The St Petersburg phenomenon. The paradoxes of probability theory. - page 4
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In this thread please post any paradoxes of probability theory you can find.
Gabor Sekei. Paradoxes in probability theory and mathematical statistics.
http://baguzin.ru/wp/wp-content/uploads/2016/05/Габор-Секей.-Парадоксы-в-теории-вероятностей-и-математической-статистике.pdf
But what if it's 0 to 5, or 7, or any other number? You still have to divide by 32768.0. Or are there options?
This is of course the way it should be:
Novaja:
В этой теме прошу выкладывать разные парадоксы теории вероятности, все, какие можно найти.
Mathematical puzzles and fun
This book by famous American science popularizer M. Gardner has a lot of very entertaining problems and puzzles from different branches of mathematics. A successful selection of material, quite unusual form of presentation, and subtle humor of the author will bring great pleasure to a wide range of readers - maths lovers who want to usefully spend their free time.
Source:
Gardner M. 'Mathematical puzzles and fun'. \\Translated from English by Danilov Y.A., edited by Smorodinsky Y.A. - Moscow: Mir, 1971 - p.511
This is, of course, the way it should be:
Without using the remainder of the division.
Without using the remainder of division.
I think that "non-randomness" was only really noticeable when defining even, odd.
If you use a remainder of division from numbers much larger than 2, it is unlikely to be detectable.
The only correct way is to use numbers that are 2 to the power of N instead of 100, e.g. 64, 128, 256...
But rand() reduction to double from 0 to 1 is also normal, because it is easier to understand a random value and most accurate, though a bit slower. But I think we are talking about fractions of a nanosecond, although it may be nanoseconds, because processor needs to switch from int operations to double operations(FPU/ALU mixing), and Renat once said that this is far from free.
In this thread, please post any probability theory paradoxes you can find.
https://oschool.ru/files/studys/55df78bc740d76b70e8b4287/1443032585656.pdf
Have you personally seen a player make a guess but not be given the winnings? I have never seen anyone win, except their accomplices.
Those three cards or three thimbles, according to the theory of probability, are the winning ones for the organizers of the game, and then there is the sleight of hand.
There's no sleight of hand or theory of improbability. :) It's all banal and simple. I know about it firsthand. It was in the 1990s and the scheme was described to me in great detail by a man who was involved in it himself. Now people are not fooled by it; scammers mostly operate online. But the basic principles remain the same. To lure a person in, take advantage of his weaknesses and get money from him, and then, under any pretext, the money is never returned.
The terms of the game are not fully described, as usual.
Is it possible to win more than your bet? If not, there is no point in playing.
Then it should be allowed to win from the pot. So it is enough to make the minimum bet, there will be a win.
Maybe the condition is that you can play 1 time, and here we have to decide on the bet, so that there would be the maximum probability of winning.
Like all "pradlocks" - from incomplete conditions.
Solution:
There's not even a question of betting. Bet 1. No more makes sense, the probability of winning doesn't depend on the size of the bet. So 1, just to start the game.
So what's the paradox? From the assumption that you have to place a bigger bet to increase your winnings? That might be it.
The point of the game: to enter the game a deposit is needed and the coin game lasts until the first eagle appears (a one-time game), 1-doucat wins when an eagle appears in the first roll, if there was a tails and an eagle appears in the second roll, the winning amount is doubled, and so on to infinity, until an eagle appears. Probability of winning 1 ducat-0.5, 2-0.25, 4-0.125 etc., so you can win infinity, and play infinity, if the pot is infinite.
this is a non-linear function.
y=2^x.
and where does Y in a non-linear function tend if X tends to infinity? correct, to infinity.
Therefore, if the number of games = infinity, then the average gain = infinity.
And the problem is also set in such a way that his losses are a linear function (always 25 rubles each), and his winnings are a non-linear function.
You need to plot two functions. First the first graph will be higher than the second, then they will swap places.
It all depends on the number of games.
Exactly right.
But what if we need from 0 to 5, or to 7 or to any other number? We still have to divide it by 32768.0. Or, are there variants?
1. If you are not satisfied with quality of CRT (for example, with low bits), then you need to pass it through AES.
2. The remainder of the division can be taken, but only if divided by powers of 2. Otherwise, no matter how good the GSF is, the output will be nothing like a uniform distribution.
3. With division by double you can get really messed up by their comparisons :-) In most cases you get a small, but skewed output.
to get a random number from 0 to 6 (one of 7):
- consider limit=RAND_MAX - (RAND_MAX%7) ; the maximum multiple of 7 in the range 0..RAND_MAX
- use RDS until we get r<limit ; i.e. if "random number" is higher than limit, we can't do anything with it - we take
- result = r % 7 or (for better or worse) r * 7 / limit
somehow so :-) could be wrong +-1,