in reply to
Spooky math problem
Another approach: There is some additional information needed. 1) The smallest size per digit you are capable of writing the numbers, 2) the largest envelope size you are capable of acquiring, these collectively limit the number of digits the remaining number can have. For the sake of argument lets call it D. If the numbers in envelopes are M and N, with M being the one handed to you, then the probability that N < M is p=(M1)*10^(D). Because D is arbitrarily large, p is arbitrarily small.
Re^2: Spooky math problem by tilly (Archbishop) on Aug 30, 2005 at 17:17 UTC 
First of all, no additional information was needed.
Secondly you are implicitly assuming a probability distribution on the numbers in the envelopes, namely that it is evenly distributed among all possible numbers that could be written on the pieces of paper. This assumption is both wrong and unnecessary.
Thirdly note that the technique must work no matter what pair of numbers happen to be in the envelope. Creating a technique which will work for 90% of the pairs that you think could be there won't cut it. It has to be all pairs.  [reply] 

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If you want to throw around ad hominem insults, nobody can stop you.
That doesn't make them justified.
For the record, I am very aware that one cannot define things like expected values and probabilities without a probability distribution. I am also painfully aware that many things that look like they might reasonably make probability distributions (eg a uniform distribution on the real numbers), don't.
But the problem that I'm giving here can readily be precisely worded in a way that entirely avoids those issues. Here is the precise wording:
Suppose that you have 2 different numbers x and y in 2 envelopes, written down in decimal form. Let us define the following experiment. You will randomly hand me one of your two envelopes, I will look at it, and I will tell you whether I think you handed me the larger. Is there an algorithm that I can use, which guarantees that the probability of my being right, given x and y (which I do not know) and my algorithm, is strictly better than 50%?
Note the following critical details:
 The numbers x and y are part of the experiment. How they came to be is not part of the question asked, and therefore questions about how to choose them do not enter into the problem.
 I lack sufficient information to calculate the probability of being correct. In particular I don't know what x and y are.
 The algorithm that I use must work no matter what x and y happen to be. If my technique depends on assumptions about x and y, then I have not succeeded because you might have a pair that my technique does not work for.
If you relax any of these conditions even slightly then the result tends to quickly becomes either false or undefined. The result is very fragile and precise.
Now I won't go through the full reasoning again here. But if you're interested, this was discussed extensively on sci.math over a decade ago. For a particularly clear explanation, see this post by Bently Preece.
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