Fantome, can you clarify how the magician is selecting which number to report to each of the blindfolded people? Is he deciding the number beforehand and rolling until he gets at least one die with that number? Or is he rolling the dice once, then picking one of those numbers? For the seventh person, is he doing the same thing, or is he doing it differently?
What you’re missing (or, more accurately, what you’re looking at differently than other people) is the *methodology *the magician uses to report a numeral.
If the magician flips two coins, then randomly picks a coin and reports its value, then you get the scenario outlined above.
If, however, the magician simply reports whether or not a “1” shows up at all (i.e., he’s not reporting on a *particular *coin), then you get different probabilities, because the information the magician is reporting has different meaning.
Most everybody else here is translating English into probabilese in the second fashion, rather than the first, that’s all.
If the magician in zut’s case is always flipping the coins once, then picking one of the numbers shown, the first two subjects are wrong also. The probability of flipping both coins and getting a total of 3 is always 50 percent. If the magician is selecting which number he is going to tell them is present ahead of time, then flipping the coins until he gets at least one of that number, then all three subjects are right.
I don’t know. Is there a totally fair and equal way to do so?
The point is, it seems the second group of 100 people have about 50 that have coins that equal 3, and the first group has about 67 that have coins that equal 3 simply by waiting for someone to talk.
If this can be resolved in a way to prove that all 100 people in the second group actually have a 1/2 probability of having coins that equal 3 even after hearing the magician, then the same would have to be said for all those that have answered “1/3” in the Boy/Girl problem. They must be wrong and the probability of a couple that has exactly two children having two girls after being told at least one is girl is 1/2 (presumably because there’s no fair and equal way for the person who supplied the information that at least one child is a girl to do so).
But that leads to absurdity as per your own previous post – two fair coins ought to sum to 3 in about 50 of the 100 cases, since there’s two ways for them to do so, and two for which they don’t, both equiprobable. I don’t see how one group having the magician’s information is supposed to change this.
The magician eliminates one case with his information – the one where both coins are the same, and the opposite of what he has called. Three quarters of all possibilities remain valid; of those, two thirds sum to three, which, as would be expected, equals half of all cases. It’s actually exactly the Monty Hall problem – to wit: you have three doors, pick one; the host opens one of the others, which contains no prize, and asks you if you want to switch or not. What you are saying is essentially that it doesn’t matter, since the probability is now 50/50 between the two remaining doors; however, the correct thing to do would be to switch, since the other door does now have a 2/3rds chance of winning.
Look at it this way: You have four doors, two of which hold a price, and pick one. Monty opens another, behind which there is no prize – equivalent to the magician telling you a number, since that eliminates one of the two options that don’t sum to three. Now, you’re arguing that every door holds a 1/3 probability of not having a price behind it; however, just like in the Monty Hall problem, the door you’re holding actually has a 50% chance of being the wrong one.
…I think. I’ve now got to make some preparations for tonight’s partying, so I’ll likely see you all next year!
This, I would say, is the crux of it. If you solely had the information “At least one of the dice is a 4” or whatever, then the conditional probability of a sum of 7 is 2/11. But this is NOT the only information you have after the magician makes his statement. You have further information. Simply knowing that at least one of your dice is a 4 is not enough to to conclude that the magician tells you at least one of your dice is a 4; knowing the latter is knowing more than the former. And whenever you gain new information, probabilities (conditioned on your knowledge) shift. Exactly how it shifts depends on the probability distribution guiding the magicians’ choice of what to tell you.
If you can start from “Hm, no matter what the magician says to me, the posterior probability will be the same”, then you can indeed conclude “Thus, that value is the current probability as well”. But in setups where you can do so, the correct invariant response given to the magician after his statement won’t be “2/11”. That response will be “1/6”.
Alternatively, he could be telling each subject whether or not a “1” (or “2” in some cases) is showing. As long as he chooses whether to talk about “1” or “2” independently from the result of the flip, the first two subjects are right and the third is wrong.
To get the 2/11 as the correct answer, the magician would need to decide in advance, for each person rolling a die, which number he was going to choose to report on, and simply say whether it was present or absent.
In that case, 25/36 of the time, the magician would be telling someone, for some n, “n doesn’t appear on either die,” and the other 11/36 of the time, he’d be telling them it appears at least once.*
If he chooses a particular die to report on, whether before or after it’s rolled, then that die is determined, and the only uncertainty pertains to the unchosen die. And the probability of any given number on that unchosen die, and a total of 7 between the two dice, is 1/6.
*ETA: He’s obviously not doing this in the OP, since he was able to tell all six people that a number did appear on at least one of their dice.
As a complete aside, if my back-of-the-envelope calculations are right, if the experiment is to roll 14 dice at once, in about 42% of iterations of the experiment, at least one number from 1 to 6 will not show up at all.
Incidentally, similar trickery regarding what information is and isn’t known guides the Monty Hall and boy-girl paradoxes.
With the Monty Hall paradox, people want to say Door A (already selected by the contestant) and Door B are equally probable to win because they think they have only the information that Door C is not a winner; if this was the only information they had, they would be correct, but what they are forgetting is that they also have the further information that Monty Hall chose to reveal Door C (and not some other door). Given a suitable setup where Monty Hall randomly reveals one of the losing doors not already selected by the contestant, this further information is enough for the usual result, that Door A has 1/3 probability of winning but Door B has 2/3 probability.
Conversely, with the boy-girl problem, people reason as though they had not just the information that at least one of the family’s two children is a boy, but, furthermore, the information that some interlocutor, picking a child at random from the two whose gender to announce, happened to select a boy. With this further information, it is indeed equally probable to be a boy-and-boy or a boy-and-girl family, but lacking this further information, we have the usual result, that a boy-and-boy family has probability 1/3 while a boy-and-girl family has probability 2/3.
Paying attention to ALL of what information has and has not been given is important; probabilities generally shift entirely with the addition of new information.
There’s nothing fair or unfair except insofar as you would consider them such; there’s just different behaviors (probability distributions) we might choose to investigate; like I said, looking at the die rolls and the magicians’ statement, there are 66 total possibilities, and it hasn’t been fully specified what all their probabilities are (for every distinct A and B, it hasn’t been specified what the relative probabilities of the two possible statements the magician could make are after rolling A then B).
The point is, no matter how we flesh this out to a full specification of a probability distribution (that is, a full specification of how the magician chooses what to announce), not everyone will be able to respond “2/11”. The reasoning that you should be able to take a weighted average of these possible responses to get a probability even before the magician says anything, well, that is sound reasoning; the problem is in assuming the correct responses are all 2/11 to begin with. They are only 2/11 if you don’t take into account the extra information provided by knowing what the magician has chosen to tell you about.
My branch, which is composed of ten statisticians and a secretary, is having its belated holiday party on Tuesday. I’m going to present this little paradox when lunch is winding down, just to liven things up. I’ll let you guys know what the reaction is. 
It is possible for everyone to be able to correctly respond 2/11, even the last person. All that is required is for the magician to select in advance which number he will tell the person is present, and to then roll the pair of dice repeatedly until he gets that number. (The blindfolded people also have to know that that is what he is doing. Otherwise, all they can correctly answer is “I don’t know.”)
Are you going to have answers to the questions I asked in post 41 lined up, or are are you just going to say “I don’t know” if they ask?
Indistinguishable, I’ve read your and others responses and it’s getting a little off track and I’m getting confused. Could you please respond to post #38? I’ll repeat it below:
And post #44:
Fine, that’s a suitably modified setup where that’s true. But in that case, as you note, one CAN correctly say “The probability of my dice adding to 7 is 2/11” without waiting for the magician to say anything. I did not think that was in the spirit of the problem, which I assumed was meant to stay within normal “All dice rolls are equally probable and independent, so the usual reasoning about dice probabilities remains sound, including the fact that having two add to 7 has probability 1/6” conventions [tossing away and redoing dice rolls in certain cases effectively departs from that]. But it’s worth noting what happens in that case nonetheless, as you do.
Anyway, all I was saying was that, if we stick to “The magician rolls your dice exactly once, with no redos, and then somehow chooses a number present to tell you about”, then you can’t have the correct response in all cases being 2/11. In this setup, 1/6 WILL be a weighted average of all the possible correct responses (specifically, weight the correct response to “At least one of your numbers is n” by the probability that the magician will say so)
As Indistinguishable said, it’s not a question of fair or unfair. It is possible to set it up so that 2/11 is the correct answer for the first six people, if that’s what you mean. Do what I said in my post just above. Have the magician select in advance which number he will tell the person is present, and then roll the pair of dice repeatedly until he gets that number.
If he does the same thing with the seventh person as well, the seventh person is correct when he answers 2/11 before the magician tells him the number.
If instead, the magician treats the seventh person differently than the first six, by just rolling the dice once and telling the seventh person one of the numbers that comes up, then it shouldn’t be surprising that the correct answer is something other than 2/11.
Either way, there is no paradox.
It depends on how the magician chooses to reveal his information.
A) Randomly choosing a coin to reveal
is different from
B) Choosing a statment to make a priori, then flipping the coins to make the statement true
is different from
C) Revealing the presence or absence of a particular number
(And there’s probably other ways to interpret the magician’s actions.)
Yes, the first group will contain about 50 with coins that equal 3. As others have mentioned, what’s important is how the magician decides what information he provides.
Consider:
Everyone in group 1 will belong to 1 of 4 subgroups (and be evenly distributed between these groups):
A: Those with 1-1
B: Those with 1-2
C: Those with 2-1
D: Those with 2-2
If the magician tells everyone in B,C, and D ‘you have one 2’ then people told ‘you have one 2’ have a 2/3 probability of having coins that total 3, and people told ‘you have one 1’ have a 0/3 probability of having coins that total 3.
Alternatively, if the magician tells everyone in C and D ‘you have one 2’ then people told ‘you have one 2’ have 1/2 probability of having coins that total 3.
If the magician chooses some other means of selecting what information to provide, or gives an answer other than ‘you have one 1’ or ‘you have one 2’ then these probabilities will change accordingly.
Given no information on the magician’s method, ‘2/3’ is an answer based on an unreasonable assumption, and the better answer is the prior, ‘1/2’.
In zut’s example, if the ONLY information the first blindfolded person knew was that “At least one of my coins comes up 1”, then he can correctly say “There is probability 2/3 that my total is 3”.
But that is not the only information he has. He has further information. He has also the information that the magician said “At least one of your coins comes up 1”, rather than saying something else. And this further information will change the probabilities. [How it changes them depends on the probabilistic rules governing what the magician chooses to say, which have been left unspecified]
Similarly for the second blindfolded person. Taking this into account, there will be no problem for the third blindfolded person; the third blindfolded person can correctly reason “The probability, right now, that my coins total to 3 is ‘What I would say if you told me at least one was a 1’ * ‘The probability that you’ll tell me at least one is a 1’ + ‘What I would say if you told me at least one was a 2’ * ‘The probability that you’ll tell me at least one is a 2’”. This is all correct. And it will come out to the value 1/2 that we know and love (at least, in the setup as I am thinking of it, though alternative setups could be provided). The exact details beyond that depend on the probabilistic rules governing what the magician chooses to say.
The important thing is that knowing “The magician chose to say at least one of my coins is a 1” is having more information than simply knowing “At least one of my coins is a 1”.
If you wouldn’t mind a TINY bit of mathematical formula, an invaluable tool in probability is Bayes’ Theorem:
The probability of A given that B holds = (the probability of B given that A holds)*(the probability of A)/(the probability of B).
In this case, “The probability of a total of 3 given that the magician says at least one is a 1” = (the probability of the magician saying at least one is a 1 given that the total is 3) * (the probability that the total is 3)/(the probability that the magician says at least one is a 1) = (the probability of the magician saying at least one is a 1 given that the total is 3) * 0.5/(the probability that the magician says at least one is a 1). Beyond that, the details depend on the rules governing what the magician says, but note that it probably won’t come out to 2/3 (at least, if it does in this case, then it won’t in the case where the magician announces at least one is a 2).
What is 2/3 is “The probability of a total of 3 given that at least one is a 1”. But this is not really the relevant probability to know; the above is the relevant one.
Hm… there are things I want to say but I don’t know where they belong.
One thing I want to say is that much of the problem in these “probability paradoxes” is that so much of our probabilistic language relies on leaving things unsaid, to be reconstructed by implicit convention. The paradoxes depend on us reconstructing them a different way than the questioner actually intends. Were the questioner more explicit, there would be less difficulty.
For example, I flip a coin. What’s the probability that it lands heads?
Who knows? It’s up to me; I can assign probabilities any way I want (subject to very minimal constraints). But by convention, we want heads and tails to be equiprobable, so I should answer 1/2.
My buddy rolls a die, out of sight. He tells me it comes up even. What’s the probability that it came up as 2?
Who knows? It’s up to me; I can assign probabilities any way I want (subject to very minimal constraints). But by convention, every die face is equiprobable; furthermore, by convention, you do not want the a priori probability that the die comes up as 2 (which would be 1/6, on the equiprobability convention), but rather the probability conditioned on the event that my friend says it comes up even. But this isn’t quite right, actually; the superseding convention in this case is that you don’t want to condition on my friend saying it comes up even (maybe my friend’s a liar? Maybe he’s not a liar, but he only says it comes up even when it comes up on 4. Maybe all kinds of things), but, instead, you actually want to condition on it actually coming up even. Fair enough; now the answer is 1/3.
Monty Hall hides a prize behind one of three doors. I select a door. What’s the probability I selected the right door? I can assign probabilities however I want… Convention says each door is equally probable to have originally housed the prize. But this isn’t enough to answer the problem; maybe I have an unnatural affinity for selecting doors with prizes behind them, for all I know. To solve the problem, we need to rely on the further convention that the placement of the prize is probabilistically independent of the choice I make. With all these conventions in place, the answer is 1/3. But none were given explicitly…
Monty Hall hides a prize behind one of three doors. I select door B. Then Monty Hall reveals door C, and shows that it’s not a prize-winning door. What’s the probability that door A is the winner? We need to drag in all the conventions from before… but this still isn’t enough to answer the problem. Perhaps Monty Hall only reveals door B when door A is the winner, or perhaps he reveals door B every time, or perhaps something else. Perhaps he flips a weighted coin. Much turns on what has here been left unsaid.
Making everything explicit makes it easier to clear up misunderstandings and “paradoxes” in probability.
In the OP’s problem, in particular, we’ve seen many, many different formalizations in this thread, reaching different answers. And the thing is, most of these answers are correct… they’re just to different questions. Different probabilities being calculated, based on different probability distributions being investigated. The OP’s problem did not explicitly pick any one thing to look at. There are conventions guiding how to reconstruct what was being asked, but sometimes, people disagree about what convention demands, or stray from convention without realizing it. Much is cleared up by being explicit. Anyway, yeah.