How not to get published by Jonny Blamey
Some time ago I promised to send my two envelope solution to a journal and post the referees reports on this blog. Here’s the whole story
Last May I gave a talk at a conference and someone told me afterwards that he thought my thesis would have some bearing on the Two Envelope paradox. I saw the solution straight away. Using Ramsey’s measure for degree of belief, the paradox becomes a simple disagreement in probability assignments. I was looking for a problem with which to demonstrate my new techniques in decision theory and here was one that fell into my lap.
I put my solution on the blog and received a huge amount of negative comments. As one of the labels was “rigid designator” I got one comment saying that I obviously didn’t understand the two envelope paradox since “rigid designator” was irrelevant to the problem. I got some other comments along the lines that I didn’t understand probability, maths, philosophy, the principle of indifference, Bayes theorem, the nature of argument and anything at all. My lack of knowledge is on Socratic proportions.
At another conference I met someone who had published work on the two envelope paradox. Friends with him was Sorin Bangu who gave a talk on the principle of indifference. I asked Dr Bangu to post on Bloggin the question. By November an article came out in Mind on the two envelope paradox using the concept of rigid designation by a pair of scholars from Dr Bangu’s university. Does this make my use of rigid designation retro actively valid?
Meanwhile someone on the blog challenged me to send my solution to a journal, specifically the BJPS and when it was rejected, to publish the referees reports on the blog. Instead of a flat rejection the BJPS asked me to rewrite in response to the referees reports. The referees reports said my solution was an ingenious contribution to the literature, but had a flaw. If I could respond to the comments they would consider publishing.
This I did, but in response to something that David Papineau said I was working on trying to express the solution in more natural terms. My Dad came over to visit and we spent an afternoon drawing graphs of possible pairs of envelopes, pigs, people etc. who fit the description. We found that the more densely packed and widely spread the envelopes, the more closely the graph fits 1/x for the sum of the pair and 1/SQRx for the second envelope given the first. If you normalise this then for any N you can show that the probability that swapping will double your money is 1/3. What no one in the literature had grasped was that the envelopes necessarily come in pairs. Once you accept this fact it is possible to find the probability distribution. It was assumed in the literature that you can have any prior probability distribution you like, which is absurd. Why should you act on an arbitrary probability distribution? And how can the probability that an envelope contains 2x P(2x) be greater than P(4x) + P(x)? And if you accept two end points below and above which there can be no envelope, then an equal distribution is impossible. P(2x, x) = 1/3 and P(1/2x, x) = 2/3 is the only pair of conditionals that works.
I tried to put this into easy accessible language and resubmitted to the BJPS. They were late coming back to me, saying that they had to wait ‘til after Christmas for one referee. Finally on my birthday they sent me a rejection with only an extract from one referee’s report.
Begin Quoted Text----------------------------------------------------------Sorry to be a bit slow on this. I have read the paper, and don't think it's really good enough to recommend. There is an interesting central idea, but (a) the paper has too much irrelevant detail in the first six pages and (b) I don't think the main contention of the paper is strongly enough supported. There are also some minor (but distracting) inaccuracies.----------------------------------------------------------End Quoted Text----------------------------------------------------------
I asked them to sent me the full reports, and they said they would, but it turned out that one referees report was lost and the other referee did not want his comments exposed to judgement.
Meanwhile I sent off a really accessible version for the Jacobsen essay prize and did get to see some of the examiners reports from that one. So I’ll paste those below as a substitute.
EXAMINERS REPORTS
3. Envelope stall.
This short essay simply presents a well-known ‘paradox’. No references are given and there is nothing original said about the issues that it raises.
I expected to warm most to essay 3. But, aside from the fact that it makes no reference at all to the literature, and the fact that the answer surely (at least partly) lies in the fact that no one could really think, in anything like a real situation, that there is a uniform prior over all possible pay-offs (s/he assumes this away), I just couldn't see what basis there was for what seems to be the crucial premise namely the distinction between probabilistic and 'financial' terms (p.3).
3. Sam’s Envelope Stall
Lively, but the suggested solution is not cogent. And this is a topic on which there is a huge and sophisticated literature which the author simply ignores.
END
I hope this story is useful to those researchers hoping to get published.
Last May I gave a talk at a conference and someone told me afterwards that he thought my thesis would have some bearing on the Two Envelope paradox. I saw the solution straight away. Using Ramsey’s measure for degree of belief, the paradox becomes a simple disagreement in probability assignments. I was looking for a problem with which to demonstrate my new techniques in decision theory and here was one that fell into my lap.
I put my solution on the blog and received a huge amount of negative comments. As one of the labels was “rigid designator” I got one comment saying that I obviously didn’t understand the two envelope paradox since “rigid designator” was irrelevant to the problem. I got some other comments along the lines that I didn’t understand probability, maths, philosophy, the principle of indifference, Bayes theorem, the nature of argument and anything at all. My lack of knowledge is on Socratic proportions.
At another conference I met someone who had published work on the two envelope paradox. Friends with him was Sorin Bangu who gave a talk on the principle of indifference. I asked Dr Bangu to post on Bloggin the question. By November an article came out in Mind on the two envelope paradox using the concept of rigid designation by a pair of scholars from Dr Bangu’s university. Does this make my use of rigid designation retro actively valid?
Meanwhile someone on the blog challenged me to send my solution to a journal, specifically the BJPS and when it was rejected, to publish the referees reports on the blog. Instead of a flat rejection the BJPS asked me to rewrite in response to the referees reports. The referees reports said my solution was an ingenious contribution to the literature, but had a flaw. If I could respond to the comments they would consider publishing.
This I did, but in response to something that David Papineau said I was working on trying to express the solution in more natural terms. My Dad came over to visit and we spent an afternoon drawing graphs of possible pairs of envelopes, pigs, people etc. who fit the description. We found that the more densely packed and widely spread the envelopes, the more closely the graph fits 1/x for the sum of the pair and 1/SQRx for the second envelope given the first. If you normalise this then for any N you can show that the probability that swapping will double your money is 1/3. What no one in the literature had grasped was that the envelopes necessarily come in pairs. Once you accept this fact it is possible to find the probability distribution. It was assumed in the literature that you can have any prior probability distribution you like, which is absurd. Why should you act on an arbitrary probability distribution? And how can the probability that an envelope contains 2x P(2x) be greater than P(4x) + P(x)? And if you accept two end points below and above which there can be no envelope, then an equal distribution is impossible. P(2x, x) = 1/3 and P(1/2x, x) = 2/3 is the only pair of conditionals that works.
I tried to put this into easy accessible language and resubmitted to the BJPS. They were late coming back to me, saying that they had to wait ‘til after Christmas for one referee. Finally on my birthday they sent me a rejection with only an extract from one referee’s report.
Begin Quoted Text----------------------------------------------------------Sorry to be a bit slow on this. I have read the paper, and don't think it's really good enough to recommend. There is an interesting central idea, but (a) the paper has too much irrelevant detail in the first six pages and (b) I don't think the main contention of the paper is strongly enough supported. There are also some minor (but distracting) inaccuracies.----------------------------------------------------------End Quoted Text----------------------------------------------------------
I asked them to sent me the full reports, and they said they would, but it turned out that one referees report was lost and the other referee did not want his comments exposed to judgement.
Meanwhile I sent off a really accessible version for the Jacobsen essay prize and did get to see some of the examiners reports from that one. So I’ll paste those below as a substitute.
EXAMINERS REPORTS
3. Envelope stall.
This short essay simply presents a well-known ‘paradox’. No references are given and there is nothing original said about the issues that it raises.
I expected to warm most to essay 3. But, aside from the fact that it makes no reference at all to the literature, and the fact that the answer surely (at least partly) lies in the fact that no one could really think, in anything like a real situation, that there is a uniform prior over all possible pay-offs (s/he assumes this away), I just couldn't see what basis there was for what seems to be the crucial premise namely the distinction between probabilistic and 'financial' terms (p.3).
3. Sam’s Envelope Stall
Lively, but the suggested solution is not cogent. And this is a topic on which there is a huge and sophisticated literature which the author simply ignores.
END
I hope this story is useful to those researchers hoping to get published.
Labels: literature, publishing, references
posted by bloggin the Question at 9:45 AM
10 Comments:
Ho hum. Better luck next time Johnny. I didn't read the original post, but you are absolutely right that the problem is about rigid designation. In particular, about the assignment to variables which needs to remain constant.
Hi JB, I sympathise with your reaction to referees' words (at least they said it was interesting though); it seems to be pretty random what such people say (unless you write unoriginally). On the other hand, I think (well, guess that) your solution is false... Surely the probability of your having picked the envelope with the smaller amount in it was 50%, so the probability that your money would, if you swapped, be doubled is 50%... So why not solve the problem by noting (who's solution is this?) that only the smaller amounts get doubled, and only the larger amounts get halved?
Dear Who,
That's a brilliant way of putting it, I guess that is my solution, at any rate it is *the* solution. (Only the smaller amounts get doubled and only the larger amounts get halved). What is the conclusion in terms of action? Indifference! What degree of belief does someone have who is indifferent that they will double their money? 1/3. So what degree of belief should you have that you will double your money? 1/3. If that is how we are to interpret probability then the paradox is solved.
The argument you give that my solution is false:
"Surely the probability of your having picked the envelope with the smaller amount in it was 50%, so the probability that your money would, if you swapped, be doubled is 50%"
Is a tough nut to crack, but I have explained in about 15 different ways why this reasoning is erroneous, and there are plenty of published papers that also explain why it is (the most recent of which, in Mind Nov 2007, pretty well used ideas I'd posted on this blog). The point most simply put is: Fine, if you want to say that you have a 50% chance of doubling your money, then swap, and 50% of the time you will double your money, but over a large number of trials, you'll make the same money either way. i.e. you won't double your money at probability 0.5 and halve it at probability 0.5. So if doubling your money is want you want to do, don't play the two envelope game. You'll only double 1/3 of your money, and halve 2/3 of your money. The probability of doubling your money then is *really* only a 1/3.
Here is another scientific way of looking at it. How can we examine the problem in hindsight so we can test the hypothesis over a number of trials? Easy, just shuffle the envelopes and pick again. Record the results in two columns, 1. the amount in the first envelope, 2. The amount in the second envelope. The two hypothesise are
H1 P(double your winnings on swapping) = 1/2 P(halving your winnings on swapping) = 1/2
H2 (Jonny's hypothesis) P(double your winnings on swapping = 1/3, p(halve your winnings on swapping) = 2/3.
H1 predicts that, with an increasing number of trials, the sum of column 1 divided by the sum of column 2 = 0.8
H2 predicts that the sum of column 1 divided by the sum of column 2 = 1.
Try it! I think it is pretty obvious that H2 is going to be better at predicting the results of a large number of trials.
Hi again. I'm afraid I've just been trying Lemsip this (long) weekend; but I looked at what would happen with coin tosses, doubling on heads and halving on tails (and repeating with whatever I end up with) and the halvings do seem to cancel out the doublings.
I'm puzzled by the last examiner's reference to a huge sophisticated literature, though; is that supposed to be a philosophically good thing?
We know that someone had put 3 (of some units) into the 2 envelopes. The amount in the envelope you picked is X, say (and the amount in the other is Y). So (X + Y = 3, and) either (50%) X = 1 (= Y/2), or else (50%) X = 2 (= 2Y).
After you picked an envelope, either X was definitely (whether you looked at the amount or not) 1, or else X was definitely 2.
Suppose you got 1. Call that possible world A. You would definitely gain X = 1 on switching, but if you would have lost you would have been in possible world B, with 2 in your envelope, and you would have lost 1/2 of that.
Now suppose you're in B. Then you would definitely lose X/2 = 1 on switching, but if you would have gained you would have been in A, with 1 in your envelope, and you would have gained that amount.
The problem is to calculate your expected gain on switching. It is 50/50 which world you are in. If you are in A you gain X = 1, and if you are in B you lose X/2 = 1. So your expected gain is 50% of X or 1 (or Y/2) + 50% of X/2 or 1 (or Y) = 0.
For my sins, whilst I can just about see the relevence of your rigid designation (although philosophy of language is very obscure to me) I fail to see how any huge literature on that last equation could be sophisticated in a good way!
Hello, enigamn. That's great, its not the experiment I had in mind, (I don't think) But of course, even if you continued swapping at 50% chance either way, over a long period of trials you going to end up with what you started with. This is the reality underlying the two envelope paradox.
The reason why "rigid designators" is needed is because we have some algebraic reference problems in the formation of the paradoxical thinking.
Suppose you choose an envelope and it contains £10. According to your original post P(£10 = the highest amount) = 1/2. But now suppose £10 is the highest amount, (the other envelope contains £5), One ought to be able to substitue co refering terms. So P(£10 = £10) = 1/2. Sheer nonsense. No one, what ever their interpretation of probability has the probability of a tautology at less than 1. So what is the probability that £10 is the highest amount? There is more than one way to interpret this. If the highest amount is a non rigid designator, then it is possible that £10 is not the highest amount, but if it is a rigid designator, then it is impossible that £10 is not the highest amount.
We have the erroneous reasoning:
1. P(I picked the envelope with the highest amount) = 1/2
2. I picked the envelope with £10.
Therefore
3 P(£10 = the highest amount) = 1/2.
This argument is invalid because "the highest amount" is a rigid designator in 1 and a non rigid designator in 2.
Without showing this argument to be invalid, the paradox retains it's force.
Yeah, upon reflection I now find this paradox fairly deep! For what it's worth I've noted my failure to understand it here.
I'm trying to contact Jonny Blamey. If he would be willing to write me back (and perhaps delete this comment), I could give him specifics about why I want to get in touch with him
Sounds intriguing Keith, but I don't know how to write back to you. Give me your email address.
Curiosity killed the cat. This is Jonny Blamey, what do you want with me Keith?
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