The Puzzle of Probability Itself

[marplots]

Probability in her various guises comes up a lot at ISF. We cite statistics and percentages, mention things like relative risk, and point to studies with p values to justify our positions. We invoke Bayes and Occam. But I'm often troubled by just what probability is supposed to mean. Is it, for example an estimate of epistemic ignorance only? Or does it say something meaningful (and real) about the actual status of the world? Maybe probability is something more bizarre, like a kind of clock which separates the past (certainty) from the future (uncertain), or maybe it measures entropy. Or maybe it's all just a happy mathematical game without any larger meaning.

If I read last week that Hillary has a 60% chance of winning the election, and I read this week that her chances have improved to 80%, it feels to me like that's a real difference - a factual, measured difference. And yet, after the election, when I know the outcome, it's hard to look back with any seriousness and think those numbers meant much other than I didn't know what the outcome would be.

Here's a problem making the rounds about probability. I mention it not to get the "correct" answer, but as a target to invite conversations about what we should take probability to mean.

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?

1) Either zero or one.
2) 1/2
3) 1/6
4) 1/11
5) 1/12
6) 1/36
7) Make up another answer or even reject the premise.

I think there are reasonable arguments for each of those answers. Here are a few and I'll leave the rest of the answers for others to champion.

1) The outcome is already determined and in fact, was determined long ago as each cause led to a subsequent effect until a combination of material events plus the laws of nature resulted in the pre-determined outcome. The results are fixed and must be either zero or one, depending on those prior causes. The fact that we don't know the details is irrelevant - only adding an unnecessary subjective element. In the actual world the dice have already been rolled and their state is a matter of certain, historical fact.

2) True, the results were determined, but the question is about my own estimation and state of knowledge. Since I cannot determine between zero or one and have no insight into the prior determinants, I must average the two and state "1/2".

[3) through 5) left for others]

6) (I picked this one before I found it lacking.) The relevant randomizing event is the original roll. Anything after is tainted by agency (Sally's choices). We know that rolling two die will generate 36 possibilities, only one of which is a pair of sixes. Therefore, keeping only the original roll "pristine" we get 1/36.

 

[Jack by the hedge]

The puzzle, as so often, is not so much "what is the answer?" but rather "what is the question?". A more carefully stated question might say "Given what you know, what is the probability ...". In this case that gives answer (3).

 

[Aepervius]

Actually the probability that the OTHER dice is a 6 , is 1/6 , you are confusing probability of two dice being a 6. If the die are fair and independent it 1/number of face. That's the only good answer. (ETA: remember the question is "What is the probability that the other one is a six" it only question the other die, the 6 of the first die is irrelevant)

As for making up explanation for wrong answer, with a bit of imagination you can make up everything.

I am not sure what you post is about.

 

[Thermal]

Answer:3, 1/6. The value of the first die is known, taking it out of the probability calculation. The unknown die can only show one of six outcomes.

 

[psionl0]

The answer is 3).

Don't make the mistake of trying to find so much meaning in the concept of probability that it ends up having no meaning.

 

[marplots]

Actually the probability that the OTHER dice is a 6 , is 1/6 , you are confusing probability of two dice being a 6. If the die are fair and independent it 1/number of face. That's the only good answer.

As for making up explanation for wrong answer, with a bit of imagination you can make up everything.

I am not sure what you post is about.

It's about a lot of stuff, but the easiest to see is the tension between probability as a property which exists in the real world - that is, objectively and without humans - or whether it simply measures my ignorance about what exists in the world.

If I roll a dice the outcome is immediately fixed at whatever number comes up. If I look, I am certain (probability 1). If I do not look, then I assign a probability of 1/6 to each of the possible outcomes. So, it seems I am merely pointing at my own ignorance, not something fundamental outside of myself.

However, I also think that a die somehow "captures" a real set of events that has nothing to do with my state of knowledge, that the probabilities are somehow built into the die's construction - because of symmetry. (This also comes up in physics, the dice in the thought experiment only make it easier to talk about.)

 

[marplots]

The puzzle, as so often, is not so much "what is the answer?" but rather "what is the question?". A more carefully stated question might say "Given what you know, what is the probability ...". In this case that gives answer (3).

Answer:3, 1/6. The value of the first die is known, taking it out of the probability calculation. The unknown die can only show one of six outcomes.

The answer is 3).

Don't make the mistake of trying to find so much meaning in the concept of probability that it ends up having no meaning.

Here is an argument for 4) being correct.

Once you see that one of the dice is a six, you know the set we are dealing with is: {1,6; 2,6; 3,6; 4,6; 5,6; 6,6; 6,1; 6,2; 6,3; 6,4; 6,5}

No other rolls allow Sally to show us a six. Showing us a 6 has given us more information and we now know the subset listed above. We don't know which pair in that set was rolled, so we set the probability as 1/11 - because there are eleven items in the set.

 

[Jack by the hedge]

Here is an argument for 4) being correct.

Once you see that one of the dice is a six, you know the set we are dealing with is: {1,6; 2,6; 3,6; 4,6; 5,6; 6,6; 6,1; 6,2; 6,3; 6,4; 6,5}

No other rolls allow Sally to show us a six. Showing us a 6 has given us more information and we now know the subset listed above. We don't know which pair in that set was rolled, so we set the probability as 1/11 - because there are eleven items in the set.

That's not correct. There are two 6,6 outcomes; one where Sally show us the first six and one where she shows us the second. So the probability becomes 2/12 or 1/6.

 

[Aepervius]

Here is an argument for 4) being correct.

Once you see that one of the dice is a six, you know the set we are dealing with is: {1,6; 2,6; 3,6; 4,6; 5,6; 6,6; 6,1; 6,2; 6,3; 6,4; 6,5}

No other rolls allow Sally to show us a six. Showing us a 6 has given us more information and we now know the subset listed above. We don't know which pair in that set was rolled, so we set the probability as 1/11 - because there are eleven items in the set.

Only if you confuse the dice. But you cannot, one has already been thrown and is six. So a whole subset of your set fall out , the one where dice 1 is not a 6 and you get {1,6; 2,6; 3,6; 4,6; 5,6; 6,6} of which only 1 is 6 so 1/size of set = 1/6.

You cannot simply say 6;3 is equivalent to 3;6

ETA: typing with a cat trying to get food on my lap is hard

 

[Garrette]

It seems sort of like Monty Hall in that a key component -- which is left out -- is whether Sally knows both dice before she shows a 6 (and whether she had pre-decided that she would only show a 6) or she showed a die and it happened to be 6.

 

[Jack by the hedge]

That's why the question must be about what we know rather than what we can imagine. The probability would change if we knew that Sally was choosing what to show us based on some rule.

If we knew Sally used a rule "show the smallest number that comes up" then we know the probability of another six is 1.
If we knew the rule was "always show an odd number if there is one" then the probability becomes 1/3.

 

[Thermal]

Only if you confuse the dice. But you cannot, one has already been thrown and is six. So a whole subset of your set fall out , the one where dice 1 is not a 6 and you get {1,6; 2,6; 3,6; 4,6; 5,6; 6,6} of which only 1 is 6 so 1/size of set = 1/6.

You cannot simply say 6;3 is equivalent to 3;6

ETA: typing with a cat trying to get food on my lap is hard

Right, if you confuse the dice, the inverted outcomes (1,6 as opposed to 6,1) have no meaning. If you do not confuse them, the first is known and the second goes back to 1/6 (barring the Schrodinger-ish correctness of zero or one).

 

[Aepervius]

It seems sort of like Monty Hall in that a key component -- which is left out -- is whether Sally knows both dice before she shows a 6 (and whether she had pre-decided that she would only show a 6) or she showed a die and it happened to be 6.

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

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

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.

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.

 

[bruto]

Here is an argument for 4) being correct.

Once you see that one of the dice is a six, you know the set we are dealing with is: {1,6; 2,6; 3,6; 4,6; 5,6; 6,6; 6,1; 6,2; 6,3; 6,4; 6,5}

No other rolls allow Sally to show us a six. Showing us a 6 has given us more information and we now know the subset listed above. We don't know which pair in that set was rolled, so we set the probability as 1/11 - because there are eleven items in the set.

Is the set something that affects the dice, or is it something you make up after the fact? How would your set differ if the dice were separated in time or space? If you roll a die here, and another person rolls a die in Spain, will they still be a set? If you roll a die here, and the person in Spain does not know about yours, will they still be a set?

Is not the set you have named dependent on the first roll as well as the second? Half the probability of the set's outcome is already completed. Is it not an error to burden the second roll with the probability of the first?

 

[marplots]

It seems sort of like Monty Hall in that a key component -- which is left out -- is whether Sally knows both dice before she shows a 6 (and whether she had pre-decided that she would only show a 6) or she showed a die and it happened to be 6.

That's why the question must be about what we know rather than what we can imagine. The probability would change if we knew that Sally was choosing what to show us based on some rule.

If we knew Sally used a rule "show the smallest number that comes up" then we know the probability of another six is 1.
If we knew the rule was "always show an odd number if there is one" then the probability becomes 1/3.

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.

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.

 

[Aepervius]

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.

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.

Hu. No. How you ask the question should not change the answer, unless you are monkeying with people and try to cheat/play with their mind them (e.g. riddle, tv games, advertising, marketing , polls, politics etc...) which is why certain domain get bad reps.

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

 

[Jack by the hedge]

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.

 

[marplots]

Is the set something that affects the dice, or is it something you make up after the fact? How would your set differ if the dice were separated in time or space? If you roll a die here, and another person rolls a die in Spain, will they still be a set? If you roll a die here, and the person in Spain does not know about yours, will they still be a set?

Is not the set you have named dependent on the first roll as well as the second? Half the probability of the set's outcome is already completed. Is it not an error to burden the second roll with the probability of the first?

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.)

 

[Aepervius]

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.

And if somebody want to look it up, look up monte carlo and maybe also random walk.

 

[marplots]

Hu. No. How you ask the question should not change the answer, unless you are monkeying with people and try to cheat/play with their mind them (e.g. riddle, tv games, advertising, marketing , polls, politics etc...) which is why certain domain get bad reps.

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

No fooling, honest. But this is in the philosophy section for a reason. It's meant to be meta. The problem is an illustration of a type. (Although it's a fun puzzle to chew on anyhow.)