= 27d567bc2d48d9f40e9ccc20ea370b15)
timtyler comments on Beauty quips, "I'd shut up and multiply!" - Less Wrong
You are viewing a comment permalink. View the original post to see all comments and the full post content.
= 783df68a0f980790206b9ea87794c5b6)
Loading…= b715d02e85f7b3b4ae65ca3d3e450868)
Subscribe to RSS Feed
Powered by Reddit
= f037147d6e6c911a85753b9abdedda8d)
You are viewing a comment permalink. View the original post to see all comments and the full post content.
Comments (335)
Re: "But if we specify that the money will be put into an account (and she will only be paid one winning) that she can spend after the experiment is over, which is next week, then she will find that 1/2 is the "right" answer"
That seems like a rather bizarre way to interpret: "What is your credence NOW for the proposition that our coin landed heads?" [emphasis added]
NOW. One bet.
Again, consider the scenario where at each awakening we offer a bet where she'd lose $1.50 if heads and win $1 if tails, and we tell her that we will only accept whichever bet she made on the final interview.
If her credence for heads on an awakening, on every awakening (she can't distinguish between awakenings), really was 1/3, she would agree to accept the bet. But we all know accepting the bet would be irrational. Thus, her credence for heads on an awakening is not 1/3.
So: you are debating what:
"What is your credence now for the proposition that our coin landed heads?"
...actually means. Personally, I think your position on that is indefensible.
This would make it clear exactly where the problem lies - if not for the fact that you also appear to be in a complete muddle about how many times Beauty awakens and is interviewed.
We both know what question is being asked. We both know how many times she awakens and is interviewed. I know what subjective probability is (I assume you do too). I showed you my math. I also explained why your ratio of expected frequencies does not correspond to the subjective probability that you think it does.
Does it not concern you even a little that the Wikipedia article you linked to quite clearly says you are wrong and explains why?
I started by reading the wikipedia page. At that point, the 1/3 solution made some sense to me, but I was bothered by the fact that you couldn't derive it from probability laws. I then read articles by Bostrom and Radford. I spent a lot of time working on the problem, etc. Eventually, I figured out precisely why the 1/3 solution is wrong.
Is Wikipedia a stronger authority than me here? Probably. But I know where the argument there fails, so it's not very convincing.
I think we are nearing the end here. Someone just wrote a whole post explaining why the correct answer is 1/3: http://lesswrong.com/lw/28u/conditioning_on_observers/
It's fascinating to me that you won't tell me which probability is wrong, p(H)=1/2, P(monday|H)=1
It's also interesting that you won't defend your answer (other than saying I'm wrong). You are in a situation where the number of trials depends on outcome, but are using an estimator that is valid for independent trials. Show me that yours converges to a probability. Standard theory doesn't hold here.
Probabilities are subjective. From Beauty's POV, if she has just awakened to face an interview, then p(H)=1/3. If she has learned that is Friday and the experiment is over, (but she has not yet been told which side the coin came down), then she updates on that info, and then p(H)=1/2. So, the value of p(H) depends on who is being asked - and on what information they have at the time.
It's the first one - P(H)=1/2 is wrong. Before going any further, we should adopt Jaynes' habit of always labelling the prior knowledge in our probabilities, because there are in fact two probabilities that we care about: P(H|the experiment ran), and P(H|Sleeping Beauty has just been woken). These are 1/2 and 1/3, respectively. The first of these probabilities is given in the problem statement, but the second is what is asked for, and what should be used for calculating expected value in any betting, because any bets made occur twice if the coin was tails.
How can these things be different, P(H|the experiment ran) and P(H|Sleeping Beauty has just been woken)?
Yes, a bet would occur twice if tails, if you set the problem up that way. But the question has to do with her credence at an awakening.
The 1/3 calculation is derived from treating the 3 counts as if they arose from independent draws of a mulitinomial distribution. They are not independent draws. There is 1 degree of freedom, not 2. Thus, the ratio that lead to the 1/3 value is not the probability that people seem to think it is. It's not clear that the ratio is a probability at all.
What's this about a multinomial distribution and degrees of freedom? I calculated P(H|W) as E(occurances of H&&W)/E(occurances of W) = (1/2)/(3/2) = 1/3.
Yes, exactly. That would be a valid probability if these were expected frequencies from independent draws of a multinomial distribution (it would have 2 degrees of freedom). Your ratio of expected values does not result in P(H|W).
It might become clear if you think about it this way. Your expected number of occurrences of W is greater than the largest possible value of occurrences of H&W. You don't have a ratio of number of events to number of independent trials.
Picture a 3 by 1 contingency table, where we have counts in 3 cells: Monday&H, Monday&T, Tuesday&T. Typically, a 3 by 1 contingency table will have 2 degrees of freedom (the count in the 3rd cell is determined by the number of trials and the counts in the other cells). Standard statistical theory says you can estimate the probability for cell one by taking the cell one count and dividing by the total. That's not the situation with the sleeping beauty problem. There is just one degree of freedom. If we know the count the number of coin flips and the count in one of the cells, we know the count in the other two. Standard statistical theory does not apply. The ratio of count for cell one to the total is not the probability for cell one.