Paul
Posts:
676
Registered:
7/12/10


Re: Beating the Odds?
Posted:
Feb 2, 2013 4:02 PM


On Saturday, February 2, 2013 3:16:27 PM UTC, David C. Ullrich wrote: > On Fri, 1 Feb 2013 10:21:54 0800 (PST), Paul <pepstein5@gmail.com> > > wrote: > > > > >On Friday, February 1, 2013 2:47:43 PM UTC, David C. Ullrich wrote: > > >> On Thu, 31 Jan 2013 14:17:38 0800 (PST), pepstein5@gmail.com wrote: > > >> > > >> > > >> > > >> >On Wednesday, January 30, 2013 3:14:00 PM UTC, David C. Ullrich wrote: > > >> > > >> >> On Wed, 30 Jan 2013 00:29:09 0800, William Elliot <marsh@panix.com> > > >> > > >> >> > > >> > > >> >> wrote: > > >> > > >> >> > > >> > > >> >> > > >> > > >> >> > > >> > > >> >> >There is a fair coin with a different integer on each side that you can't > > >> > > >> >> > > >> > > >> >> >see and you have no clue how these integers were selected. The coin is > > >> > > >> >> > > >> > > >> >> >flipped and you get to see what comes up. You must guess if that was the > > >> > > >> >> > > >> > > >> >> >larger of the two numbers or not. Can you do so with probability > 1/2? > > >> > > >> >> > > >> > > >> >> > > >> > > >> >> > > >> > > >> >> Of course not. Seeing one side gives you no information about > > >> > > >> >> > > >> > > >> >> what's on the other side. > > >> > > >> > > > >> > > >> >I don't completely agree with this answer. The concept of a number being selected such that "you have no clue how it was selected" doesn't translate readily into mathematics. > > >> > > >> > If it's mathematics, you need to specify the procedure or specify the proability space (or set of possible probability spaces) etc. > > >> > > >> > > >> > > >> True. > > >> > > >> > > >> > > >> >Since this isn't mathematics, the best I can do is use intuition and commonsense. Surely, if one of the numbers was > 10 ^ 1000, the most intelligent guess is that the number on the reverse side is smaller. > > >> > > >> > > >> > > >> Why in the world would that be? Most positive integers are larger than > > >> > > >> that. In fact all but finitely many positive integers are larger. > > >> > > > > > >1) Most positive integers are larger, but the context is integers not positives. 2) Imagine that you did see 10^1000. > > >There are two choices  guess that the reverse is larger, or guess that the reverse is smaller. If you disagree so strongly that > > >I'm wrong to guess that the reverse is smaller, please could you explain how large the number would have to be for you to guess that the reverse is smaller? > > > > Assuming that you mean me when you say "you": Did you read what > > I wrote in my first reply? I can't imagine why you think there is > > such a thing as "how large the number would have to be" for > > me to guess it was the larger  that's missing the whole point. > > We have no informationabout the other number. > > > > >Or would that always be a bad guess? > > > > Any guess is as good or as bad as any other. > > > > >3) Having researched the correct answer, > > > > > > What??????????????????????? > > > > >I'm right about 10^1000 even if you restrict the domain to positives. > > > > >The answer involves selecting your own probability measure and then randomizing a choice according to your own selection. In practice, if you did that on the domain of the positive reals, it would be most unlikely that you would pick a probability measure such that the integral from 10^1000 to infinity is greater than 0.5. > > > David,
This links to the problem and intended solution. It's a very slightly different problem because the domain is reals rather than integers. http://blog.xkcd.com/2010/02/09/mathpuzzle/commentpage1/#comments
The solution in the blog is one of a huge family of answers. All of these answers (meaning similar answers given from standard probability distributions and given by other mathematicians) would be likely to involve guessing that B > A if B is the observed coin and has value 10^1000
I don't regard this intended solution as uncontroversial but I don't think that Fred's point that there exists no positive epsilon such that a strategy exists for winning with probability > 0.5 + epsilon is an argument against the intended solution, or intended family of solutions.
I gave a simpler problem involving an exam with N questions to explain why I didn't accept Fred's argument against the solution.
I would prefer it if you gave some comments about the problem and intended solution rather than focusing on what my own limitations might be by saying things like "You didn't read this. You don't seem to get your head around that. You're doing this wrongly. You're doing that wrongly."
It's not about me. It's a very common problem with a standard intended solution, and the standard intended solution is different from yours. Could you please explain what is wrong with the intended solution?
Paul Epstein

