tag:blogger.com,1999:blog-3891434218564545511.post7181702410234561566..comments2021-05-15T08:44:54.382-05:00Comments on Alexander Pruss's Blog: Uncountably infinite fair lotteryAlexander R Prusshttp://www.blogger.com/profile/05989277655934827117noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-3891434218564545511.post-9376360385907209362020-08-06T05:07:59.238-05:002020-08-06T05:07:59.238-05:00Here is a countable version that seems to avoid so...Here is a countable version that seems to avoid some of the side issues.<br /><br />Start with a countable fair infinite lottery on 0, 1, 2, 3 …, defined by the property that for any finite set of possible outcomes, given that the outcome is in the set, all members are equally likely. (Leave aside how or whether this could be achieved.) Independently roll a fair 3-sided die, with faces marked 0, 1, 2. Multiply the lottery outcome by 3 and add the die outcome. The resulting lottery is fair in the sense given above. It has the extra property that P(0 mod 3) = P(1 mod 3) = P(2 mod 3) = 1/3.<br /><br />Suppose that all the tickets in the new lottery are held by a countable infinity of players, one ticket per player. As in your example, Alice offers each player a two-tickets-for-one swap. All the players accept. Alice makes the swaps Hilbert’s Hotel style: in return for ticket N, she gives tickets 3N+1 and 3N+2. She keeps the tickets with numbers divisible by 3 for herself.<br /><br />It seems that each player has improved her chances of winning, or a least made them no worse. (Given that one of N, 3N+1 and 3N+2 win, N has probability 1/3, 3N+1 or 3N+2 has probability 2/3.) But collectively, the chance of any player winning has fallen from 1 to 2/3.IanShttps://www.blogger.com/profile/00111583711680190175noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-64435667379787886802020-08-05T10:17:40.771-05:002020-08-05T10:17:40.771-05:00Ian:
Good question.
Everyone:
I really screwed ...Ian:<br /><br />Good question.<br /><br />Everyone:<br /><br />I really screwed up in my description of this lottery. I start off by saying that there are infinitely many coin flips. But then I don't say anything about the mechanics of making sure that everyone gets a *different* ticket. Generating such mechanics is nontrivial, and I don't know how to do that!<br /><br />What would have worked would be to *stipulate* for a reductio that there is a fair lottery with continuum many tickets, which can then be labeled with H-T sequencees. But if I did that, I wouldn't be entitled to assume that that the probability that the first four letters in a winning sequence are HHHH is only 1/16.<br /><br />In other words, the whole post is just plain confused.Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-66509408924561404202020-08-04T12:42:47.055-05:002020-08-04T12:42:47.055-05:00Another nice paradox, although I'd guess that ...Another nice paradox, although I'd guess that this boils down to a Hilbert's Hotel-style paradox with the tickets. It's also like the St Petersburg paradox: people should not play Alice's game, even though it seems like it makes sense. I fail to see why it means that the lottery would not be fair, though.Martin Cookehttps://www.blogger.com/profile/11425491938517935179noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-60196754522193032392020-08-04T03:04:12.969-05:002020-08-04T03:04:12.969-05:00I see I am just echoing Benci et al (2018).I see I am just echoing Benci et al (2018).David Duffyhttps://www.blogger.com/profile/12142997170025811780noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-73855516419542353332020-08-04T01:14:12.070-05:002020-08-04T01:14:12.070-05:00Is the lottery really fair? Grant the usual strong...Is the lottery really fair? Grant the usual strong independence of the flips. Then the lottery is visibly fair between any two sequences that differ at only a finite number of places, in the sense that given that given one or the other, each is equally likely. To see this, condition on the flips for which the sequences match. But for sequences that differ at an infinite number of places, this does not work. What, for example, is P(HHH… | HHH… or TTT….)? As far as I can see, it does not follow from the setup.<br /><br />The holder of HHH… gets her ticket back, and new one. That’s clearly a good deal. Same for HHHHT repeating. Some people get tickets that differ in only finitely many places from their original one. They, too, have a visibly good deal. But for most people, both new tickets differ from the original at infinitely many places. It does not seem obvious that this is a good deal. At least, it does not seem to follow from the fairness and independence of the flips.IanShttps://www.blogger.com/profile/00111583711680190175noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-23957054916968658522020-08-03T18:44:48.466-05:002020-08-03T18:44:48.466-05:00In Nelson's approach, at least, I think these ...In Nelson's approach, at least, I think these kind of paradoxes are blocked by the prohibition on illegal set formation. Charlie Geyer summarizes it as "All objects are internal. Only<br />properties can be internal or external". I think it is correct to say that since a finite number of tickets in such lotteries still only offer an infinitesimal chance of winning, it is rational to neglect them.<br /><br />Re conditional probabilities, when I look at<br />https://www.sciencedirect.com/science/article/pii/S0888613X07001223<br /><br />which uses a nonstandard arithmetic, "Axiom 9 was added in order to conform with the useful<br />practice of assuming conditional probability to be 1, whenever the condition has the probability 0". I don't know if that holds for infinitesimal probabilities.<br />David Duffyhttps://www.blogger.com/profile/12142997170025811780noreply@blogger.com