Suppose that we rearrange the order of operations. First, we eliminate the WW box. Then, we choose one of the other boxes. Question 1. What is the probability that the box we chose has two Black stones in it? Then we draw a stone from the box, and it is a Black stone. Question 2. Now what is the probability that the box had two Black stones in it?

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

Visualize whirled peas.

Everything with love. Stay safe.

Monty Hall was interviewed after he retired, and the interviewer gave him the Monty Hall problem as it is usually posed. (That's important.) Monty had never heard of the problem, but he gave the correct answer. He said that he could manipulate contestants to make the wrong guess.

To get the standard answer to the Monty Hall problem you have to restrict Monty's choices.

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

Visualize whirled peas.

Everything with love. Stay safe.

perceval,

I borrow the "The little green woman" from Monty Hall Problem:

(0) 3 bowls: ( ), ( ), and ( ).
(1) Suppose you randomly pick a bowl, then take one of the 2 stones out from the bowl.
(2) This stone turns out to be .
(3) What is the probability that the second stone in the bowl is also ?

Between (2) and (3), I now insert two important steps:
(2a) We get rid of the 2 un-chosen bowls;
(2b) a little green alien appears from a UFO.
She did not see (0), (1), or (2). In fact, we don't even tell her anything about what's inside the (singular) bowl.
From her perspective, this is what she sees:

(3) We take something out from the bowl (OUR perspetive: the same bowl in (1); HER perspective: the ONLY bowl).
(Remember, she has no knowledge of what's inside the bowl -- we could've picked a cow out of it, as far as she knew.)

This is the crux of the matter: if we repeat this thought experiment 1 million times --
that is, we invite the same little green alien to do this 1 million times --
for the little green alien, she will see approx. half a million times; and approx. half a million times.
If we ask her, what is the probability of seeing come out of the bowl based entirely on what she has seen,
she would say approx. 50%.

So perceval, my question is, in your original wording of (3), what do you mean by "what IS the probability..." ?

(And clearly, this is NOT what TJ's code simulated -- the perspective of the little green alien.)

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?

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

Visualize whirled peas.

Everything with love. Stay safe.

Bill, exactly -- this is also my question (confusion): P = X/Y. In the original wording of " (3) What is the probability of...? ",
I want to know what is X and what is Y for that wording of (3) ?

( Because, from the perspective of the little green alien, X ~= half a million, and Y = 1 million exactly; so for her P ~= 50%. )

Actually, the fact that the stones are distinct matters. Probability with fermions, which are distinct, and bosons, which are not, is different. Coins and go stones are distinct.

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

Visualize whirled peas.

Everything with love. Stay safe.

Bill, for this question, clearly we must assume the stones and coins OF THE SAME COLOR are like bosons, completely indistinguishable from one another.

This gets into what is called pragmatics. Humans are good communicators, and one reason for that is that we do not state everything in each sentence. The pragmatic meaning of "probability" in (3) is the "probability, given (0), (1), and (2)".

That is why the little green alien misinterprets (3). It was not meant for her. She is unaware of the context.

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

Visualize whirled peas.

Everything with love. Stay safe.

I see what you mean. Suppose that one of the Black stones is chipped, and we do not know in which box it is. Then if the two boxes contain different number of stones, and we pick the chipped stone, the probability of each box is proportional to the number of Black stones in the box.

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

Visualize whirled peas.

Everything with love. Stay safe.

Yes, I see; this is key:

If we start counting the 1 million from after (2) -- her perspective -- X will be ~= half a million.
If we start counting the 1 million at (0) -- our perspective -- X will not be ~= half a million. (TJ's simulation).

I now rephrase the question, completely removing "pragmatics" (for my own sanity and clarity):

(0) 3 bowls: ( ), ( ), and ( ).
(0a) We start our counter X = 0.
(1) Suppose you randomly pick a bowl, then take one of the 2 stones out from the bowl.
(2) This stone turns out to be .
(3a) IF second stone in the bowl is also , we increment X.
(4) Go back to (1) -- NOT (0) -- repeat this 1 million times. What is the approx. value of X after 1 million iterations ?

Now I see it clearly -- (2) SUBSTANTIALLY lowers X from ~= half a million.

(I still think the original wording is not as good as in the Monty Hall problem --
there, they ask "should the contestant switch" instead of "what is the probability of..." ?
I think this way of asking "should the contestant switch" is less ambiguous than the "pragmatics" of "what is the probability of...". )

( Sanity check for myself: if WE do the experiment 1 million times, the little green alien
will not be invited approx. 333,333 times. This is the part SHE does not know.
Conversely, for HER to be invited 1 million times, WE would have to repeat 1.5 million times. )

OK. Here is an similar problem. Suppose, first, that the number of boys born is equal to the number of girls born. Suppose also that you know that I have two children, but you do not know their genders. One afternoon you (randomly) run into my wife, along with one of my children, who is a girl. What is the probability that my other child is also a girl?

Edit: To be clear, my wife is equally likely to be with one child as the other.

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

Visualize whirled peas.

Everything with love. Stay safe.

Ehee i was going to ask that one next, that one drives me crazy . Its the same, except with children and family we instinctively rebel against the idea of retrying until we get the one we want... to be more clear : to erase the ambiguity in the bowl + stones case, , we explain with an algorithm to pick the stones, and state that we only accept to count a subset of the experiments: the one where indeed we drew a Black stone.
With your 2 children we refuse to make this experience of thought as you have 2 children and you are stuck with them

i am still not clear but that one gives me headaches

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In theory, there is no difference between theory and practice. In practice, there is.

There are indeed three remaining options, but the first of the three is twice as likely as the other two. The probability is therefore 50/50.

edit: ...actually, looking at it again I think a case can be made for 60%. But I still think it's 50%.

edit 2: The 60% argument goes as follows. Think about it this way. Let's say there's four families: one with two girls, another with two boys, and two families with one of each. You've met one of the girls, so the all-boy family is eliminated. So there's six kids left altogether, four girls and two boys. You're talking to one of them, a girl. So the "missing" five must be three girls and two boys.

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Poka King of the south east.

there is a whole page on it on wikipedia (spoiler, obviously):

http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

_________________
In theory, there is no difference between theory and practice. In practice, there is.

How about a variation on the original?

We have 3 boxes,
the first contains an unknown number of black stones
the second contains an unknown number of black and white stones,
the third contains an unknown number of white stones

We pick a box, pull a stone and it's black,

What is the probability that if we pull another stone from the box, that it will also be black?

_________________
Tactics yes, Tact no...

Box 1:
Box 2:
Box 3:

If the stone you picked is ____ then the other stone will be _____:

If then
If then
If then
if then
if then
if then

You know you have in your hand a white stone, you do not know whether the white stone is 1, 2, or 3, therefore one of the following is true.

If then
If then
If then


2 in 3 chance to have two white stones.

this is how i would explain it :]

well if we do not know anything its hard to compute something.

With N B stones in the all black box, M Black + K white in the B and W box, and L W sotnes in the W bowl -then we can compute something:

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In theory, there is no difference between theory and practice. In practice, there is.

Bill,