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Topic: Beating the Odds?
Replies: 35   Last Post: Feb 6, 2013 3:44 PM

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Paul

Posts: 467
Registered: 7/12/10
Re: Beating the Odds?
Posted: Feb 2, 2013 9:36 AM
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On Saturday, February 2, 2013 12:17:37 PM UTC, 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.
>
...
>
> > 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?


1. This is a common problem in maths circles. The research was to google around and find the standard intended solution. Probably about five minutes research. I never meant to imply that this was a significant research project.

2) The standard intended solution (which may indeed be problematic) tells the solver to name any probability measure whatsoever (for example the standard normal distribution if the domain is the reals). (Since we're dealing here with integers, this standard approach will need to be modified but only slightly). Since standard accounts don't suggest the probability measure should be intricate or esoteric, and since it's good mathematical technique to pick the simplest possible examples that are needed to prove your point, it would be unlikely from a psychological point of view for the integral from 10^1000 to infinity of the function to be > 0.01 because the convergence would be quicker for most standard examples.

3) With the standard intended solution, you don't need a method for selecting probability measures and you don't assign a probability to each possible probability measure. By "likely", I meant "likely from a practical point of view". That use of "likely" can be illustrated by the following claim of mine: When demonstrating the existence of irrational numbers, a maths teacher is more likely to use sqrt(2) as an example than sqrt(57134.23)

Paul Epstein




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