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I am coming from an information theoretic standpoint. Let me start over: imagine the number writer chooses ANY two numbers completely at random. Knowing what on

I am coming from an information theoretic standpoint. Let me start over: imagine the number writer chooses ANY two numbers completely at random. Knowing what one number is gives NO information about the second number since they were both chosen completely randomly. Therefore, the chance that the second number is larger is equivalent to the number of larger numbers divided by the total (i.e. 50% because a constant number is infinitesimally small compared to the infinity that lies on both sides). Therefore, to suggest that the probability can be swayed one way or another has to violate an assumption that I made in this argument. The assumption that the article makes which violates my argument is the assumption that larger numbers are less likely to be chosen by the number writer than small numbers. It is that crucial piece of knowledge that allows the probability to sway from 50%. That is the whole purpose of the coin flip or the gaussian equation: to reflect that piece of knowledge in your guesses. To take the idea to the extreme, if I know that there are only two numbers then I can win 100% of the time. In conclusion, the point is that this method doesnt work in all, or even most cases; only in cases in which you KNOW that larger numbers are less likely to be chosen than small numbers. Furthermore, how much better than 50% you can do depends completely on how well your approximation function (the gaussian function or flipping coins) accurately reflects the smaller probability of larger numbers. Now if you want to deal with people, the number of ways in which people generate random numbers is so large that I am hard pressed to believe that you can find an accurate enough approximation function that will work across the population. For example, try playing this game with your 12 year old nephew who just learned about million, billion, and googleplex and you are likely to do no better than 50%.

logicallee 12y ago
http://pastebin.com/akQRtDH3 http://pastebin.com/akQRtDH3
- chacham15 12y ago
  Your "base case" is correct, no argument there and so is your secondary example. The error comes in by looking at specifics instead of the whole. The probability of success using a strategy is the sum of (the probability of success in a case multiplied by the probability of that case) for every case. Now, since the set of integers is infinite, the sum probability of having the first number you see be less than one googleplex if the number is chosen completely at random is 0 (not just "really small" like 1/2^googleplex, so small that it is zero in the sense that lim(x->inf)1googleplex/x IS zero). Therefore the chance that you will get numbers which will lead you to have a probability over 50% is 0.

logicallee 12y ago

http://pastebin.com/akQRtDH3 http://pastebin.com/akQRtDH3

chacham15 12y ago
Your "base case" is correct, no argument there and so is your secondary example. The error comes in by looking at specifics instead of the whole. The probability of success using a strategy is the sum of (the probability of success in a case multiplied by the probability of that case) for every case. Now, since the set of integers is infinite, the sum probability of having the first number you see be less than one googleplex if the number is chosen completely at random is 0 (not just "really small" like 1/2^googleplex, so small that it is zero in the sense that lim(x->inf)1googleplex/x IS zero). Therefore the chance that you will get numbers which will lead you to have a probability over 50% is 0.