17 ms·
Well, yeah :) The problem is still in assuming something about the distribution without having any good reason to. In original "paradox" reasoning it was imposs
by bluecalm 13y ago
Well, yeah :)
The problem is still in assuming something about the distribution without having any good reason to.
In original "paradox" reasoning it was impossible uniform distribution. What you show is that even assuming possible distribution could still lead nowhere.
It still doesn't mean there is any reason to assume it's P($2^i) = 1/(2^i). I may just as well assume it's exactly 100$-200$ and that switching from 100$ gives me guaranteed payoff. It would be as baseless as assuming uniform distribution or the one given by you.
I think the main lesson from this puzzle is that you can't assume stuff without good reason to.
- jamii 13y agoI'm not assuming that that is the distribution. I'm saying that the problem as presented is underspecified, but even if you give this specific distribution in the problem you can still cause confusion. Assuming stuff is bad but the original 'paradox' still doesn't go away if you nail everything down. It's a useful problem for education people about the subtleties of infinite sums.
- bluecalm 13y agoWell, there are distributions which don't have well defined expected value and I agree that it's valuable lesson. I would argue though that this is not the best example to show it. The point of two envelope problem comes much earlier and specifically giving distribution defined by you would make it completely different problem and assuming it in original form would already be a mistake. I think there are better examples to show how infinite sums and relying on expected value based on those might leads to problems. Like this one for example: http://plato.stanford.edu/entries/paradox-stpetersburg/ http://plato.stanford.edu/entries/paradox-stpetersburg/
- jamii 13y agoThe expected value in the St Petersburg game is actually well-defined - it converges to positive infinity. That is a subtly different result than not converging at all. Both problems are useful.