This is false, beacuse:



So under my indexical-free definition of "Mon" and "Tue", P(Mon)=1 and P(Tue)=1/2, hence P(Mon)+P(Tue)=3/2, not 1. Mon and Tue are not mutually exclusive.



I have not switched to frequentist model - my model is Bayesian. My understanding is that a frequentist model defines probability as the limit of the relative frequency of the event as the number of trials tends towards infinity. I am not assuming a large number of trials in this model - my model is valid for 1 trial.

I have slighly abused notation. All probabilities are, to a Bayesian, relative. Let J be SB's knowledge at the start of the experiment:
"A fair coin will be tossed. I will be put to sleep ...[etc.]"

Then the probability of the events Mon, Tue, Heads, Tails are all conditional on J.

As I stated in my earlier post, Mon and Tues are not mutually exclusive. Hence



is false - the sum equals 3/2.



I agree with equations 1-3 above. However, I don't agree with equation 4 since this is only true when Mon and Tue are mutually exclusive. Under my definition of Mon and Tue, they are not mutually exclusive.

I agree with 5,6,7,8 and 9. I diasgree with 10 since it's based on the flawed equation 4. I disagree with 11 since P(Mon)=1 under my model. I suspect to derive this you've subsistited the various probabilites you've derived so far into the equation



Using the correct equation where the sum of the probablities equal 3/2 gives P(Mon)=1 as required.

We know there will be an interview on Monday, so it is not helpful to talk about P(Mon), P(Mon|X) or P(Y|Mon). You are just confusing yourself.


With that definition of Mon, this is like saying P(heads|True)=1/3, which of course you won't get.

As I stated, you still have to deal with cases. So I was not clear. Sorry.

For each possible interview, given "Mon or Tue" and "heads or tails", . . .

No single interview can be on both Monday and Tuesday. That is a given of the problem, and therefore of the solution.

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Sorry to repeat, but I was wrestling with the cat, and only addressed the key point.




OK. "For each possible interview event, SB was interviewed on Monday or SB was interviewed on Tuesday, so P(SB was interviewed on Monday) + P(SB was interviewed on Tuesday) = 1" That is true, because each possible interview occurs on only one day.





Well, when you switched to talking about events instead of what Sleeping Beauty knew, you sounded like a frequentist.



And you are assuming that these probabilities are the same when Beauty is being interviewed, that Beauty's knowledge of being interviewed is irrelevant?

So let me ask you again. With my variation where Beauty wakes up on both days but does not know whether it is Monday or Tuesday, and is interviewed on Tuesday if and only if the coin came up tails, do you think that Beauty should have the same belief on Tuesday, whether she is interviewed or not?

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

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When the inspector is present at the interview, the probability of heads is 1/3. What is the probability when the inspector is not present at the interview?

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

We are agreed that P(heads | Mon) = 1/2. A typo?

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

In only one of three boxes there is a price hidden. You get one choice, you choose one box but the operator always ( you know beforehand ) opens another box to show you it is empty and allows you to change your choice. Do you change?

Very nice problem. At first I got it wrong as most people (%50).

I fully understand the reasoning behind the solution. For example daal's explanation is very simple and nice.




For this explanation you think in terms of stones (3 stones one of which is black).

For the explanation that leads to the wrong answer (%50), you think in terms of bowls (2 possible bowls one of which will surely contains the black stone while the other surely does not).


What I don't see is the following: At which part of the explanation does wrong answer fail? I understand that the two bowls compared are not equal (one contains one stone, the other contains two). But why does it matter? Once you chose one of the two bowls, the stone you will pick up is fully determined. Then why does thinking in terms of bowls lead to a wrong answer ?

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There are a couple of matters to address, I think.

First, determinism. There are people who think this way. If I toss a fair coin, before I toss it the probability that it will come up heads is 50%. Suppose that you toss the coin out of my sight and then without disturbing it put a small box over it, by agreement. Now, there are those who say that the probability that it came up heads is not 50%; it is 100% or 0%, because it is determined, it is no longer a matter of chance. Such people are frequentists, but not all frequentists think that way. Most people are willing to say that the probability is 50%. For us, probability is not about events, but about what we know.

Now, if probability is about what we know, then we need to account for everything that we know that is relevant. (Including irrelevant knowledge does not hurt, except by making our task harder.) The problem with thinking in terms of bowls is not thinking in terms of bowls. It is thinking only in terms of bowls. It is not taking account of all the knowledge that we have. We also know about the proportion of B stones to W stones in the bowls. When we include that information, we get the right answer. The probability of picking a B stone out of the BB bowl is 100%; the probability of picking a B stone out of the BW bowl is 50%. The odds of picking the BB bowl vs. the BW bowl, without including the knowledge about the proportions of stones, is 1:1. When we include that knowledge, the odds change to 2:1. daal just took a shortcut.

As others have pointed out, the shortcut works only because there are the same number of stones in each bowl. If there were, say, 5 black stones in one bowl and only two in the BW bowl, the right answer would be the same, but the shortcut would give the wrong answer.

¿Es claro?

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

i didnt want to beat a dead horse but since the thread is still active, here is a generalisaion of the problem and the calculation:

Suppose you have N bowls, each containing B_i black stones and W_i Whits stones.

You randomly pick a Bowl, then extract a stone. if the stone turns out to be Black, what is the probability of having picked bowl i (1<i<=n)?
ie we want P(i| )
(its the setup propose by shape and aji above, but with a different question)
Note that the original problem is a special case of this one,
with N=3
B_1=0,W_1=2
B_2=1,W_1=1
B_2=0,W_1=2

The question "if we pick a black stone, what the probability that the remaining one is black" is equivalent to looking for P(1| )
here shortcuts mentionned above are no use, so we need to rely on bayse theorem

P(i| ) =P( |i ) P(i)/P( )

Now obviously we have P(i)=1/N (probability of picking bowl i)
P( ) is interesting
we can view the "event "we picked a black stone" as the union of the disjoin events " we picked a black stone form bowl k"

So
P( )= sum (k) P(:black:|k)P(k)
with P(k)=1/N

finally for any bowl P(:black:|k) =B(k)/(B(k)+W(k))
so

P(i| ) =P( |i )/(sum_K P( |k))


with P( |k)) = B(k)/(B(k)+W(k))

Here we find some nice properties and can answer various questiosn that were a bit unclear on the 3 bowl example:
-the sum overall i of the P(i| ) is 1 (of course)
-The result does not depend directly of N so you can add as many bowl with P(:black:|k)= 0 without changing the result ( case of removing the 3rd bowl in hte original problem)- adding B stones to a bowl containing only black stones changes nothing: its goes with the intuition.

- Of course we retrieve our special 3 bowl case:
P(1| ) = 1/(1+0.5+0)=2/3

something that may be counter intuitive at first sight: P(1| )+ P(1|:white: ) is not 1 actually i am a bit puzzled by this



Now we understand what really makes the "stone counting" argument works, and how to correct the formula when it does not:
What matters is B(k)/(B(k)+W(k)), the proportion of black stones in each bowl, NOT the total number of B stones in the bowl alonne.

The simplification in Daal argument above is that, IF every bowl have the same number of stones, ie B(k)+W(k) is the same for each bowl, you can simplify the general formula above and get:

P(i| ) = B(i)/Sum(B(k))
Here you directly get again , for hte original problem:
P(1| )= 2/(2+1+0)= 2/3
i think that when one starts to have some intution about a problem, putting some numbers done in a general case helps refine the intuition

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