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

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.

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

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@drwmc,

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.

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

Visualize whirled peas.

Everything with love. Stay safe.

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?

The second bet is a sure loss for me, and we both know that. To ask what odds I require is ridiculous. You are forcing me to give 2:1 odds on a 50:50 bet. (Or you are forcing me to make a sure loss bet if the coin comes up tails. Either way you look at it, I am the one giving odds.)

Edit: To be clear. I want 50:50 odds on the bet as a whole, instead of my laying 2:1. Or, if we are separating the bets, I want to bet 0 on the sure loss bet, with 50:50 odds on the coin toss bet.

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

Visualize whirled peas.

Everything with love. Stay safe.

I have a question about the first puzzle. Why have a box with two silver coins? It doesn't add anything right?

It does. Each of the 3 boxes matters. See Post #50.

Yeah what the hell was that? English dude english

That was no higher than 9th grade English, and between 4th and 9th grade math,
depending on which country's math education.
"increment" here meant "add 1 to"; "iterations" here meant "repetitions."

Let's label the 3 bowls:
A:
B:
C:

If we remove B, then the first bowl randomly picked has a 100% chance of it being ( either A or C ).
If we keep B, then the first bowl randomly picked has only a 2/3 chance of it being ( either A or C ).

So whether B is there or not affects the outcome of the first bowl picked, which also affects the outcome of the second stone.

in fact here removing the bowl WILL give the same result:


applying Bayes therorem on the 3 bowl problem:
P(we picked bowl A | the extracted stone was ) = P(we picked a stone | A).P(A)/P(we picked a stone) = 1. 1/3/(1/2) = 2/3.
because: P(A)=1/3 (obvious)
P(we picked a stone | A) = 1 (A contains only B stones)
P(we picked a stone) = 1*P(A)+1/2*P(B)+0.P(C)=1/3+1/2*1/3=3/6=1/2
Now with only bowls A and B:

P(we picked bowl A | the extracted stone was ) = P( | A).P(A)/P( ) = 1. 1/2/(3/4) = 2/3. !!!
because:
P(A)=1/2 (obvious)
P(we picked a stone | A) = 1 (A contains only B stones)
P(we picked a stone) = 0.5*P(A)+1/2*P(B)+0.P(C)=1/2+1/4=3/4

different computation, same result

If we try to change the numbe of stones in each bowl, we can see what is needed for the 2 problems to give the same answer:

we need tha P( in the 3 bowl problem)/P( in the 2 bowl problem)=2/3. (ie the proba of picking a stone must cancel out the change of proba in picking bowl A whne we go from 2 bowl to 3 bowls)

This is true as long as there are as much stones in the A and the B bowl, and if the bowl B contains as many and stones.

The number of stones in the bowl C is irrelevant (which make sense as if we pick a stone we know we did not pick it)!!

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

As perceval already pointed out, it doesn't. What it affects is the probability of picking a white (or black stone) initially, but not the probability being asked, which was (roughly) "how likely is it that the second stone in the same bowl will be of the same color". If you first pick a white stone, it doesn't matter how many bowls with only black stones there were to begin with (or, if you pick a black stone, it doesn't matter how many bowls with only white stones there were to begin with).

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One thing that surprised me about the Sleeping Beauty problem is the claim that the Bayesians think that Sleeping Beauty should believe that the probability that the coin came up heads is 1/2 , while the frequentists think that she should believe that the probability is 1/3. I don't believe that there should be any difference, but if there was one, I would have thought that it would be the other way around. After all, the frequency of heads is 1/2, and the application of Bayes Theorem gives you a probability of 1/3. Frankly, I am gobsmacked.

Here is a simplification of the problem that illustrates the main point. A fair coin is tossed. If it comes up tails, Sleeping Beauty is asked what is the probability that it came up heads. If it comes up heads, she is not asked. Sleeping Beauty knows all of this. What should her answer be?

Well, duh! This is not even worth hiding. It is hardly worth stating. The probability that the coin came up heads is 0.

But why is that? The key answer is that for Sleeping Beauty the fact that she is being asked what the probability is is evidence about the probability. (That is unusual in real life, which is part of what makes this a problem. ) By Bayes Theorem the posterior odds that the coin came up heads versus tails is the product of the prior odds of the same times the odds that Sleeping Beauty will be asked about the probability if the coin came up heads versus that she will be asked if the coin came up tails. The prior odds are 1, the odds about being asked are 0, and the posterior odds are 0. That makes the probability 0. Bingo!

In the more complex problem Sleeping Beauty's amnesia guarantees that the conditions of each question are the same. (Again, not like real life. ) We condition on what she knows, not on what has happened. Thus Sleeping Beauty's answer should be the same for each question. She is asked once if the coin came up heads and twice if it came up tails. So the odds that she will be asked if the coin came up heads versus if it came up tails are 1:2. By Bayes Theorem, the posterior odds that the coin came up heads versus tails are also 1:2, and so the probability that the coin came up heads is 1/3. Bingo, again.

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

Visualize whirled peas.

Everything with love. Stay safe.

perceval and TJ, thanks --

Two simulations:

Simulation 1:
(0) 3 bowls: ( ), ( ), and ( ).
(0a) Set X = 0.
(1) 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 , increment X.
(4) Go back to (1); repeat this 1 million times. What is the approx. value of X after 1 million iterations ?

Simulation 2:
(0) 102 bowls: ( ) and ( ); plus 100 ( ).
(0a) Set Y = 0.
(1) 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 , increment Y.
(4) Go back to (1); repeat this 1 million times. What is the approx. value of Y after 1 million iterations ?

Re: Bill's "pragmatics" -- I apologize because I'm easily confused by the wording of "What's the probability of..." --
could someone run the above two (simple) simulations?
I'd like to know the values of X and Y. Thanks.

I'd also like to know, in the original question: " What is the probability that the second stone in the bowl is also ? ",
is it asking what is the value of (X/N) as N tends to infinity ? (where N is the number of interations.)

And, to understand BigBadBuu's quesiton, will (X = Y) as N tends to infinity ?

Do you mean for (2) If this stone turns out to be .

In that case you are asking how often you pick the bowls with .

Or do you want to ask, given that the chosen stone is , how often is the other stone ? If so, you need to count how often the chosen stone is .

E. g.:

(0) 3 bowls: ( ), ( ), and ( ).
(0a) Set X = 0.
(0b) Set Z = 0.
(1) Randomly pick a bowl, then take one of the 2 stones out from the bowl.
(2) If this stone turns out to be , increment Z.
(3a) IF second stone in the bowl is also , increment X.
(4) Go back to (1); repeat this until Z = 1 million. What is the approx. value of X?



Same deal. Do you want to know how often you pick the bowl? Or do you want to know how often the second stone is , given that the chosen stone is ?

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

Visualize whirled peas.

Everything with love. Stay safe.