Life In 19x19

HermanHiddema wrote:

Bantari wrote: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.

Bantari wrote:

HermanHiddema wrote:

Bantari wrote: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 wrote:

Bantari wrote:

HermanHiddema wrote:
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?

Bantari wrote:

Bill Spight wrote:

Bantari wrote: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.

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?

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 ?

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

jts wrote: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 wrote: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 wrote:

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

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

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

Bill Spight wrote: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.

jts wrote:@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 wrote:

speedchase wrote:

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.

Bill Spight wrote: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.