tag:blogger.com,1999:blog-3891434218564545511.post7876634527723948288..comments2018-06-20T17:45:39.878-05:00Comments on Alexander Pruss's Blog: Fun with St PetersburgAlexander R Prusshttp://www.blogger.com/profile/05989277655934827117noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-3891434218564545511.post-10955518110271951872017-05-09T21:14:53.746-05:002017-05-09T21:14:53.746-05:00Does this really avoid the swamping problem? If yo...Does this really avoid the swamping problem? If you can afford to lose $2A, having just received $A, you must have had at least $A to start with. And A could be arbitrarily large. So, for the setup to work, you must have started with infinite assets. This will swamp any finite loss.<br /><br />But does swamping matter? Isn’t the paradox about the value of the deal? Judged by expected return, <i>conditional on knowing your winnings</i>, it seems to have infinite value. But by the pairing, you can force a sure total loss.IanShttps://www.blogger.com/profile/00111583711680190175noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-784714987651288082017-05-08T13:22:02.131-05:002017-05-08T13:22:02.131-05:00Unless I'm missing something, you lose $2 each...Unless I'm missing something, you lose $2 each time, but the team also gets two St Petersburg payoffs which swamp the loss.<br /><br />Here's a way to make it just a solid loss. You and Delenn play against the house. The house offers you a free St Petersburg payoff and offers Delenn an independent St Petersburg payoff, also for free. Then the house pays you and Delenn, but you don't see what the other got. Then the house offers you and Delenn this second deal: if you pay in twice what you won, plus a dollar, you get the other's winnings. Now it's a solid $2 loss to the team if you both go for the swap. <br /><br />Here's a possible solution, though.<br /><br />This problem is most compelling if both you and Delenn know that the other is getting this deal--but it seems you're still each rational in going for the swap.<br /><br />But now it's not so clear. Let p be the probability of Delenn swapping and let A and B be what you got in the first round. Then here are the overall payoffs for your choice:<br /><br /> SWAP: With probability p, get -1. With probability 1-p, get (1/2)(-1-A+2B).<br /> NOT: With probability p, get (1/2)(-1-B+2A). With probability 1-p, get (1/2)(A+B).<br /><br />Since you know A, EA is finite; but you don't know anything about what B is, so EB is infinite. Hence, E(SWAP)=infinity, and E(NOT)=infinity. Now it's not so clear that you should go for the swap.Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-9553405890825358232017-05-01T23:18:20.877-05:002017-05-01T23:18:20.877-05:00Accepting my deal will always lose Team ‘You’ exac...Accepting my deal will always lose Team ‘You’ exactly $2. I would have thought that any such loss, however small, would suffice to make the point. But you can easily make it more dramatic: ‘half my winnings for all yours’, or ‘half my winnings less $x for all yours’ etc.<br /><br />As a point of exposition, you would do well to say explicitly that Team ‘You’ know that Team Pruss make their offers before they see their winnings, or else that they are bound to make their offers in any case. Team Pruss must not have the option of offering or not, depending on their winnings, and Team ‘You’ must know this.IanShttps://www.blogger.com/profile/00111583711680190175noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-27340084379075861372017-05-01T15:32:33.231-05:002017-05-01T15:32:33.231-05:00On reflection, Ian, I like your simpler version a ...On reflection, Ian, I like your simpler version a lot and I may use it in print, but it has the disadvantage that (right? I haven't checked) eventually the dollars will likely be swamped and insignificant.Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-30307207269876580932017-04-30T19:38:37.906-05:002017-04-30T19:38:37.906-05:00A different line: Conditional on any particular wi...A different line: Conditional on any particular winnings, the deal in the OP has infinite expected value. But unconditionally, its expected value is indeterminate. The maths is fine, if a bit unintuitive. But what do we make of it? One answer: Don’t conditionalize. Don’t accept that we have to act only on the latest conditional probabilities, and that nothing else is relevant. In normal cases, it can be proved as a theorem that it is indeed safe to ignore everything else. But is weird cases like this, it need not be. Note, non-conglomerability raises similar issues.<br /><br />Thinking in this way, ‘you’ (and Delenn) could adopt a strategy like this: accept the offer on ‘low’ winnings (i.e. less than some number N) and reject otherwise. Adopting such a strategy would give you infinite expected gains without the risk of unbounded losses.IanShttps://www.blogger.com/profile/00111583711680190175noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-20236491179124047692017-04-29T23:17:56.372-05:002017-04-29T23:17:56.372-05:00Wouldn’t a simpler deal ask for $1 to swap winning...Wouldn’t a simpler deal <i>ask for $1 to swap winnings </i> make the same point, albeit less dramatically?<br /><br />On causal finitism: Think about original St Petersburg (the coin is flipped until it lands heads). You have to <i>be prepared</i> for the coin to land tails every time, even though it almost certainly won’t. Then you will need an infinite number of flips. Maybe this makes the setup count as causally infinite.<br /><br />Put differently: To run St Petersburg ‘synchronously’ with coins, you would need an infinite number of them.<br /><br />An obvious objection to this idea is that there is no problem with a probability distribution like P(n) = 2^(-n) in itself. The problem comes from the probabilities and the rapidly increasing payouts together. A possible response: Strictly, causal finitism would rule out <i>all</i> infinite distributions. But normal cases could be approximated for practical purposes by a sufficiently large number of coins, and infinite expectation cases could not. [uniform convergence]<br /><br />But what about just rolling a suitably biased die with a countably infinite number of faces? I’m not entirely convinced by any of this.IanShttps://www.blogger.com/profile/00111583711680190175noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-54090759104396906892017-04-29T13:31:56.214-05:002017-04-29T13:31:56.214-05:00The expected utility will be infinite but the actu...The expected utility will be infinite but the actual utility will be finite after a finite number of runs. So doubling that finite payoff seems worth it.Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-32035192307751152802017-04-29T12:13:04.851-05:002017-04-29T12:13:04.851-05:00This is unnecessarily complicated. The expected we...This is unnecessarily complicated. The expected wealth of all parties at the end will be infinite, but with the distribution much worse for some than for others.<br /><br />What the argument actually proves, is that it is foolish to use an unbounded utility function.Anonymousnoreply@blogger.com