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

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Jim Burns

Posts: 1,061
Registered: 12/6/04
Re: Beating the Odds?
Posted: Feb 2, 2013 7:17 AM
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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.

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?





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