4 ms·

I'm not entirely convinced that cases (1) and (2) have the 50% probability stated. We have X, the number revealed to us; Y, the hidden number; and our random n

I'm not entirely convinced that cases (1) and (2) have the 50% probability stated. We have X, the number revealed to us; Y, the hidden number; and our random number, R. Let's assume we have case (2), so X > R and Y > R, but the relationship between X and Y is not defined. So, X and Y lie somewhere in the range (+∞, R). We then partition that range with X: (+∞, X) and (X, R). Y lies in one of these two ranges. By the rules defined, we state that we expect Y to be less than X. But the interval less than X is finite, while the interval above X is infinite. I doubt that this range distribution results in a 50% chance. I think that by picking R, we are modifying the probability that Y is less than or greater than X in cases (1) and (2). However, since we cannot distinguish between those cases and case (3), all the winnings gained in (3) are actually lost to in cases (1) and (2). EDIT: I guess what I'm saying is that X and Y are no longer independent in cases (1) and (2), but are conditioned by R. That conditioning means we can no longer claim a 50% probability for those cases as if we had never picked R.

alister 11y ago
You have an interesting thought and I was wondering the same thing. OP says that the probability of winning in the 3 cases are: (1) 50% win, (2) 50% win, (3) always win If I understand correctly, you're saying that the probabilities might perhaps be: (1) 50% win minus some probability of case 3 happening, (2) 50% win minus some probability of case 3 happening, (3) always win Then the three cases even out to 50% and there is no winning strategy. But wouldn't your argument break down where you write, "But the interval less than X is finite, while the interval above X is infinite."? Since we're dealing with reals, then both (+∞, X) and (X, R) have an equally large and infinite set of numbers (i.e., the latter is not a finite set)[1]. So, Y is as likely to be found in (+∞, X) as in (X, R). I'm still wondering if there's a way to make your argument work. [1] https://proofwiki.org/wiki/Closed_Interval_in_Reals_is_Uncountable https://proofwiki.org/wiki/Closed_Interval_in_Reals_is_Uncou...
- jdmichal 11y ago
  > But wouldn't your argument break down where you write, "But the interval less than X is finite, while the interval above X is infinite."? Since we're dealing with reals, then both (+∞, X) and (X, R) have an equally large and infinite set of numbers (i.e., the latter is not a finite set)[1]. So, Y is as likely to be found in (+∞, X) as in (X, R). Yes, of course. I realized this about 5 minutes ago in the shower and was hoping no one had called me out on it yet! I think the real argument is actually a combination of this and landryraccoon's post, depending on the problem definition. [1] The problem is either defined in such a way that the (X, R) interval is finite, and therefore falls under my argument. -OR- the problem defines (X, R) to be infinite, and P(case 3) is zero. [1] https://news.ycombinator.com/item?id=10647331 https://news.ycombinator.com/item?id=10647331

alister 11y ago

You have an interesting thought and I was wondering the same thing. OP says that the probability of winning in the 3 cases are: (1) 50% win, (2) 50% win, (3) always win If I understand correctly, you're saying that the probabilities might perhaps be: (1) 50% win minus some probability of case 3 happening, (2) 50% win minus some probability of case 3 happening, (3) always win Then the three cases even out to 50% and there is no winning strategy. But wouldn't your argument break down where you write, "But the interval less than X is finite, while the interval above X is infinite."? Since we're dealing with reals, then both (+∞, X) and (X, R) have an equally large and infinite set of numbers (i.e., the latter is not a finite set)[1]. So, Y is as likely to be found in (+∞, X) as in (X, R). I'm still wondering if there's a way to make your argument work. [1] https://proofwiki.org/wiki/Closed_Interval_in_Reals_is_Uncountable https://proofwiki.org/wiki/Closed_Interval_in_Reals_is_Uncou...

jdmichal 11y ago
> But wouldn't your argument break down where you write, "But the interval less than X is finite, while the interval above X is infinite."? Since we're dealing with reals, then both (+∞, X) and (X, R) have an equally large and infinite set of numbers (i.e., the latter is not a finite set)[1]. So, Y is as likely to be found in (+∞, X) as in (X, R). Yes, of course. I realized this about 5 minutes ago in the shower and was hoping no one had called me out on it yet! I think the real argument is actually a combination of this and landryraccoon's post, depending on the problem definition. [1] The problem is either defined in such a way that the (X, R) interval is finite, and therefore falls under my argument. -OR- the problem defines (X, R) to be infinite, and P(case 3) is zero. [1] https://news.ycombinator.com/item?id=10647331 https://news.ycombinator.com/item?id=10647331