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].
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.
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.
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!
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%.
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.
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.
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.
For me, the fun part of articles like this is convincing myself it could never work, then re-reading the article a few times and discovering that it does.
That being said, when they talk about the marriage problem I wish they wouldn't talk about it in terms of "at most 100 candidate spouses". Not even mathematicians start life saying "I will have 100 potential partners in life".
The "marriage" problem is a strategy for selecting the best item amongst k items. The strategy for finding the best item full stop hinges on how the k items are selected. Both in marriage and hiring (the two obvious places to apply this sort of thing), where the pool is drawn from has a much bigger impact than getting the best person in the pool.
Take hiring as a better example than marriage (because with jobs it often the case that there are a known, finite number of applicants). Most of them will be people who are difficult to employ, floating around in the job market. If you wanted the real 'best', you need an aggressive poaching strategy rather than a best-in-the-pool strategy.
Has anyone tried gambler's ruin in real life? Theoretically, you have the best odds of doubling your money if you put all your money down on one hand.
Given most casino games, your odds of winning a single game is less than 50%. So lets say you encounter the best case; your odds of winning are 49%. Even then, for every game you play, you are more likely to lose money.
Probability of winning = .49
N = number of times you play
.49^N will become a smaller number as N gets bigger.
This is described in "the Newtonion Casino" (an amazing book about a bunch of people who build 8080 computers into their shoes to beat roulette wheels).
Roulette is rigged in such a way that the best strategy is to not play. But, after that, you take all the money you would ever play on roulette and put it on a single number.
They talk about some of the other systems. They also talk about the skill of the croupiers to throw the ball such tha they can predict roughly (enough to beat the odds) where it'll land.
