## A probability(?) problem

**Moderators:** gmalivuk, Moderators General, Prelates

### A probability(?) problem

I'm not sure if this is technically probability or not, but I have been puzzling about it with my friend to no avail.

Known:

-There is a dice with a finite number of sides. This number may or may not be random.

-Every side of the dice has a finite positive number on it. These numbers may or may not be random.

-The only time the dice has ever been rolled, it landed on a 2

Based on what is known, what is the most likely number for it to land on next? I think it would be 2, because you know that at the very least the dice has a 2. If you were to guess a number randomly then there is an infinitesimally small chance that you would even guess a number that is on the dice, whereas you know that there is a measurable chance that the dice will land on a two on the second roll. My friend believes that there is not enough information to make a practical conclusion.

Known:

-There is a dice with a finite number of sides. This number may or may not be random.

-Every side of the dice has a finite positive number on it. These numbers may or may not be random.

-The only time the dice has ever been rolled, it landed on a 2

Based on what is known, what is the most likely number for it to land on next? I think it would be 2, because you know that at the very least the dice has a 2. If you were to guess a number randomly then there is an infinitesimally small chance that you would even guess a number that is on the dice, whereas you know that there is a measurable chance that the dice will land on a two on the second roll. My friend believes that there is not enough information to make a practical conclusion.

### Re: A probability(?) problem

Ask your friend this:

If she or he only had $10 dollars, and needed $1000 to save her or his mother’s life, and a bookie was offering 100:1 odds for guessing the result of one single roll of that die, and that was the only possible way to get enough money in time, what number would your friend bet on? Why?

If she or he only had $10 dollars, and needed $1000 to save her or his mother’s life, and a bookie was offering 100:1 odds for guessing the result of one single roll of that die, and that was the only possible way to get enough money in time, what number would your friend bet on? Why?

wee free kings

- Xanthir
- My HERO!!!
**Posts:**5413**Joined:**Tue Feb 20, 2007 12:49 am UTC**Location:**The Googleplex-
**Contact:**

### Re: A probability(?) problem

I don't think you can put an initial probability distribution on the die, because it requires a uniform distribution over the integers, which doesn't exist.

But you're right, after having seen a 2, your revised distribution must at least increase the probability of seeing a 2. Since all integers were previous equal, 2 is now the most likely number, and so guessing a 2 is the choice most likely to be correct.

But you're right, after having seen a 2, your revised distribution must at least increase the probability of seeing a 2. Since all integers were previous equal, 2 is now the most likely number, and so guessing a 2 is the choice most likely to be correct.

(defun fibs (n &optional (a 1) (b 1)) (take n (unfold '+ a b)))

### Re: A probability(?) problem

A frequentist approach definitely wouldn't work, since you don't have enough information about the system to build a workable model - the best you could do would be if you knew an upper bound to the number of sides, or something about the distribution of the number of sides or values on them. From a Bayesian viewpoint, though, you can start with an uninformative and even physically impossible prior, and the single data point given will help collapse that slightly - to the point where, yes, you will have slightly more confidence in the die's chance of landing on a 2 than anything else.

pollywog wrote:I want to learn this smile, perfect it, and then go around smiling at lesbians and freaking them out.Wikihow wrote:* Smile a lot! Give a gay girl a knowing "Hey, I'm a lesbian too!" smile.

### Re: A probability(?) problem

"It landed on a 2"

So we don't know what number(s) is(are) showing?

Yeah yeah just kidding. I vote 2.

So we don't know what number(s) is(are) showing?

Yeah yeah just kidding. I vote 2.

Why not just say "If you guess correctly you save your mother's life - now what number do you guess?"Qaanol wrote:Ask your friend this:

If she or he only had $10 dollars, and needed $1000 to save her or his mother’s life, and a bookie was offering 100:1 odds for guessing the result of one single roll of that die, and that was the only possible way to get enough money in time, what number would your friend bet on? Why?

-Adam

- gmalivuk
- GNU Terry Pratchett
**Posts:**26767**Joined:**Wed Feb 28, 2007 6:02 pm UTC**Location:**Here and There-
**Contact:**

### Re: A probability(?) problem

While it's true that you can't assign specific probabilities without more restrictions on the distributions, and so it might not (depending on your interpretation of probability) make sense to say any number is the most likely result, it still seems that 2 is the most rational choice.sgfw wrote:I'm not sure if this is technically probability or not, but I have been puzzling about it with my friend to no avail.

Known:

-There is a dice with a finite number of sides. This number may or may not be random.

-Every side of the dice has a finite positive number on it. These numbers may or may not be random.

-The only time the dice has ever been rolled, it landed on a 2

Based on what is known, what is the most likely number for it to land on next? I think it would be 2, because you know that at the very least the dice has a 2. If you were to guess a number randomly then there is an infinitesimally small chance that you would even guess a number that is on the dice, whereas you know that there is a measurable chance that the dice will land on a two on the second roll. My friend believes that there is not enough information to make a practical conclusion.

As Qaanol put it, if you *had* to make a prediction, you'd pick 2, even if you couldn't quantify how much more likely that is than any other option.

### Re: A probability(?) problem

sgfw wrote:[b]

My friend believes that there is not enough information to make a practical conclusion.

You're both right. There's an infinitesimally tiny higher probability of a 2. Practically speaking, that's nothing. All depends on how you look at it.

### Re: A probability(?) problem

Qaanol wrote:Ask your friend this:

If she or he only had $10 dollars, and needed $1000 to save her or his mother’s life, and a bookie was offering 100:1 odds for guessing the result of one single roll of that die, and that was the only possible way to get enough money in time, what number would your friend bet on? Why?

I'm not sure I like this approach. I think most people would be very strongly inclined to go with their intuition in that situation, which would (for instance) lead to them answering wrong on the Monty Hall problem.

### Re: A probability(?) problem

sgfw wrote:-Every side of the dice has a finite positive number on it. These numbers may or may not be random.

Does this mean that you aren't assuming that the values on the die are 1, 2, ..., n for the unknown value of n? It could be that there are 7 sides and the values on the die are 2, 29, 23451234, 78, 79, 100, 1010101101?

If that's the case then I don't know how you could guess anything other than 2.

double epsilon = -.0000001;

### Re: A probability(?) problem

Here's another reason why you might say 2:

Change the question slightly to: let X be a finite set of natural numbers. Choose an element randomly from X with uniform probability (can be done since X is finite). This element is 2. What does this tell you about X? Well, one obvious thing that it tells you that X almost certainly has fewer than ten elements in it, and probably has about three or four elements in it.

Coming back to the die question, if we assume the faces are all different (or else merge the faces) then the fact that 2 is one of them, and it was rolled, means that it's highly likely that the number of faces is small.

Change the question slightly to: let X be a finite set of natural numbers. Choose an element randomly from X with uniform probability (can be done since X is finite). This element is 2. What does this tell you about X? Well, one obvious thing that it tells you that X almost certainly has fewer than ten elements in it, and probably has about three or four elements in it.

Coming back to the die question, if we assume the faces are all different (or else merge the faces) then the fact that 2 is one of them, and it was rolled, means that it's highly likely that the number of faces is small.

- Xanthir
- My HERO!!!
**Posts:**5413**Joined:**Tue Feb 20, 2007 12:49 am UTC**Location:**The Googleplex-
**Contact:**

### Re: A probability(?) problem

I don't see how that follows at all. It seems like you're implicitly relying on some other distribution information to make that statement about the size of X?

(defun fibs (n &optional (a 1) (b 1)) (take n (unfold '+ a b)))

### Re: A probability(?) problem

Xanthir wrote:I don't see how that follows at all. It seems like you're implicitly relying on some other distribution information to make that statement about the size of X?

That was pretty much my thought too. Unless you are implicitly assuming that X takes a form similar to {1, 2, ..., n} then I don't see how the magnitude of the element you chose relates to the size of the set itself.

double epsilon = -.0000001;

### Re: A probability(?) problem

DavCrav wrote:if we assume the faces are all different

Heck of an assumption to make, but it lets us get useful information out of this.

You know, if the die were a standard n-sided die numbered from 1 to n for some unknown value of n, and it was rolled once and came up 2, then we're dealing with the German Tank Problem, and we might guess that n is no more than 4.

Sure, we can't do better than guessing 2, that's fairly obvious, because 2 is the only number that we know exists on the die. (Possible followup question: You can't name a number that you've already seen the die roll, what do you guess will come up next?) But what do we know about the distribution of the numbers on the die, and the size of the die? "Random" isn't a distribution. Can we guess, for example, that the number of sides on the die was chosen by some human writing down a number and going with it (possibly with the aid of a PRNG)? Because that's actually a fairly strong statement - we know the die is unlikely to have more than a googool sides, for one thing.

The preceding comment is an automated response.

### Re: A probability(?) problem

Dason wrote:Xanthir wrote:I don't see how that follows at all. It seems like you're implicitly relying on some other distribution information to make that statement about the size of X?

That was pretty much my thought too. Unless you are implicitly assuming that X takes a form similar to {1, 2, ..., n} then I don't see how the magnitude of the element you chose relates to the size of the set itself.

The natural numbers are well ordered. So our faces are a subset of {1,...,n} for some n, and so the worst case is when they actually are {1,...,n}.

### Who is online

Users browsing this forum: No registered users and 6 guests