well known proba problem

[Bantari]

Interpretation #2:
You FIX the situation so that B WILL COME FIRST ON EACH DRAW (the wording of the question MIGH SEEM TO stipulate this) - like you throw away the WW box and attach a string to A BLACK STONE in each remaining box ensuring it to be the first draw. In which case it is only the question which of the two boxes you pick and the answer is 50%.

Well, it seems to me that this interpretation is a bit strange. The original question says nothing about attaching strings to stones or anything, it just describes what happens.

I know, the string is my idea... pretty clever, eh?
But seriously - its just one of the possible ways to ENSURE that the first stone picked is B. Another way would be to place A BLACK STONE on top of the box... Or whatever, I'm sure there are other ways Think in practical terms: If you were to conduct such experiment physically, with actual boxes and stones, how can you ENSURE that your first pick is a B stone?

[HermanHiddema]

Interpretation #2:
You FIX the situation so that B WILL COME FIRST ON EACH DRAW (the wording of the question MIGH SEEM TO stipulate this) - like you throw away the WW box and attach a string to A BLACK STONE in each remaining box ensuring it to be the first draw. In which case it is only the question which of the two boxes you pick and the answer is 50%.

Well, it seems to me that this interpretation is a bit strange. The original question says nothing about attaching strings to stones or anything, it just describes what happens.

I know, the string is my idea... pretty clever, eh?
But seriously - its just one of the possible ways to ENSURE that the first stone picked is B. Another way would be to place A BLACK STONE on top of the box... Or whatever, I'm sure there are other ways Think in practical terms: If you were to conduct such experiment physically, with actual boxes and stones, how can you ENSURE that your first pick is a B stone?

Well, the thing is, why would you want to? The original problem uses the phrase: "This Coin turns out to be a Gold coin". It doesn't ensure anything, it just describes what happens. To me, the idea of "ensuring" looks like a far-fetched attempt to make the problem fit an intuitive but mistaken answer.

In practise, the color doesn't matter, due to symmetry. If you were to conduct the experiment, you would ask anyone grabbing a black stone to estimate the probability that the other stone in the bowl is also black, and you would ask anyone grabbing a white stone to estimate the probability that the other stone is also white. Both cases are exactly equivalent to the original problem, both have the same answer, and 100% of test subjects can be asked the question.

[Bantari]

Well, it seems to me that this interpretation is a bit strange. The original question says nothing about attaching strings to stones or anything, it just describes what happens.

I know, the string is my idea... pretty clever, eh?
But seriously - its just one of the possible ways to ENSURE that the first stone picked is B. Another way would be to place A BLACK STONE on top of the box... Or whatever, I'm sure there are other ways Think in practical terms: If you were to conduct such experiment physically, with actual boxes and stones, how can you ENSURE that your first pick is a B stone?

Well, the thing is, why would you want to?

Ugh... I don't know what to say to that. I guess I had it coming.
Anyways - nice talking to you, bud.

[Bill Spight]

So, the whole question, paraphrased, simplified, and without confusing and unnecessary details, boils down to this:

- You have TWO boxes, Bb and Bw.
- Pick one at random.
- What is the chance of having picked Bb? (since this is the only scenario in which the second/hidden stone is also B)

The answer is clearly 50%.

If you pick one of those two boxes at random, indeed, the probability is 50% that one of the boxes has two Black stones in it. But we have not used all the information that we got from drawing a Black stone from a box. Suppose now that you draw a stone from the box that you picked, and it is Black. Now what is the probability that the box has another Black stone in it?

Yeah, sleep is a wonderful thing... I should not get into these types of discussions in the middle of the night.
Now I see that you can view the question in two different ways, maybe its the wording... and I think I got fixated on the wrong interpretation. Or maybe everybody else has... Anyhow... I think this is the bottom of the issue here:

Interpretation #1:
You can see it like the TJ's code - you have three boxes, draw a stone randomly, and out of the cases when B comes up first, what is the statistical probability that a second B will come second too? You have 3 possible cases in the results set: (B1,B2), (B2,B1), and (B3,W1) - and out of the 3 in 2 B will also be on second draw. Thus the answer in this case is 2/3, as demonstrated. Or:

Interpretation #2:
You FIX the situation so that B WILL COME FIRST ON EACH DRAW (the wording of the question MIGH SEEM TO stipulate this) - like you throw away the WW box and attach a string to A BLACK STONE in each remaining box ensuring it to be the first draw. In which case it is only the question which of the two boxes you pick and the answer is 50%.

I assume that interpretation #1 is perceived as 'correct' by the math types out there... although I don't really see why that should be without knowing more about the circumstances of the initial problem.

Anyways - does the above two interpretation seem like a likely source of the confusion?
It makes it a non-issue for me, mathematically... everything's clear other than the problem itself.

I went back and took a look at your first post in this thread. Here is what you were responding to.

Choices at first:

- BB
- BW
- WW

I choose a box and take out a B stone, so the WW option has just been eliminated, hasn’t it? How should then a 2/3 chance remain? There is no /3 anymore, so it should be 2/2 -> 1:1 chance, no?

You are right that taking out a B stone is ambiguous. But here is an earlier statement of the problem.

Suppose you have 3 Go bowls:
- One with 2 Shell stones
- One with 2 Slate stones
- One with 1 Shell and 1 Slate stone

Suppose you randomly pick a bowl, then take one of the 2 stones out from the bowl.
This stone turns out to be . What is the probability that the second stone in the bowl is also ?

The color of the stone changed, but that is immaterial.

Here you did not deliberately pick a , that's just how it turned out. Therefore an interpretation that guarantees that that the picked stone is is incorrect. (Sure, somebody may have rigged the choice, but probability is about what what we know. Somebody could have rigged the choice the other way, as well. )

[kivi]

Actually you can change the last line of the puzzle to be:

This stone turns out to be a particular color. What is the probability that the second stone in the bowl is also same color?

in order to avoid the fixation about fixing the situation

[drmwc]

An interesting variant is the Sleeping Beauty Problem.

On Sunday, Sleeping Beauty is given a drug to make her sleep. Immediately after she falls asleep, a fair coin is tossed.
1) If it is heads, she is woken on Monday. She is interviewed on Monday. and the experiment ends.
2) If it tails, she is also woken on Monday and interviewed. However, she is then put back to sleep and given an amnesia-inducing drug which causes her to forget the previou interview. She is woken on Tuesday and interviewed again. Then the experiment ends.

At each interview, she is asked "What is your credence to the propoosition that the coin landed heads?"

The problem is to determine the best answer. This is suprisingly non-trivial.

[speedchase]

I think she should say it is monday. There is a 75% chance she is correct

|,,/\
|,m,/\
|,,m,,t
each branch is 50/50.

edit: whitespace fail. give me a sec to figure it out.
edit2: commas are whitespace

[jts]

Speedchase, she's being asked about the coin flip, not the day (if I'm reading the question correctly).

Beforehand she expects a fair coin comes up heads 50%, and after the flip she receives no information that would differentiate heads from tails, so her credence should still be 0.5.

It's a little different if you reward her for being more right, though; in that case, depending on her risk aversion she should give a higher credence to tails, to pick up a larger reward twice if tails came up.

[drmwc]

Speedchase, she's being asked about the coin flip, not the day (if I'm reading the question correctly).

Beforehand she expects a fair coin comes up heads 50%, and after the flip she receives no information that would differentiate heads from tails, so her credence should still be 0.5.

It's a little different if you reward her for being more right, though; in that case, depending on her risk aversion she should give a higher credence to tails, to pick up a larger reward twice if tails came up.

As a Bayesian, I personally agree with your answer. However, frequentists disagree.

They argue that if the experiment was run 1,000,000 times, then she is asked the question 500,000 times when the coin was heads, and 1,000,000 times when the coin is tails. Hence the odds on the coin being heads is 1/3.

A related question is if, during the interviews, Sleeping Beauty is asked what odds she would require to bet on the coin being tails .

[Bill Spight]

An interesting variant is the Sleeping Beauty Problem.

On Sunday, Sleeping Beauty is given a drug to make her sleep. Immediately after she falls asleep, a fair coin is tossed.
1) If it is heads, she is woken on Monday. She is interviewed on Monday. and the experiment ends.
2) If it tails, she is also woken on Monday and interviewed. However, she is then put back to sleep and given an amnesia-inducing drug which causes her to forget the previou interview. She is woken on Tuesday and interviewed again. Then the experiment ends.

At each interview, she is asked "What is your credence to the propoosition that the coin landed heads?"

The problem is to determine the best answer. This is suprisingly non-trivial.

Suppose that, instead of an interview, she is offered an even money bet for 100 yen where she bets on heads. On Monday she breaks even, on Tuesday she loses. So she should not take the bet.

But if she is offered 2:1 odds, the bet is fair. If the coin came up heads, she wins 200 yen on Monday. If the coin came up tails, she loses 100 yen on Monday and 100 yen on Tuesday.

I do not see a frequentist vs. Bayesian conflict here. The Bayesian question for Sleeping Beauty is what are the odds that I am being interviewed (offered a bet) given the result of the coin toss.

[perceval]

what kind of psycho prince would do that to sleeping beauty instead of kissing her ?
answer: geeky prince of course

[jts]

@drwmc,

Eh, by asking her about her credence, don't you prejudge that we're asking the Bayesian question? If you asked her, "out of a hundred thousand trials, how many time would we flip heads?" or "... how many times would we interview you after flipping heads?" those would be well-defined questions with frequentist answers.

[drmwc]

I think she should say it is monday. There is a 75% chance she is correct

|,,/\
|,m,/\
|,,m,,t
each branch is 50/50.

edit: whitespace fail. give me a sec to figure it out.
edit2: commas are whitespace

I'll come back to the probability of it being a speicfic day. It raises some intersting points.

Suppose that, instead of an interview, she is offered an even money bet for 100 yen where she bets on heads. On Monday she breaks even, on Tuesday she loses. So she should not take the bet.

But if she is offered 2:1 odds, the bet is fair. If the coin came up heads, she wins 200 yen on Monday. If the coin came up tails, she loses 100 yen on Monday and 100 yen on Tuesday.

I do not see a frequentist vs. Bayesian conflict here. The Bayesian question for Sleeping Beauty is what are the odds that I am being interviewed (offered a bet) given the result of the coin toss.

Gambling odds are not the same as probability. The two are related, but they are not identical. Suppose a fair coin is tossed, and you bet on the coin being heads. If it is heads, you bet once. However, if it is tails you are required to bet the same amount twice. What odds should you accept, and what is the probability of the coin being heads?

This is a similar set-up as in the Sleeping Beauty Problem. The fact the gambling odds differ from the probability reflec the number of bets going in.

@drwmc,

Eh, by asking her about her credence, don't you prejudge that we're asking the Bayesian question? If you asked her, "out of a hundred thousand trials, how many time would we flip heads?" or "... how many times would we interview you after flipping heads?" those would be well-defined questions with frequentist answers.

It is possibly designed to do that. However, frequentists often insist that even with this set-up, the probability is 1/3.

[Bill Spight]

I think she should say it is monday. There is a 75% chance she is correct

|,,/\
|,m,/\
|,,m,,t
each branch is 50/50.

edit: whitespace fail. give me a sec to figure it out.
edit2: commas are whitespace

I'll come back to the probability of it being a speicfic day. It raises some intersting points.

Suppose that, instead of an interview, she is offered an even money bet for 100 yen where she bets on heads. On Monday she breaks even, on Tuesday she loses. So she should not take the bet.

But if she is offered 2:1 odds, the bet is fair. If the coin came up heads, she wins 200 yen on Monday. If the coin came up tails, she loses 100 yen on Monday and 100 yen on Tuesday.

I do not see a frequentist vs. Bayesian conflict here. The Bayesian question for Sleeping Beauty is what are the odds that I am being interviewed (offered a bet) given the result of the coin toss.

Gambling odds are not the same as probability. The two are related, but they are not identical. Suppose a fair coin is tossed, and you bet on the coin being heads. If it is heads, you bet once. However, if it is tails you are required to bet the same amount twice. What odds should you accept, and what is the probability of the coin being heads?

This is a similar set-up as in the Sleeping Beauty Problem. The fact the gambling odds differ from the probability reflec the number of bets going in.

The correct odds are 50:50. Half the time I win X, one quarter of the time I break even, and one quarter of the time I lose 2X. X/2 - 2X/4 = 0.

Odds aside, a key difference to the Sleeping Beauty Problem is what Sleeping Beauty knows. Her bets are not independent, and she knows that. That knowledge affects the probability, according to Bayes Theorem.

[drmwc]

I may have phrased the alternative badly.

We bet on a fair coin. If it's tails, you win £1. I it's heads, I win £2 since you are forced to place the wager twice.

What odds do you require?