How to gamble if you must – the mathematics of optimal stopping
americanscientist.org
americanscientist.org
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...
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.
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!
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%.
"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.
http://play.golang.org/p/E1L9spOjBk
Cool.
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.
http://www.theatlantic.com/health/archive/2012/08/better-tha...
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.
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.