Friday, September 19, 2014

Sleeping beauty problem III

Previous post in this series, next post in series

I am writing a third post about the sleeping beauty problem partly because I have seen Sean Carmody's post  (26th August) in which he describes the problem and expresses his uncertainty about it, and promises an opinion in a future post. We are waiting, Sean, for your ideas on the matter!

$\bf Note\; about\; comments\; on\; this\; blog $
I have never taken very seriously responses to posts at this blog because (apart from spam) there have been very few. After the sleeping beauty posts I did receive quite a few comments as a result of Lubos Motl's reposting  on his blog. I allowed several, mostly anonymous, but then I decided I prefer to have some idea of the identity of persons with whom I am discussing. I made it a bit harder to respond to posts. I may reverse that policy since the blog has returned to normal, however in general I will not accept anonymous responses. On popular blogs one sees hundreds of anonymous comments of very variable quality. It is not what I have in mind for this blog. Of course I am interested in serious responses which may persuade me to change my opinions.

$\bf Remarks\;  about\; the \; sleeping\; beauty\; problem$
I  want to make a couple of remarks about the sleeping beauty problem. Most people who take the thirder position are not talking about probability but rather the average amount of time spent in a state, or similar concepts such as how much one could win by betting on a state. This is certainly not the probability of arriving in a state at a specified time or times, and in my opinion is not the question under consideration.
The other curious feature of discussion of this clearly stated problem is that everyone wants to consider modifications of the problem in trying to persuade their opponents. I am going to commit a similar sin in this post.

But first a comment on a strange aspect of the thirder position. Conditional probability classically is about restricting the possible (atomic) events or actions, and results in some events having zero probability, while the remaining events have $\bf increased$ probability. Instead, in the thirder "solution" an action which has probability $1/2$ is changed to have a $\bf lesser$ probability (of $1/3$) as a result of a condition. Strange!

$\bf A\; true\; story:\; The \; Tirano-Torino \; Problem$
$NOT\; the\; sleeping\; beauty\; problem$
(names changed to protect the innocent)
One day I was running to catch a train at Milano Centrale, and I had no time to check the platform. I found two trains which were just beginning to depart, one for Torino and one for Tirano, but I did not know which was which. I had to choose which I did with no information. Now it was a dark and stormy night, and the stations are badly signed so after one, two and three stations I was still unable to identify which station we were passing. What was my opinion as to whether I had caught the right train? I had not a clue. I estimated $50\%$.

Now let's take the position that I could bet at each station which train I was on. There are more stations from Milan to Tirano than to Torino so I should bet that I am on the Tirano train.

Or alternatively, as suggested by one commenter, I should imagine 1000 computers simulating the train voyages repeatedly starting at different times, and obviously I would see that more of the trains were on the Tirano line.

Great arguments but I had no idea which train I was on, until of course I recognized a station.



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9:35 AM  
Blogger LuboŇ° Motl said...

Dear Robert, it's a nice exercise. But I wouldn't endorse the answer 50% to the Tirano-Torino problem unconditionally.

If some train leaves very very rarely, then it *is* very very unlikely that it was the train you jumped at, isn't it?

If your seeing of two trains would have proven that both trains - one for Torino and one for Torano or what the names are - were leaving at the moment, then the odds would return to 50%. But if the trains could go anywhere (and perhaps both to the same destination), then the odds are given by the relative frequency of the trains, I think.

I don't see in what sense this is analogous to the Sleeping Beauty, and if it is, why the analogy is the way you constructed it. I would think that the analogy goes in the opposite direction. The thirders want to assign "equal probabilities" to every option that "looks the same qualitatively", and that's exactly what you're doing when you decide that the answer is 50% for Torino and 50% for Torano.

But in reality, the odds are almost always non-equal - they're non-equal if one isn't very lucky and if there is no symmetry argument making them the same. The odds may be anything and the ratio of frequency of trains is what I would pick here.

8:14 AM  
Blogger Robert Walters said...

Sorry Lubos. I knew I was taking a risk to tell a new story but I was persuaded to do so because it was a true story, of course not the sleeping beauty problem. Of course I agree with you about the frequency of trains being relevant to the situation I found myself in.
However the story begins after that. I knew two trains were leaving from from adjacent platforms at the same time, one for Tirano one for Torino but I didn't know which was which. So I tossed a fair coin to decide which of the moving trains to enter. That's where the 50% comes from. I was trying to point out that the lengths of the voyages gave me no information in the first part of the trip until I could recognize a station.
The sleeping beaty has another complication. I am afraid from the posts and replies to Sean Carmody's blog that he is not making clear progress.

11:03 AM  

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