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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: 1,200 Registered: 12/6/04
Re: Beating the Odds?
Posted: Feb 2, 2013 7:20 AM
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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?
>
>

Date Subject Author
1/30/13 William Elliot
1/30/13 forbisgaryg@gmail.com
1/31/13 William Elliot
2/1/13 forbisgaryg@gmail.com
2/1/13 Dan Heyman
1/30/13 David C. Ullrich
1/30/13 Steve Oakley
1/31/13 Paul
1/31/13 Frederick Williams
1/31/13 David Petry
2/1/13 Richard Tobin
2/1/13 David C. Ullrich
2/1/13 Paul
2/2/13 Jim Burns
2/2/13 Jim Burns
2/2/13 Paul
2/2/13 David C. Ullrich
2/2/13 David C. Ullrich
2/2/13 Paul
2/3/13 David C. Ullrich
1/31/13 Frederick Williams
1/31/13 Butch Malahide
1/31/13 JohnF
2/1/13 ArtflDodgr
2/1/13 Butch Malahide
2/2/13 William Elliot
2/2/13 Paul
2/3/13 quasi
2/3/13 Paul
2/3/13 Paul
2/3/13 quasi
2/3/13 quasi
2/3/13 Butch Malahide
2/6/13

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