http://www.isaiadis.com/2007/02/27/mathematical-paradoxes-twin-coin-flipping/ Discussions on science, skepticism, pseudoscience; trips on the borders of religion and science; brief adventures in hi-fi; this is a Journey through a Burning Mind... Sat, 19 May 2012 16:47:51 +0000 http://wordpress.org/?v=2.7 hourly 1 http://www.isaiadis.com/2007/02/27/mathematical-paradoxes-twin-coin-flipping/comment-page-3/#comment-110530 Lane Dillon Thu, 15 Mar 2012 20:50:53 +0000 #comment-110530 Hiya, I'm really glad I've found this info. Today bloggers publish only about gossips and web and this is actually annoying. A good web site with exciting content, this is what I need. Thank you for keeping this web site, I'll be visiting it. Do you do newsletters? Cant find it. London Escorts Agency, 020 3011 7825 http://www.isaiadis.com/2007/02/27/mathematical-paradoxes-twin-coin-flipping/comment-page-3/#comment-69360 Arthur Fri, 18 Mar 2011 20:30:00 +0000 #comment-69360 davidh has a good point in his post 86. The question is badly written. It should be more like, "if you flip two coins simultaneously and at least one of the coins shows heads, what is the probability that both coins show heads?" The original question implies (because it refers to "the other coin") that you are looking at a particular coin and seeing that it's showing a head. The other coin can be showing a head or a tail with a probability of 0.5 for each case. Nice problem though.
The question is badly written. It should be more like, “if you flip two coins simultaneously and at least one of the coins shows heads, what is the probability that both coins show heads?”

The original question implies (because it refers to “the other coin”) that you are looking at a particular coin and seeing that it’s showing a head. The other coin can be showing a head or a tail with a probability of 0.5 for each case.

Nice problem though.]]> http://www.isaiadis.com/2007/02/27/mathematical-paradoxes-twin-coin-flipping/comment-page-3/#comment-61785 Chris Sun, 02 Jan 2011 19:38:09 +0000 #comment-61785 ...hmmm. Post 110 was sufficiently well written that I suspect that it may be from an impostor. But #1. The strange sentence "I feel no compunction to have to explain what I say when I know that you fully inderstand what I am saying" is consistent with the real davidh. i.e. I can't imagine why it was written. I have only asked for calculations to be done, not for the words to be explained. I'll admit that I have said that a lot of what davidh has written is ambiguous and vague; thus causing me difficulty in nailing him down. But #2: The incorrect results for the die are consistent with the real davidh, as is the failure to say what magical ritual was used to summon them up. I'll go for a probability of 1/3 that post 110 is from a fake davidh.
But #1. The strange sentence “I feel no compunction to have to explain what I say when I know that you fully inderstand what I am saying” is consistent with the real davidh. i.e. I can’t imagine why it was written. I have only asked for calculations to be done, not for the words to be explained. I’ll admit that I have said that a lot of what davidh has written is ambiguous and vague; thus causing me difficulty in nailing him down.

But #2: The incorrect results for the die are consistent with the real davidh, as is the failure to say what magical ritual was used to summon them up.

I’ll go for a probability of 1/3 that post 110 is from a fake davidh.]]> http://www.isaiadis.com/2007/02/27/mathematical-paradoxes-twin-coin-flipping/comment-page-3/#comment-61781 Chris Sun, 02 Jan 2011 17:30:47 +0000 #comment-61781 Hi davidh, you'll like this one. You have 1025 coins in a bag. One of them is double-headed, the rest are normal. You pick one at random and flip it 10 times and get 10 heads. What is the probability that you had selected the double-headed coin? Hint: it's what you thought the answer was to the posted problem. PS In the original problem, I had assumed that the coins weren't biased. With the right bias, you would have been right. The bias would have been such that P(H) = 2/3 and P(T) = 1/3 (for identical coins). I expect that the coins would look very strange though.
You have 1025 coins in a bag. One of them is double-headed, the rest are normal. You pick one at random and flip it 10 times and get 10 heads. What is the probability that you had selected the double-headed coin?

Hint: it’s what you thought the answer was to the posted problem.

PS In the original problem, I had assumed that the coins weren’t biased. With the right bias, you would have been right. The bias would have been such that P(H) = 2/3 and P(T) = 1/3 (for identical coins). I expect that the coins would look very strange though.]]> http://www.isaiadis.com/2007/02/27/mathematical-paradoxes-twin-coin-flipping/comment-page-3/#comment-61772 Chris Sun, 02 Jan 2011 14:41:14 +0000 #comment-61772 Hi davidh. The die: the correct answers are 1/5 (for not a 5), 1/4 (for not a 4 or 5), 1/3 (for not a 3,4 or 5), 1/2 (for not a 2,3,4 or 5) and 1 (for not a 1,2,3,4 or 5). That's 20% (not 33%), 25% (not 50%), 33% (not 67%), 50% (not 83%) and 100% (you got that one right). e.g. for not a 4 or 5. We are left with 4 equally likely possibilities: 1,2,3 and 6. We must have one of them, so that's a 25% probability for each case, and 25% for the 6 in particular. I was going to introduce a tetrahedral die and associate 1 with HH, 2 with HT, 3 with TH and 4 with TT. I can see that I'd have got out of the frying pan and fallen into the fire. I don't know why you are so adverse to what you term my "probability notation". It is extremely easy to understand, and far more succinct than long-winded descriptions. Although I'd heard of Bertrand's Paradox, I didn't know what it was, so I looked it up. I'm surprised that it is famous; it's manifestly obvious that the probability distribution must be specified. I have been fully aware of that since my teenage years (over 40 years ago). Practical joker! Nice try, LOL. Anyway, you finally understood the original problem, so there's a result.
e.g. for not a 4 or 5. We are left with 4 equally likely possibilities: 1,2,3 and 6. We must have one of them, so that’s a 25% probability for each case, and 25% for the 6 in particular.

I was going to introduce a tetrahedral die and associate 1 with HH, 2 with HT, 3 with TH and 4 with TT. I can see that I’d have got out of the frying pan and fallen into the fire.

I don’t know why you are so adverse to what you term my “probability notation”. It is extremely easy to understand, and far more succinct than long-winded descriptions.

Although I’d heard of Bertrand’s Paradox, I didn’t know what it was, so I looked it up. I’m surprised that it is famous; it’s manifestly obvious that the probability distribution must be specified. I have been fully aware of that since my teenage years (over 40 years ago).

Practical joker! Nice try, LOL. Anyway, you finally understood the original problem, so there’s a result.]]> http://www.isaiadis.com/2007/02/27/mathematical-paradoxes-twin-coin-flipping/comment-page-3/#comment-61766 davidh Sun, 02 Jan 2011 09:32:43 +0000 #comment-61766 OK, I'll answer your question. First, I've been playing with you all along. I'm a practical joker by heart and have been known to drag out a joke for several years. I know full well that the probability of a second head appearing if one coin is heads is 33% (rounded). The paradox isn't that the other coin can only be heads or tails but that the other coin is twice as likely to be tails than heads. As for the die; if a thrown die is announced to not be a 5, the probability of it being a 6 is 33%. Not a 4 or 5, the probability is 50%. Not a 3, 4 or 5, 67%. Not a 2, 3, 4 or 5, 83%. Finally, not a 1, 2, 3, 4 or 5, 100%. (All % rounded to whole numbers). For the record, in some of my previous responses, I used terms and expressed results for which you demanded an explanation. I speak English, as do you. The words I use have common meanings which are well known to the majority of speakers. You prefer to use probability notation; I prefer not to. The descriptive words I use convey the same meaning as the notations you use. I feel no compunction to have to explain what I say when I know that you fully inderstand what I am saying. What's your opinion of the Bertrand paradox?
First, I’ve been playing with you all along. I’m a practical joker by heart and have been known to drag out a joke for several years.

I know full well that the probability of a second head appearing if one coin is heads is 33% (rounded). The paradox isn’t that the other coin can only be heads or tails but that the other coin is twice as likely to be tails than heads.

As for the die; if a thrown die is announced to not be a 5, the probability of it being a 6 is 33%. Not a 4 or 5, the probability is 50%. Not a 3, 4 or 5, 67%. Not a 2, 3, 4 or 5, 83%. Finally, not a 1, 2, 3, 4 or 5, 100%. (All % rounded to whole numbers).

For the record, in some of my previous responses, I used terms and expressed results for which you demanded an explanation. I speak English, as do you. The words I use have common meanings which are well known to the majority of speakers. You prefer to use probability notation; I prefer not to. The descriptive words I use convey the same meaning as the notations you use. I feel no compunction to have to explain what I say when I know that you fully inderstand what I am saying.

What’s your opinion of the Bertrand paradox?]]> http://www.isaiadis.com/2007/02/27/mathematical-paradoxes-twin-coin-flipping/comment-page-3/#comment-61739 Chris Sun, 02 Jan 2011 04:22:55 +0000 #comment-61739 Hi davidh. I've been away for a week, and see you have been unable to answer my questions. Perhaps you've been away too. I've just being refreshing my memory of where we had got to and noticed the following from your post 99: "At no point have you offered a simple answer, in the English language, to the simple question - if one coin shows heads, what are the three states that you claim the second coin can have landed? I say (1/2) either heads or tails, you say (1/3) heads, tails or ? Heads, tails or tails? This is patently absurd." - I have never said that one coin can have three states. Is this a deliberate misrepresentation of what I've said? Whatever, the three states that I have been referring to are for the <strong>pair of coins</strong>, after discarding the TT case. To wit: HH, HT and TH. You also imply that you had asked me to provide that as an answer to a question of yours - where did you ask it? The initial 4 states, HH,HT,TH,TT are known to anyone who has done more than an hour or two of a probability/statistics course, and it's also evident that you are fully aware of it (see your own post 31).
I’ve just being refreshing my memory of where we had got to and noticed the following from your post 99: “At no point have you offered a simple answer, in the English language, to the simple question - if one coin shows heads, what are the three states that you claim the second coin can have landed? I say (1/2) either heads or tails, you say (1/3) heads, tails or ? Heads, tails or tails? This is patently absurd.” - I have never said that one coin can have three states. Is this a deliberate misrepresentation of what I’ve said? Whatever, the three states that I have been referring to are for the pair of coins, after discarding the TT case. To wit: HH, HT and TH. You also imply that you had asked me to provide that as an answer to a question of yours - where did you ask it?

The initial 4 states, HH,HT,TH,TT are known to anyone who has done more than an hour or two of a probability/statistics course, and it’s also evident that you are fully aware of it (see your own post 31).]]> http://www.isaiadis.com/2007/02/27/mathematical-paradoxes-twin-coin-flipping/comment-page-3/#comment-60913 Chris Sun, 26 Dec 2010 02:54:08 +0000 #comment-60913 Hi again davidh. Just in case you think I've ignored you, I agree with what you say in your last post (103) i.e. everything before your ----. I'll ignore everything after the ----. I don't understand why you think that I disagree with what you say. To be clear, any individual trial ends with a particular result. I have never had any doubt about that. In the case of the coins problem, you aren't given all of the information. You only know that one of the coins is heads, so the best you can do (to the best of my knowledge) is to use probabilistic concepts to determine the state of the other coin. In my mind the problem as given is a little false - after all if you flipped the two coins, you wouldn't somehow only see that at least one of them is a heads, you'll see both of them. But I understand the gist of what the question was about, and to make it more sensible, I semi-unconsciously reinterpret it as, "out of your sight, someone flips two coins and announces to you that at least one of them is a heads. What is the probability that the other one is a head also?". At least that has become a plausible situation. I automatically consider what the outcome would be if the trial was repeated many times, with the understanding that the other guy will keep on flipping the coins until at least one is a heads. While I may have got rather silly in some of my responses to you, at heart I'm really a nice bloke and just love some of these seemingly counter-intuitive problems. I also love the fact that such simple problems have so many different ways of being analysed. The fact that I keep on corresponding with you is that I love the challenge and because I'm a sad git. OK now I've been completely straightforward with you, please answer my die problems, or acknowledge the 1/3 result for my suggested interpretation of the original problem.
I don’t understand why you think that I disagree with what you say. To be clear, any individual trial ends with a particular result. I have never had any doubt about that. In the case of the coins problem, you aren’t given all of the information. You only know that one of the coins is heads, so the best you can do (to the best
of my knowledge) is to use probabilistic concepts to determine the state of the other coin.

In my mind the problem as given is a little false - after all if you flipped the two coins, you wouldn’t somehow only see that at least one of them is a heads, you’ll see both of them. But I understand the gist of what the question was about, and to make it more sensible, I semi-unconsciously reinterpret it as, “out of your sight, someone flips two coins and announces to you that at least one of them is a heads. What is the probability that the other one is a head also?”. At least that has become a plausible situation. I automatically consider what the outcome would be if the trial was repeated many times, with the understanding that the other guy will keep on flipping the coins until at least one is a heads.

While I may have got rather silly in some of my responses to you, at heart I’m really a nice bloke and just love some of these seemingly counter-intuitive problems. I also love the fact that such simple problems have so many different ways of being analysed.

The fact that I keep on corresponding with you is that I love the challenge and because I’m a sad git.

OK now I’ve been completely straightforward with you, please answer my die problems, or acknowledge the 1/3 result for my suggested interpretation of the original problem.]]> http://www.isaiadis.com/2007/02/27/mathematical-paradoxes-twin-coin-flipping/comment-page-3/#comment-60896 Chris Sat, 25 Dec 2010 23:24:53 +0000 #comment-60896 Hii davidh. To expedite the proceedings, my next set of questions is for the die scenario, but the roller will announce that it's not a 5 or a 4. I'll then go on to say he announces that it's not a 5,4 or 3; and then in another scenario that it's not a 5,4,3 or 2 and finally it's not a 5,4,3,2 or 1. Obviously the roller might have to roll many times before he can make those announcements. In each case, what is the probability that a 6 has been rolled? Altogether, that's 5 separate scenarios. Please play with me a little longer and answer those 5 questions, this whole business can be put to rest in a few more posts.
Please play with me a little longer and answer those 5 questions, this whole business can be put to rest in a few more posts.]]> http://www.isaiadis.com/2007/02/27/mathematical-paradoxes-twin-coin-flipping/comment-page-3/#comment-60890 Chris Sat, 25 Dec 2010 21:12:39 +0000 #comment-60890 davidh. Here's one for you to try. Someone throws a perfectly ordinary die, out of you sight. They tell you that it's not showing 5. What is the probability that it's showing a 6? Someone throws a perfectly ordinary die, out of you sight. They tell you that it’s not showing 5. What is the probability that it’s showing a 6?]]> http://www.isaiadis.com/2007/02/27/mathematical-paradoxes-twin-coin-flipping/comment-page-3/#comment-60887 Chris Sat, 25 Dec 2010 21:01:37 +0000 #comment-60887 Hi davidh. You didn't show how you calculated 1/6 - you simply used a well known result. That's not logic or reasoning (although it is sensible). Here's a simple question for you. Do you agree that if you flip two normal and fair coins, that the probability that they will land HH is 1/4, TT is 1/4 and mixed HT is 1/2? Obviously discard the cases where the coins land on their sides or any other such ducking and diving. On the asumption that you agree, then you must accept that if we were to flip a pair of coins 4 times, then <strong>on average</strong>, we will get 1 case where we got HH, 1 where we got TT and 2 where we got a mixed HT. Now back to the original problem. We know we haven't got a TT. So we must have HH or mixed HT. In 1 case out of the remaining cases, we get HH and in 2 we get mixed HT. So the probability (relative frequency) of getting HH is 1/(1+2) = 1/3 QED.
Here’s a simple question for you. Do you agree that if you flip two normal and fair coins, that the probability that they will land HH is 1/4, TT is 1/4 and mixed HT is 1/2? Obviously discard the cases where the coins land on their sides or any other such ducking and diving.

On the asumption that you agree, then you must accept that if we were to flip a pair of coins 4 times, then on average, we will get 1 case where we got HH, 1 where we got TT and 2 where we got a mixed HT.

Now back to the original problem. We know we haven’t got a TT. So we must have HH or mixed HT. In 1 case out of the remaining cases, we get HH and in 2 we get mixed HT. So the probability (relative frequency) of getting HH is 1/(1+2) = 1/3 QED.]]> http://www.isaiadis.com/2007/02/27/mathematical-paradoxes-twin-coin-flipping/comment-page-3/#comment-60880 davidh Sat, 25 Dec 2010 18:50:27 +0000 #comment-60880 That should be "solve", not "slove" in the last paragraph. http://www.isaiadis.com/2007/02/27/mathematical-paradoxes-twin-coin-flipping/comment-page-3/#comment-60879 davidh Sat, 25 Dec 2010 18:47:44 +0000 #comment-60879 Once again (why do I torture myself like this?) - the following may be stupid in your mind, but the concept is central to my reasoning. If you roll a die, the probability of a 6 coming up is 1 in 6. After the die has been rolled, if a 6 is showing, it is a 100% 6, not 1/6 of a 6. There is no longer a 5/6 probability that it is not a 6. If you are jay-walking in a large city you may have a 1 in 1000 probability of getting hit by a car. However, if you successfully make the crossing, it's not 999/1000 of you standing on the curb, you have beat the odds and made the 100% cross. If you play the 3-digit lottery your chances of picking the winning number are 1 in 1000. After the number has been selected, if you have the winning number they pay you 100% of your winnings, not 1 one-thousandth. The point of all this is what you do not want to see; the odds of what may or can happen are not the same as what did happen. Odds exist only before the event; certainty exists after the completion of the event. There are four ways that two coins might land if filpped, but after the flip there is only one way that they have already landed. If one head is showing, it may be a head coin of the head-head pair, or it may be the head coin of one of the two head-tail pairs. But no matter which of these it may be, if one of the coins IS heads, the other one can only BE either heads, or tails. --------------- You want to tout your intelligence and your superior understanding of probabilities, while denegrating my use of logic and reasoning. I have several Mensa-level puzzle books full of question and puzzles, each of which have a single answer. I'd gladly let you use your probability knowledge to "slove" each of the questions, while I use logic and reasoning. Which of us do you think would win that race?
If you roll a die, the probability of a 6 coming up is 1 in 6. After the die has been rolled, if a 6 is showing, it is a 100% 6, not 1/6 of a 6. There is no longer a 5/6 probability that it is not a 6.

If you are jay-walking in a large city you may have a 1 in 1000 probability of getting hit by a car. However, if you successfully make the crossing, it’s not 999/1000 of you standing on the curb, you have beat the odds and made the 100% cross.

If you play the 3-digit lottery your chances of picking the winning number are 1 in 1000. After the number has been selected, if you have the winning number they pay you 100% of your winnings, not 1 one-thousandth.

The point of all this is what you do not want to see; the odds of what may or can happen are not the same as what did happen. Odds exist only before the event; certainty exists after the completion of the event.

There are four ways that two coins might land if filpped, but after the flip there is only one way that they have already landed. If one head is showing, it may be a head coin of the head-head pair, or it may be the head coin of one of the two head-tail pairs. But no matter which of these it may be, if one of the coins IS heads, the other one can only BE either heads, or tails.

—————

You want to tout your intelligence and your superior understanding of probabilities, while denegrating my use of logic and reasoning. I have several Mensa-level puzzle books full of question and puzzles, each of which have a single answer. I’d gladly let you use your probability knowledge to “slove” each of the questions, while I use logic and reasoning. Which of us do you think would win that race?]]> http://www.isaiadis.com/2007/02/27/mathematical-paradoxes-twin-coin-flipping/comment-page-3/#comment-60837 Chris Sat, 25 Dec 2010 12:39:18 +0000 #comment-60837 Merrry Christmas everyone. davidh. Re your post 99. You say I haven't evene once explained why I say there are 3 states. Well check my posts 30,41,50,56,58,63,74 and 76. Also much discussion of that very topic permeates this page. I have also defended and commented throughout. You yourself have written these states down several times (so you have acknowledged them as being correct). I used H, H' and brown and silver coins in order to establish the states and two reject you coin melding stuff (HT is TH from the other side of the table). You have tried to argue that TH and HT are the same, well I've exploded that one. But I have allowed that description, but only if you accept that mixed state is twice as likely as the non-mixed states. You say I haven't explained why the probability is 1/3 - well that's essentially all I've been doing - you must have overlooked most of my posts. In your post 99, you mention that if one of the coins is heads, then the other is or isn't heads. I've never contradicted you on that point - why do you imply that I have? In fact I have proven that the probability of the other one being heads is 1/3 many times. As for doing the computer modelling. I think you suggested that once (post 90). Why do you want me to do that, that comment seems to imply that I was disputing your claim (even thoough I had agreed with it). Perhaps you neglected to read my post that agreed with you. On the other hand, Stavros and I had recommended that you try a computer model many times over an extended period before you did it. And we only did this because you kept on failing to obtain the correct result. So in this you are being unreasonable and childish. The hilarious thing is that you did get it right (for a brief moment) in your post 31.
davidh. Re your post 99. You say I haven’t evene once explained why I say there are 3 states. Well check my posts 30,41,50,56,58,63,74 and 76. Also much discussion of that very topic permeates this page. I have also defended and commented throughout. You yourself have written these states down several times (so you have acknowledged them as being correct). I used H, H’ and brown and silver coins in order to establish the states and two reject you coin melding stuff (HT is TH from the other side of the table).

You have tried to argue that TH and HT are the same, well I’ve exploded that one. But I have allowed that description, but only if you accept that mixed state is twice as likely as the non-mixed states.

You say I haven’t explained why the probability is 1/3 - well that’s essentially all I’ve been doing - you must have overlooked most of my posts.

In your post 99, you mention that if one of the coins is heads, then the other is or isn’t heads. I’ve never contradicted you on that point - why do you imply that I have? In fact I have proven that the probability of the other one being heads is 1/3 many times.

As for doing the computer modelling. I think you suggested that once (post 90). Why do you want me to do that, that comment seems to imply that I was disputing your claim (even thoough I had agreed with it). Perhaps you neglected to read my post that agreed with you. On the other hand, Stavros and I had recommended that you try a computer model many times over an extended period before you did it. And we only did this because you kept on failing to obtain the correct result. So in this you are being unreasonable and childish.

The hilarious thing is that you did get it right (for a brief moment) in your post 31.]]> http://www.isaiadis.com/2007/02/27/mathematical-paradoxes-twin-coin-flipping/comment-page-3/#comment-60680 chris Sat, 25 Dec 2010 00:21:52 +0000 #comment-60680 Hi davidh. I'm sorry, I rushed that last post. I'll accept that it is reasonable to say that there are two states. But I won't accept that you can ignore the fact that that state is twice as likely as the non-mixed HH state. As I'm here again, we have a die, 5 of the sides are labelled with 1 dot, the last is labelled with 6 dots. I'll side with you and say that there are two states that you can roll, a 1 or a 6. So using your assertion that the probabilities don't matter, and your method of calculating I get the probability of throwing a 6 is 1/2 QED. Can you see that I've used precisely the same argument that you've used with the coins? i.e. ignore the probabilities for each state and simply divide the number of states we're interested in by the total possible number of states. I would argue (using similar notation to that used with the coins) that there are 6 (equally likely) states, 1, 1', 1", 1'", 1"", 1'"" and 6. So now the probability of throwing a 6 is 1/(1+1+1+1+1+1) = 1/6. Or I might economise and simply say that you are 5 times as likely to throw a 1 as a 6 and say 1/(1+5) = 1/6. That really is my last word (unless I change my mind).
As I’m here again, we have a die, 5 of the sides are labelled with 1 dot, the last is labelled with 6 dots. I’ll side with you and say that there are two states that you can roll, a 1 or a 6. So using your assertion that the probabilities don’t matter, and your method of calculating I get the probability of throwing a 6 is 1/2 QED.

Can you see that I’ve used precisely the same argument that you’ve used with the coins? i.e. ignore the probabilities for each state and simply divide the number of states we’re interested in by the total possible number of states.

I would argue (using similar notation to that used with the coins) that there are 6 (equally likely) states, 1, 1′, 1″, 1′”, 1″”, 1′”" and 6. So now the probability of throwing a 6 is 1/(1+1+1+1+1+1) = 1/6. Or I might economise and simply say that you are 5 times as likely to throw a 1 as a 6 and say 1/(1+5) = 1/6.

That really is my last word (unless I change my mind).]]> http://www.isaiadis.com/2007/02/27/mathematical-paradoxes-twin-coin-flipping/comment-page-2/#comment-60610 Chris Fri, 24 Dec 2010 21:51:02 +0000 #comment-60610 davidh. I have analysed and responded to nearly every point you have made and I have shown you several different ways of obtaining the solution. I have also shown you how to calculate the solution to every problem that you have added in trying to explain your viewpoint to me. I've broadly used the same approach each time, you don't seem to have registered that I get results that agree with your simpler ("obvious" result problems), and not see that my logic is impeccable. You almost never seem to be actually responding to any point that I've made (hence the brick wall reference). I've repeatedly asked you to show how you calculate any of your results - you have not done so once. You accuse me of not responding to your posts - I can only conclude that you haven't actually studied them (or unable to understand them). I'll give you that you may well have run a computer simulation. Yes, your result of 1/2 is patently absurd. You have yourself run a computer simulation that shows that (see your post 81 and my response 83). Again if the coins a brown and silver you can get brown heads + silver heads, brown heads + silver tails, brown tails + silver heads - that's 3 states with equal probability. If you see it as 2 states, then I can't help you; I'm not a psychiatrist. FYI there is no way to Vulcan meld those coins together. You have suceeded in demonstrating to me that you do not understand probability, simple arguments or simple logic. So yes, I give up - you are a hopeless cause.
I have also shown you how to calculate the solution to every problem that you have added in trying to explain your viewpoint to me. I’ve broadly used the same approach each time, you don’t seem to have registered that I get results that agree with your simpler (”obvious” result problems), and not see that my logic is impeccable.

You almost never seem to be actually responding to any point that I’ve made (hence the brick wall reference). I’ve repeatedly asked you to show how you calculate any of your results - you have not done so once. You accuse me of not responding to your posts - I can only conclude that you haven’t actually studied them (or unable to understand them). I’ll give you that you may well have run a computer simulation.

Yes, your result of 1/2 is patently absurd. You have yourself run a computer simulation that shows that (see your post 81 and my response 83). Again if the coins a brown and silver you can get brown heads + silver heads, brown heads + silver tails, brown tails + silver heads - that’s 3 states with equal probability. If you see it as 2 states, then I can’t help you; I’m not a psychiatrist. FYI there is no way to Vulcan meld those coins together.

You have suceeded in demonstrating to me that you do not understand probability, simple arguments or simple logic. So yes, I give up - you are a hopeless cause.]]> http://www.isaiadis.com/2007/02/27/mathematical-paradoxes-twin-coin-flipping/comment-page-2/#comment-60579 davidh Fri, 24 Dec 2010 19:31:16 +0000 #comment-60579 Chris - you feel I have ignored the points you have made; I feel, also, that you have ignored the points that I have made. You want me to do computer modeling and use probability notation. I contend that simple logic and reasoning are sufficient. I have suggested simple experiments that would produce the answer to this whole thing and you have totally ignored them. At no point have you offered a simple answer, in the English language, to the simple question - if one coin shows heads, what are the three states that you claim the second coin can have landed? I say (1/2) either heads or tails, you say (1/3) heads, tails or ? Heads, tails or tails? This is patently absurd. For the final time, once the coins have landed, the possible ways that they could have landed is no longer relevant; the only definitive thing is how they are now laying on the ground. If one is heads, the other one either is, or it is not also heads. Your descent into name calling and insults show that you have given up on the ability to use logical reasoning and are taking your ball and going home. So be it.
You want me to do computer modeling and use probability notation. I contend that simple logic and reasoning are sufficient.

I have suggested simple experiments that would produce the answer to this whole thing and you have totally ignored them.

At no point have you offered a simple answer, in the English language, to the simple question - if one coin shows heads, what are the three states that you claim the second coin can have landed? I say (1/2) either heads or tails, you say (1/3) heads, tails or ? Heads, tails or tails? This is patently absurd.

For the final time, once the coins have landed, the possible ways that they could have landed is no longer relevant; the only definitive thing is how they are now laying on the ground. If one is heads, the other one either is, or it is not also heads.

Your descent into name calling and insults show that you have given up on the ability to use logical reasoning and are taking your ball and going home. So be it.]]> http://www.isaiadis.com/2007/02/27/mathematical-paradoxes-twin-coin-flipping/comment-page-2/#comment-60561 Chris Fri, 24 Dec 2010 13:04:06 +0000 #comment-60561 Hi Stavros. davidh has already used his computer - but he is unable to understand the result. I formally concede that he is beyond my ability to reason with and I shan't try to help him to understand the problem any more. Besides I've finally become bored with talking to a brick wall - but I did enjoy making the attempt for the most part. Merry Christmas.
I formally concede that he is beyond my ability to reason with and I shan’t try to help him to understand the problem any more. Besides I’ve finally become bored with talking to a brick wall - but I did enjoy making the attempt for the most part.

Merry Christmas.]]> http://www.isaiadis.com/2007/02/27/mathematical-paradoxes-twin-coin-flipping/comment-page-2/#comment-60547 Stavros Fri, 24 Dec 2010 11:55:23 +0000 #comment-60547 Come on guys give it a rest. Chris there's no way you'll ever convince davidh. Davidh, such a simple probability teaser, you're still unable to grasp after thorough explanations and experimental verification! What else you need is beyond me... For crying out loud take a sample space of 100 flips, then on average 25 would be HH, 25 would be TT and the remaining 50 would be HT/TH. Discard the 25 TT. From the 75 flips left of which at least one is heads, how many are actually two heads?!?
Davidh, such a simple probability teaser, you’re still unable to grasp after thorough explanations and experimental verification! What else you need is beyond me…

For crying out loud take a sample space of 100 flips, then on average 25 would be HH, 25 would be TT and the remaining 50 would be HT/TH. Discard the 25 TT. From the 75 flips left of which at least one is heads, how many are actually two heads?!?]]> http://www.isaiadis.com/2007/02/27/mathematical-paradoxes-twin-coin-flipping/comment-page-2/#comment-60527 Chris Fri, 24 Dec 2010 09:59:24 +0000 #comment-60527 Hi davidh, as I expected, you are quite unnable/unwilling to address any of the important points that I've made. You have now resorted to quibbling about the wording of the question; what is actually being asked has been made more than adequately clear. It's a pathetic attempt to wriggle out of the fact that you have got the problem wrong and accused me (incorrectly) of being in error. Perhaps you have seen that your logic is inconsistent, and just aren't man enough to admit it. The meaning of flipping and landing is accessible to most people; you're the only person I've come across who doesn't understand this everyday idea. At any time whatsoever, I know what the probabilities of the two coins landing in any given configuration is - it's very simple, and no philosophy is required. I note that you did a computer model. The before discarding the TT cases you got it right (and were unnabe to see that's all that you needed). Then you did another where you were supposed to discard the TT cases; but you retained them and discarded half of the H+T flips - that was hilarious. The computer modelling stuff was suggested to you as you the evidence was that you were unnable to solve the problem using your brain. But it also turns out you can't even do the computer model right. I didn't do a computer model, I did it with real coins to show my brother the facts. Clearly you seem to believe that one of the coins is being constrained to land heads up - well that's just being silly and a blatant disregard of reality; especially as you have shown that you accept that P(HH)= 1/4 etc. I have also disposed of several of your side problems e.g. "simultaneous" flipping. I showed why that it doesn't matter if one coin is flipped before the other - you act as if you hadn't doubted it for a moment (of course a few picoseconds don't matter" - quote from memory). Yet it was you who made a big deal of it - not me. I have run rings around you and your pathetic reasoning. You are so stupid that you can't recognise that I am far more intelligent than you. And more importantly, I have demonstrated the truth of the main result, and of several other problems that you have posed. You at best resort to arm-flailing, claiming some results are obvious, but not once have you actually shown the results using mathematical modelling. Holier-than-thou - that's the pot calling the kettle black. Despite that lot, of course I hope you have an enjoyable Christmas and new year.
The meaning of flipping and landing is accessible to most people; you’re the only person I’ve come across who doesn’t understand this everyday idea. At any time whatsoever, I know what the probabilities of the two coins landing in any given configuration is - it’s very simple, and no philosophy is required.

I note that you did a computer model. The before discarding the TT cases you got it right (and were unnabe to see that’s all that you needed). Then you did another where you were supposed to discard the TT cases; but you retained them and discarded half of the H+T flips - that was hilarious. The computer modelling stuff was suggested to you as you the evidence was that you were unnable to solve the problem using your brain. But it also turns out you can’t even do the computer model right. I didn’t do a computer model, I did it with real coins to show my brother the facts.

Clearly you seem to believe that one of the coins is being constrained to land heads up - well that’s just being silly and a blatant disregard of reality; especially as you have shown that you accept that P(HH)= 1/4 etc.

I have also disposed of several of your side problems e.g. “simultaneous” flipping. I showed why that it doesn’t matter if one coin is flipped before the other - you act as if you hadn’t doubted it for a moment (of course a few picoseconds don’t matter” - quote from memory). Yet it was you who made a big deal of it - not me.

I have run rings around you and your pathetic reasoning. You are so stupid that you can’t recognise that I am far more intelligent than you. And more importantly, I have demonstrated the truth of the main result, and of several other problems that you have posed. You at best resort to arm-flailing, claiming some results are obvious, but not once have you actually shown the results using mathematical modelling.

Holier-than-thou - that’s the pot calling the kettle black.

Despite that lot, of course I hope you have an enjoyable Christmas and new year.]]>