tag:blogger.com,1999:blog-3891434218564545511.post6217002615781463922..comments2024-03-28T19:56:42.305-05:00Comments on Alexander Pruss's Blog: Shuffling an infinite deck of cardsAlexander R Prusshttp://www.blogger.com/profile/05989277655934827117noreply@blogger.comBlogger28125tag:blogger.com,1999:blog-3891434218564545511.post-77636233017721561292023-01-31T17:27:02.274-06:002023-01-31T17:27:02.274-06:00Unknown: Since God is omnirational, He cannot do a...Unknown: Since God is omnirational, He cannot do anything that results against reason, like with the infinite lottery paradox. "For maybe God's perfection would not permit him to place a person in a situation where there is a rationality paradox. God is Himself a rational being, rational beings are made in God's image, and to act irrationally is to act in some sense against God. Thus, perhaps, God couldn't put a person in a position where rationality required two incompatible courses of action." - Infinity, Causation, and Paradox p. 190.<br /><br />I was thinking about how to construct an infinite fair lottery. Let's say that there has been an infinite regress of me flipping a coin. If I get heads, I add one from the number I have. If I get tails, I subtract one from the number I have. I also skip over zero. I then take the absolute value of the result as my number. Why did I land on that number? Because I flipped a coin, and added/subtracted from my previous value based on that. Imagine if I got all heads. Then, I basically would have been counting up from infinity to my result. This might be used to justify any number. Perhaps why I got the result is that I ended on that day, for if I ended on a different day, I might have gotten a different result. The result might have no explanation at all, as this post would suggest. https://alexanderpruss.blogspot.com/2023/01/partial-and-complete-explanations.html Anyhow, this is an easy way to generate an infinite lottery based on an infinite regress.Alitheahttps://www.blogger.com/profile/07304709411526218728noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-74656294914802705242021-05-10T08:24:12.158-05:002021-05-10T08:24:12.158-05:00Alexander R Pruss
It is unclear to me how casual ...<b>Alexander R Pruss</b><br /><br />It is unclear to me how casual finitism solves this paradox.<br /><br />Imagine a setup where God just creates two cards with random natural numbers (picked randomly with equal probability from the set all natural numbers). You get to pick one freely and view the number on the card. You then get to switch cards if you think the other card is higher. (We can add money or something to spice it up I guess but I think you see the point).<br /><br />It seems to me that the logic is the same in this case as it is in the shuffled deck case. But aren't all theist committed to God being able to do all the steps required? Surely he knows the naturals numbers, can create cards, can write and present someone with an option between 2 cards? I don't see how the causes would be infinite here but the paradox stands.<br />Anonymoushttps://www.blogger.com/profile/13071348508867530518noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-51609425387915580212019-10-03T00:52:29.623-05:002019-10-03T00:52:29.623-05:00David Duffy
Now, read the first paragraph of the ...<b>David Duffy</b><br /><br />Now, read the first paragraph of the appendix in Nelson's book on page 80... especially note the last sentence of the paragraph.<br /><br />The S-N complexity space is discrete; it is not applicable.Philip Randhttps://www.blogger.com/profile/09143527524267821692noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-27661380016036231432019-10-02T19:34:22.800-05:002019-10-02T19:34:22.800-05:00Dear Philip Rand.
Nelson frequently mentions &qu...Dear Philip Rand. <br /><br />Nelson frequently mentions "illegal set formation", where one may not form subsets corresponding to external properties - this is how overspill can be used.<br /><br />Hi Professor Pruss. <br /><br />There seems to a TCS literature on the computational complexity of S_N - I have glanced at http://adsabs.harvard.edu/abs/2015arXiv150306188ADavid Duffyhttps://www.blogger.com/profile/12142997170025811780noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-35094065066681090772019-10-02T11:27:08.349-05:002019-10-02T11:27:08.349-05:00Mr Duffy:
It's hard to make sense of generati...Mr Duffy:<br /><br />It's hard to make sense of generating a "random element" of S_N. For one, S_N is not an amenable group (it has F2 as a subgroup). <br /><br />It is hard to make mathematical sense of my setup. It's an infinite process without meaningful initial conditions. BUT it intuitively seems that IF an infinite past is possible, this process could have taken place.Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-15337446952724408372019-10-01T20:20:14.484-05:002019-10-01T20:20:14.484-05:00David Duffy
Your concern of legal conditioning is...<b>David Duffy</b><br /><br />Your concern of <i>legal conditioning</i> is ill founded.<br /><br />Simply, refer to the last paragraph on page 15 of <i>Radical Elementary Probability Theory</i>.Philip Randhttps://www.blogger.com/profile/09143527524267821692noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-2134989898828434652019-10-01T03:54:27.784-05:002019-10-01T03:54:27.784-05:00I found this setup rather baffling. My first thoug...I found this setup rather baffling. My first thought was to apply the non-standard analysis that<br />Edward Nelson uses in Radically Elementary Probability Theory (1987), where he provides proofs for ordinal and cardinal Borel-Cantelli Theorems. I see Wenmackers (2011) uses the same apparatus to get hyperrational probabilities for an infinite lottery.<br /><br />If I understand Nelson correctly, the cardinal version can hold for the infinite lottery, but the ordinal version does not. Since the actual bet here relies on ordinality and tail probabilities of a nonconverging sequence I have a feeling that the conditioning cannot be legal because one is mixing internal (ie finite or "limited") and external (unlimited) notions (the difference between two draws may or may not have an expectation of 0, but does have an infinite variance). It is clear that if Alice has drawn 1, she can have sure knowledge she has lost, even though the prior probability of this ever happening is infinitesimal. If the labelling (Wenmackers) of the tickets was over the (negative and positive) integers, we would find it more "intuitive", even though it is the same size probability space.<br /><br />In passing, the shuffling of the deck is extremely well known as the group S_N, and is uncountable. I did wonder if there was a result about this setup in the guise of the length of increasing subsequences of a random walk (ie a permutation).David Duffyhttps://www.blogger.com/profile/12142997170025811780noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-38277113621165433652019-09-30T16:29:51.779-05:002019-09-30T16:29:51.779-05:00Majesty of Reason (if you are still following this...Majesty of Reason (if you are still following this thread):<br /><br />Alice was right to take the bet. She knew that the first two cards had been shuffled, so she had probability 1/2 of getting the higher one. As long as she resists the temptation to take the swap, her expected gain will be $1.<br /><br />Alice is also right to think that with probability 1, Bob’s number is greater than any <i>given</i> number (granting for the sake of argument that the infinite series of shuffles is possible and has happened). But she would be wrong to conclude that with probability 1, Bob’s number is greater than <i>her</i> number, and she would be wrong to accept the swap based on this conclusion.<br /><br />The subtle distinction is that Alice’s number is not a <i>given</i> number – it resulted from the same random process that produced Bob’s, and it could not have been given in advance. (Another example: with probability 1, Bob’s number is greater than any given number, but it is certainly less than Bob’s number + 1. The catch is that <i>Bob’s number + 1</i> is not a number that could have been given in advance.) The paradox (if indeed it is a paradox) is precisely that this distinction matters.<br /><br />What to make of this? One view (mine) is that there is no paradox. Alice is simply wrong to conditionalize on her number as if it were a number given in advance. Another view the fact that the distinction matters (note, usually it doesn’t, which is why tend to ignore it) shows that there is something wrong with the setup, or with the reasoning leading to the probabilities. In this case, Alex has pointed to the infinite causal chain. The infinite number of cards, each able to display arbitrarily large numbers, and the infinite number of arbitrarily deep shuffles also seem suspect.<br /><br />N.B. I’m strictly an amateur. Don’t take my word. There is literature on these issues, some by Alex himself. Keywords: nonconglomerable, finite additivity.IanShttps://www.blogger.com/profile/00111583711680190175noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-30758286067448358552019-09-30T12:46:46.726-05:002019-09-30T12:46:46.726-05:00There is no specific "Thomist First Principle...There is no specific "Thomist First Principle of Non-Contradiction". There is only the general Law of Non-ContradictionDominik Kowalskihttps://www.blogger.com/profile/14634739012344612398noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-132502842705296032019-09-29T07:11:27.157-05:002019-09-29T07:11:27.157-05:00Majesty of Reason
Type this search: "A053169...<b>Majesty of Reason</b><br /><br />Type this search: "<b>A053169 OEIS</b>"<br /><br />And you have a logical demonstration of your conclusion.<br /><br />Interestingly, it is also a logical demonstration that the <i>Thomist First Principle of Non-Contradiction</i> is <b>false</b>.Philip Randhttps://www.blogger.com/profile/09143527524267821692noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-30329478437608763452019-09-28T07:04:33.619-05:002019-09-28T07:04:33.619-05:00Alex:
No strategy played by both Alice and Bob ca...Alex:<br /><br />No strategy played by both Alice and Bob can do better than <i>never swap</i>. Why? Because swapping carries a penalty and if Alice and Bob play the same strategy, the expected penalty must be split equally between them.<br /><br />Note that this reasoning depends only on the symmetry. It applies equally to the finite and infinite cases. It does not depend in any way on the probabilities of the cards (as long as they are symmetric, of course). This makes the two-person setup unsuitable for testing intuitions about the probabilities of the cards. The one-person version avoids this problem.IanShttps://www.blogger.com/profile/00111583711680190175noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-30781547914223969332019-09-28T00:27:06.917-05:002019-09-28T00:27:06.917-05:00Majesty of Reason
You highlight the interesting c...<b>Majesty of Reason</b><br /><br />You highlight the interesting conclusion of the thought experiment.<br /><br />This insight of yours is especially good:<br /><br /><i>Could we derive a contradiction from infinite Dutch book cases by means of arguing that they entail a conjunction of the form A and ~A, where A is "one rationally ought to play the gambling game" and ~A is "it is false that one rationally ought to play the gambling game"?</i><br /><br />What you are saying is that the "<i>game</i>" for the <i>observer</i> is <i>zero-sum</i>...it is not a contradiction...that insight has far reaching consequences...Philip Randhttps://www.blogger.com/profile/09143527524267821692noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-44329501517953554342019-09-27T19:22:23.945-05:002019-09-27T19:22:23.945-05:00Shouldn't Alice have just reasoned beforehand ...Shouldn't Alice have just reasoned beforehand that, for any card number n, there are a finite number of cards labeled that have a value below n, whereas there are an infinite number of cards labeled that have a value above n, in which case for any card n she could potentially pick, there is an infinitesimal (else: zero) probability that she would win? But if so, then isn't it false that she should have taken the bet to begin with and that she purportedly had a 50% chance of winning to begin with?<br /><br />I hope you can help me see the paradox here.<br /><br />I am also hoping you can help explain why such situations involving Dutch books actually entail contradictions as opposed to showing perhaps inherent difficulties in the concept of rationality (like one cannot be perfectly rational). Could we derive a contradiction from infinite Dutch book cases by means of arguing that they entail a conjunction of the form A and ~A, where A is "one rationally ought to play the gambling game" and ~A is "it is false that one rationally ought to play the gambling game"?Majesty of Reasonhttps://www.blogger.com/profile/04265039708439613397noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-3450828071560287882019-09-27T12:05:18.648-05:002019-09-27T12:05:18.648-05:00I have given you plenty of info, i.e. chi-squared ...I have given you plenty of info, i.e. chi-squared test.<br /><br />Your concern technically is called <i>degrees of freedom</i> of the hypothesis. This can be handled in the chi-squared test, i.e. finite or infinite.<br /><br />Nothing enigmatic about it.Philip Randhttps://www.blogger.com/profile/09143527524267821692noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-2397119303360737782019-09-27T11:19:52.833-05:002019-09-27T11:19:52.833-05:00Mr Rand: I am afraid that too many of your comment...Mr Rand: I am afraid that too many of your comments are enigmatic and unargued-for. Please include more in the way of serious arguments in the future.Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-91302128691223501332019-09-27T08:32:32.525-05:002019-09-27T08:32:32.525-05:00Nope... you are wrong... your mistake is in your n...Nope... you are wrong... your mistake is in your notion of what is conitinuous and what is discrete....Philip Randhttps://www.blogger.com/profile/09143527524267821692noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-70663223599446065382019-09-27T08:15:57.546-05:002019-09-27T08:15:57.546-05:00One difference between the finite and infinite cas...One difference between the finite and infinite cases is that in the finite case the other player's willingness to swap carries information. But in the infinite case, it doesn't.Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-17555962402724712072019-09-27T04:30:06.840-05:002019-09-27T04:30:06.840-05:00Alex:
I’m not sure that making Carl the only sour...Alex:<br /><br />I’m not sure that making Carl the only source of utility changes things. Suppose the setup will be repeated. Then for, both Alice and Bob, the steady $1 (or equivalent utility) per round would be expected to eventually swamp the sum of the random $100 (or equivalent) wins and losses.<br /><br />Here is another way to look at it. Think about the classical version. Assume no cooperation. Alice might reason that if her number is suitably low, the odds are right to offer the swap. But then she realizes that Bob, reasoning similarly, will accept only if his number is also low. Now the odds don’t look so good. So you need game theory, not just decision theory narrowly defined. My one-person version avoids such issues.<br /><br />But here is what interests me. Think about my infinite one-person version. The obvious plan is <i>never swap</i>. I gave some plans that give an infinitesimal improvement on this. Amazingly, there seems to be a plan that that gives a genuine improvement, provided Alice has access to the other cards. She takes the next (say) 10,000 cards. She sorts them and her card into ascending order. If her card is in the first 50 (= 10,000/200), she accepts the first swap. This plan gives her an expected extra gain of about $ 0.50 /200 = $ 0.0025 (if I’m thinking straight, which I’m not at all sure of). This is small, but not zero.IanShttps://www.blogger.com/profile/00111583711680190175noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-48957111970490980312019-09-27T00:38:59.899-05:002019-09-27T00:38:59.899-05:00Pruss
Mixed phenomena makes no difference to the ...Pruss<br /><br />Mixed phenomena makes no difference to the result of the hypothesis.Philip Randhttps://www.blogger.com/profile/09143527524267821692noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-60405988616349162942019-09-26T15:30:16.475-05:002019-09-26T15:30:16.475-05:00I hadn't thought of that. But we can just stip...I hadn't thought of that. But we can just stipulate that they can't do that. Perhaps the prizes are not monetary but pleasures and pains administered by Carl. Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-44344178171040917182019-09-26T15:19:14.674-05:002019-09-26T15:19:14.674-05:00Alex:
In the classical setup, it’s true that Carl...Alex:<br /><br />In the classical setup, it’s true that Carl loses money if Alice and Bob decline the swap, as they will if their numbers are high. But if Alice and Bob agree in advance to cooperate and have the winner pay the $100 back to the loser after the event, they can <i>always</i> decline the swap and always end up with $1 each. The same applies to the infinite version.IanShttps://www.blogger.com/profile/00111583711680190175noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-28328908236118151182019-09-26T12:45:16.999-05:002019-09-26T12:45:16.999-05:00Can we argue that the thus-shuffled deck generates...<i>Can we argue that the thus-shuffled deck generates a countably infinite fair lottery, i.e., that if we pick cards off the top of the deck, all card numbers will be equally likely?</i> <br /><br /><b>No.</b> I have chi-square tested the hypothesis. It is not possible to generate a fair lottery.Philip Randhttps://www.blogger.com/profile/09143527524267821692noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-47315053249352639252019-09-26T09:10:40.650-05:002019-09-26T09:10:40.650-05:00I don't really see the Prisoner's Dilemma ...I don't really see the Prisoner's Dilemma here. Can you explain?Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-77511507918895234612019-09-26T08:43:31.167-05:002019-09-26T08:43:31.167-05:00Ian:
In the standard classical setup, Carl has no ...Ian:<br />In the standard classical setup, Carl has no guarantee of making money. Alice and Bob won't want to swap if their numbers are high.<br />Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-61384809932841161242019-09-26T02:16:45.736-05:002019-09-26T02:16:45.736-05:00The setup combines a standard Prisoners’ Dilemma w...The setup combines a standard Prisoners’ Dilemma with a probabilistic paradox. The Prisoners’ Dilemma can arise even in unparadoxical finite cases. Suppose there are 1000 cards, Alice gets ‘1’ and Bob gets ‘2’. Then it would seem to each that they should swap, but between them in total, they would do better to stick.<br /><br />I take it that your interest is in probability theory, not Prisoners’ Dilemma. You may prefer a one-person version. Alice is offered $1 to play. The first and second cards will be placed is separate sealed envelopes. Alice will be given the first envelope. If her number is smaller than the other one, she will lose $100, otherwise she will win $100. She is allowed to look at her number and reseal the envelope. For $199 she will have the option to swap envelopes. If she accepts, her memory of the first number will be erased, she will be allowed to look at her new number, and, for a further $199, she will have the option to swap back.<br /><br />It seems that Alice will accept the offer to play then accept both swaps, thereby losing for sure.<br /><br />I would give Alice my usual advice in such cases: don’t conditionalize. Make a plan in advance and stick to it. <i>Never swap</i> is a reasonable plan. It guarantees her $1. <i>Swap only on ‘1’, and never swap back</i> is slightly better. Another reasonable plan is to pick a large number N, swap if the first number is N or less, then swap back only if the second number is less than N/200 (if I’m thinking straight). Note that in all such plans, Alice will reject the first swap with probability 1 and her expected gain will be $1.IanShttps://www.blogger.com/profile/00111583711680190175noreply@blogger.com