What is the probability of getting two heads given one coin is heads? By definition, the probability of A given B is: P(AB)/P(B). P(AB) is .25.  P(B), the probability one coin is heads is .75 (1 - probability of no heads, .25).  P(AB)/P(B) is .3333... Chad Hulbert Monday, December 10, 2001 If you consider the opposite case, you can see why the odds are not 50-50, as classical independent coin-flipping logic would suggest. The question at hand can be restated as follows.. "I flip two coins and tell you that at least one of them is heads.  What are the odds that both are heads?" This wording gives away the trick, but it's equivalent to the proposed situation.  Now the opposite case... "I flip two coins and tell you that neither of them is heads.  Now what are the odds that both are heads?" This one IS painfully easy.  The odds are zero.  Obviously, if there were money at stake, I would have called foul after flipping the coins and tried again. It seems clear to me that the statement I've given you about "one coin being heads" is really a statement about BOTH coins.  Thus you can't say that the second coin is independent of the first -- I've entangled them by allowing the game to continue. Chris Farmer Tuesday, December 11, 2001 Restating the problem: I flip two coins, and then tell you that one of them is heads. What's the probability that they are both heads? Unfortunately, the answer can be anything at all, depending on why I decided to tell you that one of them is heads. Consider several cases: 1. If I decide that I'll tell you that one of them is heads whenever that's true, then the odds that both are heads is 1/3. 2. If I decide that I'll tell you than one of them is tails whenever that's true, and only tell you that one of them is heads when neither is tails, the odds that both are heads is 1. 3. If I decide to truthfully tell you either that both are heads, that both are tails, or that one is heads, then the odds that both are heads is 0. 4. By introducing some probability into my decision process, every other answer between 0 and 1 is possible. Painfully easy? No. Painful? Yes. -- Jim Lyon Jim Lyon Tuesday, December 11, 2001 "1. If I decide that I'll tell you that one of them is heads whenever that's true, then the odds that both are heads is 1/3." No, the probability that both are heads is 1/3.  The odds are 1/2.  Remember, probability is the total favorable outcomes over the total possible outcomes.  Odds are total favorable over total unfavorable. As for your other suggestions, I disagree because you're adding conditions to the problem that don't exist.  The nature of such puzzles is that what's stated is precisely and exactly correct.  No more, no less. Chad Hulbert Tuesday, December 11, 2001 I'm not sure that I get your distinction between "odds" and "probability" -- I've used the words interchangably. While there are formal mathematical definitions of probability, the simplest intuitive definition is that if we repeated the experiment many times, what fraction of the time would we get the desired outcome. And that leads to the crux of the matter. You said, "the nature of such puzzles is that what's stated is precisely and exactly correct; no more, no less." I entirely agree with this sentiment, but the puzzle as originally stated doesn't give us enough information to run the experiment multiple times. Sometimes, when I flip a pair of coins, they'll both come up tails. Therefore, sometimes I won't say that one of them is heads. Nowhere in the puzzle does it state that I'll say that one of them came up heads every time that it's true. Your attempt to use this assumption reads more into the puzzle than was there. Which is exactly my point. Jim Lyon Wednesday, December 12, 2001 I will concede that 'odds' and 'probability' are often used (incorrectly) interchangably in common speech, however in forums such as this, where we're talkin' statistics, the words are not ambiguous and have precise meanings. I'm still entirely confused about where you're going with all these hypotheticals.  What you might do or say on a different flip doesn't affect what you did do and say on the current flip, and any other outcome of the coin toss that voilates the original question necessarily *changes the question*. We don't need to know the decision making process the coin flipper used and we do have enough information to answer the question about this trial.  However, we do assume the flipper tells the truth because it would be a very uninteresting problem if he didn't.  Heck, if he's a liar then there may not even be any coins. This is a logic puzzle, not a philosophical thought exercise. Chad Hulbert Wednesday, December 12, 2001 Just goes to show how confusing probability can be. "The statement, "one of the coins came up heads," means ONLY that TT did not happen. This leaves three possibilities: HT, TH, HH." Of course, one of the coins being heads precludes TT *and* either HT or TH, leaving two possibilities. Andy West Thursday, December 13, 2001 Andy, I'm sorry, friend, but there's nothing more to be said.  The answer is 1/3, which I have tried to describe qualitatively and proven quantitatively. Go ask your statistics professor or math teacher to describe to you, as they are paid to explain such things and can no-doubt do it better than myself. Or go searching the web for more problems like this.  A common flavor of it is a mother or father making the same claim about her or his children: "One of my children is a girl.  What is the probability the other is also a girl." Good luck. Chad Hulbert Thursday, December 13, 2001 A strong argument could be said that the probability of the other coin being heads is 0.  After all the flipper said neither of "both coins came up heads" or "at least one coin came up heads" -- he said "one of the coins came up heads" -- i.e. "exactly one of the coins came up heads" by a standard pragmatic reading. A mildly interesting variant would have been if the flipper said "at least one of the coins is heads" and the question were "What is the probability of the dime being heads?" or "What is the expected number of cents which are heads?" Just my .11 worth... Jonathan Segal Thursday, December 13, 2001 Chad, there is (always) one more thing to be said. You would be correct if the question had been "I toss two coins several times, recording the results. I pick at random one toss where at least one of the coins was a head. What are the odds that the other was also a head" The actual question was (to paraphrase) "I toss two coins, one is a head, what is the probability of the other being a head" To which the answer is 1/2, since the two events are independent. I learnt when I was a student (many years ago) that you should always answer the question asked, rather than the question you assume is being asked. Andy West Friday, December 14, 2001 "The actual question was (to paraphrase) 'I toss two coins, one is a head, what is the probability of the other being a head' To which the answer is 1/2, since the two events are independent." Nope... the answer to the question above, exactly as it was asked, is still 1/3.  And the events above are not independent.  Here's a question to which the answer would be 1/2: "'I toss two coins, the first one is a head, what is the probability of the other being a head?" Paul Brinkley Friday, December 14, 2001 Andy, The answer to the puzzle is in fact 1/3.  Let me try to clear a few things up, hopefully...  ;) Earlier you wrote: "The actual question was (to paraphrase) "I toss two coins, one is a head, what is the probability of the other being a head" To which the answer is 1/2, since the two events are independent." That is where your logic is incorrect.  The two events are NOT independent, because there is only ONE event, not two.  The total sample space for the event is {TT, TH, HT, HH}.  Whether the person tossed the coins simultaneously or one at a time is irrelevant, since the results of both coins are recorded together as ONE event.  If you misinterpreted the wording in the puzzle, well... it's meant to be tricky, that's the whole point.  ;) One last thing.  Jonathan wrote: "A strong argument could be said that the probability of the other coin being heads is 0 ... he said "one of the coins came up heads" -- i.e. "exactly one of the coins came up heads" by a standard pragmatic reading." Jonathan, you're being fooled by the same tricky wording that fooled Andy.  "one of the coins came up heads" does NOT mean "exactly one of the coins came up heads."  Those are two very different statements.  By a standard pragmatic reading, "one of the coins came up heads" means just that.  The statement is not intended to reveal any information about the other coin. Hope this helps. Andy Shyne Friday, December 14, 2001 Andy & Paul, You're right of course. Thanks for explaining. To quote myself - just goes to show how confusing probability can be ;) Andy West Monday, December 17, 2001 I disagree with the posted answer. I have two coins, with which I can have THREE possible combinations of heads and tails, both being heads, both being tails, or one heads and one tails. If I already know that one is heads, that throws out the two tails combo, leaving me with either both being heads, or one being heads and the other being tails. To cut to the chase, the question was "what is the chance that the other coin also came up heads" or restated, what is the chance that a single coin toss came up heads? 50% Steve Barbour Tuesday, December 18, 2001 Arrrgh.  Here we go again.  From the top, once more... There are four possible combinations.  TT, TH, HT, HH.  Each of these is equally likely, so each has a 25% chance of occurring. The tosser reports that "one of them was heads".  In a very long thread which I can no longer locate (it was on the old editthispage site), Joel (I think it was Joel) brought up that the above wording was potentially misleading.  It could be interpreted as "exactly one of them was heads", or "at least one of them was heads".  I believe it's safe to say that the latter interpretation was intended, but for fun let's cover the former case as well, and state for the record that the probability of the other one being heads would of course be 0. Now for the latter case.  Assume that the tosser is telling the truth.  (If the tosser is lying, then the definition of "the other coin" is unclear and the question makes no sense.)  That means the TT case never happened, and we're down to three equally likely cases.  Once again, they are TH, HT, and HH.  Each has a 1/3 chance of having occurred. The chance that the other coin is heads is equal to the chance that the HH case occurred, which is 1/3.  The two coin flips cannot be considered independently, because the statement "one of them was heads" necessarily intertwines them.  A discussion of independent and dependent events is not within the scope of this post, but can be found at your nearest probability theory textbook. Now, usually I'm pretty tolerant of discussion.  However, in this one case, all paths have been treaded over and over and over.  The probability theory has been verified in Zurich three times by little probability gnomes with mathematics post-doctorate degrees, and the forms have been witnessed and stamped by eight notary publics and recognized by the UN.  -This is no longer open to debate.- If any of you think it is, it's because your understanding of probability or logic is flawed, or your line of reasoning is flawed, or you misread the question.  I know this is harsh, but it has to be so.  We gotta make a clean break, and move on.  I hope we can still be friends.  Now cue the closing credits and turn the lights off before you leave. Paul Brinkley Wednesday, December 19, 2001 Actually, I'll concede that you're right...but (you knew that was coming, right?)... There are only 3 possible combinations of our coins, since as the question is worded we have no way to distinguish the coins from each other. So the combinations are TT, HH, or HT/TH, but the TH/HT outcome is twice as likely to occur. So, my syntactical dwarfs (grudgingly) concede defeat to your probabilistic gnomes. As far as moving on...I refuse to do so until I either know I'm right, or know why I am wrong. Steve Barbour Wednesday, December 19, 2001 "There are only 3 possible combinations of our coins, since as the question is worded we have no way to distinguish the coins from each other. [...] So the combinations are TT, HH, or HT/TH, but the TH/HT outcome is twice as likely to occur." Both true.  (Well, technically, the original problem said one coin was a dime and one was a nickel, I think, so you could distinguish them if you want to.)  But you should be able to tell that it doesn't matter, since as you said, TH/HT is twice is likely.  You may as well treat them as two separate cases, TH and HT, both as likely as HH.  That's what they are, in fact, and you wouldn't be deluding yourself into anything incorrect if you did that. Even if you treat it as a single twice-as-likely case, it still means the other coin is only going to be heads 1/3 of the time.  Test it out if you like. Paul Brinkley Thursday, December 27, 2001 Yep (I thought I already agreed with that part). Just be anal, I actually did test it out before I posted it. I used two pennies and dropped them simultaneously (more or less) onto my desk from a height of about six inches. The results: TT - 15 HH - 12 Mixed - 28 I got bored after that, figuring that was close enough to confirm the 1/3 possibility. "You may as well treat them as two separate cases, TH and HT, both as likely as HH. That's what they are, in fact, and you wouldn't be deluding yourself into anything incorrect if you did that." True, but stating there were four combinations when I could only see three was what threw me off into thinking that the answer would be 1/2. And the part that irritates me...if you know which coin is heads, then the chance of the other one being heads appears to be 1/2. On the plus side, I am now looking forward to my Prob and Stats course this semester. Steve Barbour Friday, December 28, 2001 "And the part that irritates me...if you know which coin is heads, then the chance of the other one being heads appears to be 1/2." Heh.  I sympathize.  You are absolutely correct here.  Try looking at it like a panel with four lights labeled HH, TH, HT, and TT.  If you know one of the coins is heads, it's like knowing one of the lights can't have been lit; if you know which coin, it's like knowing two of the lights can't have been lit.  Knowing which coin was heads automatically knocks out a light. Good luck in the course!  It's been years since I took mine, so I don't remember a whole lot.  I do remember learning about dependent and independent events for the first time then.  Today, I still get as far as I ever did in analyzing probability problems by using the time-honored technique of counting all the possible outcomes, and computing subsets of them.  The trick here is understanding what qualifies as a possible outcome. Paul Brinkley Wednesday, January 2, 2002 Recent Topics Fog Creek Home