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I take issue with the method posed in the article when they discuss the method using a gaussian distribution to sometimes guess a number between the higher and

I take issue with the method posed in the article when they discuss the method using a gaussian distribution to sometimes guess a number between the higher and the lower. The reason I take issue is because the ability to generate a useful distribution requires knowledge of what numbers I pick. A gaussian distribution is generated via a mean and standard deviation. To defeat this system, all a knowledgeable person would need to do is randomly select a mean and a small standard deviation and choose his two numbers from there. Then the probability that the guesser will choose his number R to be within the mean and standard deviation of the one you picked will be essentially zero. To illustrate in a slightly simpler way, in order to choose your two numbers choose a random integer and the same random integer + 2 (2 to at least give the guy a chance). I could choose 1 billion and 1 billion + 2, or one and three. The chance that you would choose R to be 1 billion +1 or two, respectively, is nearly zero. In order to get above 50% you need some sort of extra knowledge. In this case that knowledge comes from the gaussian distribution. Knowledge often affects probabilities even though they dont seem like they should. Another example of this is the Tuesday Birthday Problem [1]. [1] http://scienceblogs.com/evolutionblog/2011/11/08/the-tuesday-birthday-problem/ http://scienceblogs.com/evolutionblog/2011/11/08/the-tuesday...

Someone 12y ago
What is there to take issue with, given that the article states: "Of course if the number writer knows this Gaussian strategy, he can make your winnings as close to ½ as he wants by writing numbers that are very close." It would be very close to ½, but still larger than ½.
- chacham15 12y ago
  Because that statement is backwards. It is not the case that the number writer needs to know about the gaussian distribution, but the other way around. The guesser needs knowledge of the types of numbers that the number writer is choosing in order to create the gaussian distribution.
logicallee 12y ago
Have you reviewed Svedlin's link in this thread to a related puzzle? (I think actually the exact same puzzle.) http://plus.maths.org/content/mathematical-mysteries-getting-most-out-life-part-1 http://plus.maths.org/content/mathematical-mysteries-getting... It's rigorous and from a slightly different approach - what do you think? Also it says that even knowledge of the guesser's exact strategy cannot defeat the solution mentioned there, even if the offering party knows the strategy exactly and is actively trying to break it.
- chacham15 12y ago
  The author of that article makes the same assumption / error that I discussed previously. The trick lies in the statement "Since n>m, f(m) - f(n) is positive and this probability is greater than a half." While this is true, f(m) - f(n) will almost always be very close to zero without additional knowledge. A key statement that illuminates the assumption in his argument is "The bigger the cheque, the more times you toss the coin and the more likely it is that it will come up tails at least once - in which case you keep the cheque rather than swapping." Built into that statement is the assumption that larger numbers are less likely than smaller ones.
  - logicallee 12y ago
    I don't really understand. The argument is just that you can do something to increase the probability of guessing the bigger number from being anything other than exactly, precisely 1/2. For example the author suggests that you first randomize which envelope you choose first. Then you open that envelope. You then flip a coin a number of times that matches the integer in the envelope, and if comes up all heads, then you choose the other one; otherwise you stop on the first one. Obviously if the numbers are anything other than tiny integers, e.g. if they're more htan 20 or so, then for all practical purposes it is impossible that it will come up all heads, e.g. 7,342 times in a row. But in a rigorous, exact way, that probability is not exactly 0. The probability of that many heads is lower for 7,342,352,454,356,234,753,343 than it is for 454,356,234,753,343 (where I truncated a bunch of digits) or even for 7,342,352,454,356,234,753,342 (where I changed the trailing 3 to a 2). Therefore, you are more likely to stay on 7,342,352,454,356,234,753,343 than either of those other numbers. This is purely academic. Clearly, the chances of flipping heads 7,342,352,454,356,234,753,343 times in a row is so, unbelievably 0 it isn't even funny. But it's not actually zero, mathematically speaking. So essentially, you have a greater chance of swapping if the number is smaller, because there is a greater chance that you will flip all heads - even if "greater chance" just means 1 in 2^7,342,352,454,356,234,753,343 instead of 1 in 2^7,342,352,454,356,234,753,344 chance. Do you not agree with this? Or would you still say that it's 50:50 that you will stay on 7,342,352,454,356,234,753,343 or stay on 7,342,352,454,356,234,753,344 (I changed just the trailing 3 to a 4) even if when you draw 7,342,352,454,356,234,753,343 you change your mind 1 out of 2^7,342,352,454,356,234,753,343 times and if you draw a 7,342,352,454,356,234,753,344 you change your mind 1 out of 2^7,342,352,454,356,234,753,344 times? Would you consider the probability "the same" of holding on each one? Or would you agree that in this particular example it is no longer 50:50 exactly, precisely. If you agree with me that it's not technically (pedantically) the same exact probability, then I am not understanding your argument exactly. If you just meant that "for all practical purposes" the probability is the same - then of course I agree with you whole heartedly. Do explain which one you meant!

Someone 12y ago

What is there to take issue with, given that the article states: "Of course if the number writer knows this Gaussian strategy, he can make your winnings as close to ½ as he wants by writing numbers that are very close." It would be very close to ½, but still larger than ½.

chacham15 12y ago
Because that statement is backwards. It is not the case that the number writer needs to know about the gaussian distribution, but the other way around. The guesser needs knowledge of the types of numbers that the number writer is choosing in order to create the gaussian distribution.

logicallee 12y ago

Have you reviewed Svedlin's link in this thread to a related puzzle? (I think actually the exact same puzzle.) http://plus.maths.org/content/mathematical-mysteries-getting-most-out-life-part-1 http://plus.maths.org/content/mathematical-mysteries-getting... It's rigorous and from a slightly different approach - what do you think? Also it says that even knowledge of the guesser's exact strategy cannot defeat the solution mentioned there, even if the offering party knows the strategy exactly and is actively trying to break it.

chacham15 12y ago
The author of that article makes the same assumption / error that I discussed previously. The trick lies in the statement "Since n>m, f(m) - f(n) is positive and this probability is greater than a half." While this is true, f(m) - f(n) will almost always be very close to zero without additional knowledge. A key statement that illuminates the assumption in his argument is "The bigger the cheque, the more times you toss the coin and the more likely it is that it will come up tails at least once - in which case you keep the cheque rather than swapping." Built into that statement is the assumption that larger numbers are less likely than smaller ones.
- logicallee 12y ago
  I don't really understand. The argument is just that you can do something to increase the probability of guessing the bigger number from being anything other than exactly, precisely 1/2. For example the author suggests that you first randomize which envelope you choose first. Then you open that envelope. You then flip a coin a number of times that matches the integer in the envelope, and if comes up all heads, then you choose the other one; otherwise you stop on the first one. Obviously if the numbers are anything other than tiny integers, e.g. if they're more htan 20 or so, then for all practical purposes it is impossible that it will come up all heads, e.g. 7,342 times in a row. But in a rigorous, exact way, that probability is not exactly 0. The probability of that many heads is lower for 7,342,352,454,356,234,753,343 than it is for 454,356,234,753,343 (where I truncated a bunch of digits) or even for 7,342,352,454,356,234,753,342 (where I changed the trailing 3 to a 2). Therefore, you are more likely to stay on 7,342,352,454,356,234,753,343 than either of those other numbers. This is purely academic. Clearly, the chances of flipping heads 7,342,352,454,356,234,753,343 times in a row is so, unbelievably 0 it isn't even funny. But it's not actually zero, mathematically speaking. So essentially, you have a greater chance of swapping if the number is smaller, because there is a greater chance that you will flip all heads - even if "greater chance" just means 1 in 2^7,342,352,454,356,234,753,343 instead of 1 in 2^7,342,352,454,356,234,753,344 chance. Do you not agree with this? Or would you still say that it's 50:50 that you will stay on 7,342,352,454,356,234,753,343 or stay on 7,342,352,454,356,234,753,344 (I changed just the trailing 3 to a 4) even if when you draw 7,342,352,454,356,234,753,343 you change your mind 1 out of 2^7,342,352,454,356,234,753,343 times and if you draw a 7,342,352,454,356,234,753,344 you change your mind 1 out of 2^7,342,352,454,356,234,753,344 times? Would you consider the probability "the same" of holding on each one? Or would you agree that in this particular example it is no longer 50:50 exactly, precisely. If you agree with me that it's not technically (pedantically) the same exact probability, then I am not understanding your argument exactly. If you just meant that "for all practical purposes" the probability is the same - then of course I agree with you whole heartedly. Do explain which one you meant!