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.
Re: well known proba problem
Posted: Sat Feb 16, 2013 8:58 am
by speedchase
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
Re: well known proba problem
Posted: Sat Feb 16, 2013 9:43 am
by 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.
Re: well known proba problem
Posted: Sat Feb 16, 2013 10:24 am
by drmwc
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.
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 .
Re: well known proba problem
Posted: Sat Feb 16, 2013 12:35 pm
by Bill Spight
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.
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.
Re: well known proba problem
Posted: Sat Feb 16, 2013 2:05 pm
by perceval
what kind of psycho prince would do that to sleeping beauty instead of kissing her ?
answer: geeky prince of course
Re: well known proba problem
Posted: Sat Feb 16, 2013 5:45 pm
by 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.
Re: well known proba problem
Posted: Sun Feb 17, 2013 5:07 am
by drmwc
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.
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.
It is possibly designed to do that. However, frequentists often insist that even with this set-up, the probability is 1/3.
Re: well known proba problem
Posted: Sun Feb 17, 2013 5:38 am
by Bill Spight
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.
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.
Re: well known proba problem
Posted: Sun Feb 17, 2013 10:58 am
by 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?
Re: well known proba problem
Posted: Sun Feb 17, 2013 11:50 am
by Bill Spight
drmwc wrote: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.
Re: well known proba problem
Posted: Sun Feb 17, 2013 2:56 pm
by BigBadBuu
I have a question about the first puzzle. Why have a box with two silver coins? It doesn't add anything right?
Posted: Sun Feb 17, 2013 3:16 pm
by EdLee
BigBadBuu wrote: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.
Re:
Posted: Sun Feb 17, 2013 4:54 pm
by BigBadBuu
EdLee wrote:
BigBadBuu wrote: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
Posted: Sun Feb 17, 2013 6:51 pm
by EdLee
BigBadBuu wrote: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.
