The Puzzle of Probability Itself

[Jack by the hedge]

How you ask a question should never change the answer in pure math.

Exactly. How you ask the question does not change the answer unless that change means you literally ask a different question.

 

[marplots]

I don't understand what you mean about there being something vague about the probability itself.

I see the probability as being a property of the system itself rather than some property of the human observer. You can perfectly easily write a computer program which uses random numbers to choose what to do next and over thousands of iterations the probabilities of each outcome shine through without any human intervention.

It has been done for this problem. The versions I've seen favor answer 4 (1/11). However, one suspects the programmers wrote the procedure in the way they also conceived of the puzzle (with their assumptions injected thereby) and so created a kind of self-fulfilling simulation.

 

[marplots]

How you ask a question should never change the answer in pure math.

Exactly. How you ask the question does not change the answer unless that change means you literally ask a different question.

That's the property that makes it troubling. (Which is to say, "interesting.")

 

[Jack by the hedge]

It has been done for this problem. The versions I've seen favor answer 4 (1/11). However, one suspects the programmers wrote the procedure in the way they also conceived of the puzzle (with their assumptions injected thereby) and so created a kind of self-fulfilling simulation.

In other words they made a mistake in their code. That answer is most definitely wrong.

 

[Dave Rogers]

Indeed, and this is a key point: in questions of probability, how you ask the question influences the answer you get. It's not merely deductive, the procedure makes a difference.

As Aepervius said, it's not true that the answer depends on how you ask the question; the answer depends on what the initial circumstances are. As long as the question reflects that correctly, then you'll get the correct answer. For example, taking your question: "Sally rolls two dice (6-sided, assumed fair). She shows one is a six. What is the probability that the other one is a six?" There are some things unstated here that in effect render the question unanswerable unless they are assumed. If in reality Sally picks one die at random and shows it, and if the number six is written on only one side of the other die, then the probability is 1/6, however you may want to slice and dice your other assumptions. If, however, you know when asking the question that Sally deliberately chose to reveal the die with the higher number, that may in some situations (though not this one, I think) change the answer, and if you happen to know that both dice have 6 written on all six sides then that most definitely changes it. And if you choose to conceal such information when asking the question, then the XKCD -gry rule applies.

That property is troubling all by itself. And not just troubling because of vagueness in the statement of the puzzle (or language), but something vague attached to probability itself.

Not really, no. The property is simply that, unless the initial conditions are correctly stated, then the calculated resulting probability will not be correct. That's pretty much the case for any predictive statement.

Dave

 

[Jack by the hedge]

That's the property that makes it troubling. (Which is to say, "interesting.")

I suppose I tend to find it more "annoying". Sometimes answers become obvious if people just spend a bit more effort on deciding what their actual question is.

 

[marplots]

In other words they made a mistake in their code. That answer is most definitely wrong.

Maybe not the code but the recipe the code simulates. Do you code? Apparently it can be done in Excel.

Does the answer change in your framing if one die is red and the other green or wouldn't that matter? (Just curious about how you are thinking about the puzzle.)

 

[Dave Rogers]

It is true the problem just says, "Sally rolls two dice..." and doesn't specify whether they are rolled together or separately. It is also true that the problem does not outline Sally's internal "procedure" for her subsequent actions.

If probability is about making statements about the dice (as randomizing objects) how is it that any of that matters?

It matters because Sally's actions can remove some or part of the randomizing effect of rolling the dice. For example, if Sally rolls the dice repeatedly until she gets a double 6 and then shows one of the dice, then their randomizing property is more or less irrelevant.

Dave

 

[Aepervius]

That's the property that makes it troubling. (Which is to say, "interesting.")

No that is the property which makes it far more reliable than anything else whatsoever in life. Even more than sciences.

Whoever you are, whatever you do in life, whichever time of the day it is , the binary table of truth of a XOR will always be the same. Commutation for addition is always the same in the integer ensemble (heck you can even define non commutivity over other type of ensemble).

The point is that there is only a single answer or no answer mostly which makes math so "pure" and great (I do not exclude that somebody come up to ma with a non hilbertian subspace with special property).

 

[Garrette]

Not really. In this case it is pretty simple.
Known :
* 2 dice are thrown
* 1 dice is a 6
* dice are fair so assumed independent
unknown :
* value of other dice

You've disagreed by assuming an answer, not by obviating my insistence on clarification. If it were stated as you do here, then I agree with your further analysis. My point is that it was not stated so clearly.

There is no monty halling or anything at work. It is a straightforward.

In the one interpretation, but given its wording in the OP, that interpretation is not the only legitimate one.

Whether you count all pair with 6 as above or you count all a pair where one fixed die is a six you always come to the same conclusion : 1/6 because this is by definition the value probability of any fair dice.

See above.

You could get something more complicated if :
Sally knows one die is a six.

What is the probability if you look at one of the die at random , that it is a six ?

Well you have 1/2 chance at looking at one die so 1/2 is a six
then the other die 1/6, sot the probability to SEE one six when you look at ONE die is : 1/6+1/2 = 7/12 far more interesting IMO.

But as stated , what is the probability of the other die being a six, it is straightforward.

This doesn't come naturally to me, and I'm distracted with other work, so I will admit I don't fully follow this.

 

[marplots]

I suppose I tend to find it more "annoying". Sometimes answers become obvious if people just spend a bit more effort on deciding what their actual question is.

I've run across the idea that "the question creates the answer" which sounds like it might apply. But that's another topic I guess.

 

[Aepervius]

Maybe not the code but the recipe the code simulates. Do you code? Apparently it can be done in Excel.

Does the answer change in your framing if one die is red and the other green or wouldn't that matter? (Just curious about how you are thinking about the puzzle.)

No it would not matter. Because of the property that the dices are fair. That mean they are not interacting with each other (the face of one does not depend on the other, and all faces are equiprobable).

 

[Jack by the hedge]

Does the answer change in your framing if one die is red and the other green or wouldn't that matter? (Just curious about how you are thinking about the puzzle.)

Well I was thinking of red and blue in case anyone was a red-green colourblind thinker.

In rejecting their claim for answer (4) it matters that they differentiate between 1,6 and 6,1 regarding those as two separate outcomes but they do not differentiate between 6,6 and 6,6. So there really are twelve possible outcomes, not eleven, and two of them have a six on the unseen die.

 

[Aepervius]

Well I was thinking of red and blue in case anyone was a red-green colourblind thinker.

In rejecting their claim for answer (4) it matters that they differentiate between 1,6 and 6,1 regarding those as two separate outcomes but they do not differentiate between 6,6 and 6,6. So there really are twelve possible outcomes, not eleven, and two of them have a six on the unseen die.

Yes when i was tutorring people this is the usual problem they have when seeing undifferentiated die : they forget to count the swapped probability with identical die faces (i.o.w they see only one 6;6 pair when in reality if the die are undifferentiated there is two). Which is why I prefer to use differentiated die, unless the problem call for undifferentiated dies. The way is different but the solution is for small problem, more often straightforward to get at, with differentiated dies.

 

[marplots]

As Aepervius said, it's not true that the answer depends on how you ask the question; the answer depends on what the initial circumstances are. As long as the question reflects that correctly, then you'll get the correct answer. For example, taking your question: "Sally rolls two dice (6-sided, assumed fair). She shows one is a six. What is the probability that the other one is a six?" There are some things unstated here that in effect render the question unanswerable unless they are assumed. If in reality Sally picks one die at random and shows it, and if the number six is written on only one side of the other die, then the probability is 1/6, however you may want to slice and dice your other assumptions. If, however, you know when asking the question that Sally deliberately chose to reveal the die with the higher number, that may in some situations (though not this one, I think) change the answer, and if you happen to know that both dice have 6 written on all six sides then that most definitely changes it. And if you choose to conceal such information when asking the question, then the XKCD -gry rule applies.



Not really, no. The property is simply that, unless the initial conditions are correctly stated, then the calculated resulting probability will not be correct. That's pretty much the case for any predictive statement.

Dave

I agree to an extent, but the step you took isn't the whole trip.

If I state the problem with enough specificity, I get answer 1. Why? Because enough specificity gives me all the relevant details (Sally's hand position, the forces on the dice, the constitution of the surface they land on, etc. ad nauseum) and any notions of less than certainty (probability one or zero) evaporates thereby.

What you seem to be asking for is only enough specificity to differentiate between certain answers and then stop.

If you agree with the above, then it seems the only island of refuge would be an appeal to epistemic ignorance - not knowing which procedure she chose - but if that's so, why not generate another probability to address that bit and combine it? What do we get if we set the procedures available to Sally as symmetrical and subject the whole mess to our probability machine? (Keyword "mess")

 

[marplots]

Well I was thinking of red and blue in case anyone was a red-green colourblind thinker.

In rejecting their claim for answer (4) it matters that they differentiate between 1,6 and 6,1 regarding those as two separate outcomes but they do not differentiate between 6,6 and 6,6. So there really are twelve possible outcomes, not eleven, and two of them have a six on the unseen die.

It is interesting that you differentiate a red six and a blue six from a blue six and a red six (answer 5, which no one likes yet, depends on this too). Where did the ordering element come in? Is it because Sally shows one dice and has a choice?

Suppose Sally merely stated, "At least one of the dice is a six" (but didn't show any). Would this change alter your analysis? (Keeping the colors in play.)

 

[bruto]

It is true the problem just says, "Sally rolls two dice..." and doesn't specify whether they are rolled together or separately. It is also true that the problem does not outline Sally's internal "procedure" for her subsequent actions.

If probability is about making statements about the dice (as randomizing objects) how is it that any of that matters? Surely it wouldn't matter if I were talking about the probability of other types of random events, would it?

(e.g. the chances of the San Andreas fault slipping tomorrow.)

But my point is that in fact you're not making a statement about the dice. You are making a statement about a set of dice rolling events, some of which have already occurred. The likelihood of an outcome from the set depends equally on all the rolls.

 

[marplots]

But my point is that in fact you're not making a statement about the dice. You are making a statement about a set of dice rolling events, some of which have already occurred. The likelihood of an outcome from the set depends equally on all the rolls.

I think I may have lost track of the variation you are referring to. In my conception there is only one "rolling event" the original randomizing process. By the time I get additional information, all the dice are fixed.

Forgive me if I misunderstood. (Your sig quote applies to me here.)

 

[Dave Rogers]

I agree to an extent, but the step you took isn't the whole trip.

If I state the problem with enough specificity, I get answer 1. Why? Because enough specificity gives me all the relevant details (Sally's hand position, the forces on the dice, the constitution of the surface they land on, etc. ad nauseum) and any notions of less than certainty (probability one or zero) evaporates thereby.

What you seem to be asking for is only enough specificity to differentiate between certain answers and then stop.

Yes, and that is exactly the situation that probability is supposed to describe. If the question is insufficient to specify the limits of what is unknown, then the probabilities of different outcomes cannot be calculated. The question can be rephrased as, "Given that I know [list of factors], that I do not know [second list of factors], and that I do not know [list of outcomes], what are the probabilities of the different outcomes?" Some of this is embodied in the statement that "The dice are fair," which is itself a statement about probability.

Dave

 

[Dave Rogers]

It is interesting that you differentiate a red six and a blue six from a blue six and a red six (answer 5, which no one likes yet, depends on this too). Where did the ordering element come in? Is it because Sally shows one dice and has a choice?

Wherever else there is ordering, it must at least take place when Sally shows one die and not the other. If we assume that the dice were already distinguishable - for example, by one being red and the other blue - then there are twelve possible a priori instances in which Sally shows a die reading 6, of which two are instances in which the other die also shows 6. If we assume that the dice are not distinguishable until Sally reveals one, then there are six possible a posteriori instances in which she shows a die reading 6, of which one is an instance in which the other die reads 6. In either case the probability is 1/6.

Suppose Sally merely stated, "At least one of the dice is a six" (but didn't show any). Would this change alter your analysis? (Keeping the colors in play.)

In this case, Sally has stated that there are eleven possible outcomes: 1/6, 2/6, 3/6, 4/6, 5/6, 6/6, 6/5, 6/4, 6/3, 6/2, 6/1. If we choose one of the two dice, there are 22 different outcomes, 12 of which are a 6, so the probability of one of them being a six is 6/11. However, this is a completely different situation than the original problem; knowing that at least one of the dice shows a six is a different starting condition than knowing that the other die shows a six.

Dave