
Re: Beating the Odds?
Posted:
Feb 2, 2013 7:20 AM


On 2/2/2013 7:17 AM, Jim Burns wrote: > On 2/1/2013 1:21 PM, Paul 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. > > Consider a second scenario. Instead of being flipped, the coin is > shaken in a dice cup and the cup slammed down with the coin covered. > Before the coin is revealed, the operator reaches under the cup and > turns the coin over. Now, whatever reasoning that would have led you > to guess the visible integer was larger with p > 1/2 leads you > to guess that the visible integer was larger with p > 1/2.
D'Oh. I meant "p < 1/2" the second time.
> > But, what if the operator only pretended to turn the coin over? > >>> 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? Or would that always be a bad guess? > > What is a "bad guess"? > Wrong? Well, no, you would be right 50% of the time, just like > everyone else. > Unjustified? Can you justify the sudden appearance of > a justification for your guess at some specific number > like 10^1000? It doesn't look like that, so I guess, yes, > that would always be a bad guess. > > Why are we limited to only two choices? "Not enough information" > seems perfectly reasonable to me. It's what happens most often > in real life, anyway. "What is the area of this field? It is > 87 yards wide." > > Imagine that I did see 10^1000. What are the odds that the other > side has (10^1000)! on it? > >> 3) Having researched >> the correct answer, 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. > > I am interested in this research that you have done. > > Bayesian probability calculations are utterly uncontroversial, > once one has an initial probability measure. It would be nice > to have an equally uncontroversial method of selecting > probability measures. > > Your flat assertion that you are right about 10^1000 because > of how unlikely a probability measure making you wrong would > be sounds as though you have an uncontroversial method > for assigning probabilities (or something like that) to > probability measures. > > Could you say more about that? > >

