What is it? - page 20
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Еще раз повторяю, Мы НЕ производим оценку вероятности выпадения Красного у РУЛЕТКИ по какой-то дискретной серии событий, она уже произведена ДО НАС нашими предшественниками (Лапласом, Бернулли, Байесом), нашей историей, историей выпадения Красное-Черное. Все!!! p=q=0.5 или так
They had a different roulette - a perfect, martingale one. :)
#define p 0.5 THIS IS THE POINT.
Syntax is wrong. Dots are forbidden here. ;)
I'm basing this on centuries and thousands of years of tape measure observations, and the assumption that the roulette table and wheel are perfectly manufactured and balanced. There are no zeros on my tape measure (to keep us from getting even more lost). 36 holes. 18 red. 18 black. That's 0.5 by 0.5.
What year were you born?! G:-oh.
They had a different tape measure - a perfect, martingale one. :)
The syntax is wrong. Dot is forbidden here. ;)
What year were you born?! J:-oh.
And you're still laughing and pussyfooting.... Oh, you're so young. ....
Этот вопрос меня не мучает абсолютно. Меня мучает то, что я не могу математически объяснить свой выигрыш в рулетку (по деньгам), хотя при таком объёме сыграных игр и таком отрицательном мат.ожидании ( 1/37 = зеро ) и таком стартовом капитале (депозите) Мы должны были разориться минимум 6-7 раз. Но этого не произошло.
So you won the lottery, rejoice. Or are you looking for a reason to go again? Then it's time to practice on a coin :)
Who was spinning the roulette wheel, the croupier?
А Вам всё смех..чки, да пиз..хаханьки.... Эх, молодёжжжжжж....
;)
LOL. You too :)))
Now that I've started writing this post, I'll suggest another way.
So, take your correct roulette and spin (remembering to throw the ball) many, many times. Divide ALL the results into a series of 2000 rolls. Calculate the average of the results and, if you've done a good job, get a result close to 1000. This will be the MO estimate of the number of red falls in the series of 2000 roll s. If you keep spinning to infinity, you will get an infinitely close to 1000.
But don't relax! :) Next task will be more complicated. You will have to estimate the number of red hits in 2000 series with the condition that after the first thousand there will be 600 of them. From all 2000 shots you will have to keep only those series with 600 red hits after the first thousand. And there are far fewer of them. So for a good estimate of the MO you will have to spin the roulette wheel not many, many times, but many, many times more. That's your own fault. But here you finally get a fairly large number of these series, calculate the average and... I bet it's a lot closer to 1,100 than 1,000. I'm willing to let you spin the roulette wheel until you get 1000. Or until you agree with me.
You could even practice on a simpler task first. Let it not be 2000, 1000 and 600, but 4, 2 and 2. That is, divide the results of the draws into a series of 4 and select those in which there were 2 reds after two draws. You won't need a huge number of draws for your first decent score, so you can take a coin (if you don't have roulette) and start right away. As before, you can do this until the estimate of MO is close to 2, or until you agree that the MO for that value is 3.
Do you agree?
Should a series of 4 rolls tend to your (or rather your) expectation after two red falls
Candid, you are Ours!!! Or I'm Yours! (if you accept....))
Yes. That's right. I agree with everything in principle. The misunderstanding that got us into a clash is that you are picking out some series of events << ..... That is, divide the results of the draws into series of 4 and select those in which there were 2 reds after two draws..... >>
There is no need to take anything away. (The mathematician above confirmed: There is no "if") We use everything event space. See the example with the second observer, who has no idea about the series and conditions assigned by the first.
I repeat: In my case MO=n*p, and since p=q=0.5 it means MO depends only on n.
And if we take what Avals suggested <<.... If probability of heads/tails=0.5/0.5 and when heads fall, CB=+1, tails=-1, then MO=1*0.5-1*0.5=0 ...>> then there will be no problem with understanding each other at all: MO=0. ALWAYS.
If no one has any objections, then I'll continue later....
P.S.
Well at the right university I think you can get one. Maybe you should really go and study.
Not up for discussion. I try to learn every day, and enjoy it like a child, if it really works.
And I try to learn from you too. Do you feel responsible?
I wanted to write about my "university", but why? It's trivial somehow....
And give me a link to support your assertions... What they teach in a suitable university, you can find online in two clicks. If you are very tired - tomorrow. But only necessarily.
So you won the lottery, be happy. Or are you looking for a reason to go again? Then it's time to practice on a coin :) And who was spinning the roulette wheel, the croupier?
Once again (for those who are not with us...). Our (since there's more than one person involved) result of playing roulette in a casino, cannot be equated with winning the lottery (see here https://www.mql5.com/ru/forum/122871/page14#254008).
Since the target audience on this forum is looking at Forex. I suggest this interpretation of these results:
-- open randomly buy or sell, TP=SL=700pp (5 digits), take lot = 0.1, spread = 20pp, pair let EURUSD, depo = 7,000$ (one hundred bets - one hundred suckers in a row).
-- net MO = 0, minus spread = $2 from any position. So it takes 3500 trades (on average. But "as luck would have it") to bring down this kind of deposit.
-- Was opened about 30000 positions (the real number on the roulette wheel), ie 6-7 depo or $ 60 000 expected minus.
-- result -- positive Balance ~ $10,000 - $12,000, the lot ranged from 0.1 to 0.5
..........
It's actually quite difficult to translate. There are a lot of nuances. But in essence -- didn't lie.
..........
What's the conclusion?
And for lottery buffs, the question -- When was the last time they won a lottery like this?
Как Вы эту цифру ни назовете, - матожиданием, прогнозом или еще как...
Encouraging a Babylonian pandemonium?
Well, well. Then what's left to do on a thread like this - flub and arrogantly preach?
I'll try it myself. :o))
I remember that in a proper discussion the characteristics of a random variable (a priori known all the more...) must not be substituted by their experimental counterparts.
Let me remind you ;) for probability - frequency, for MO - mean, for variance - RMS.
And replicas like.
And Michurinsky's is about as good as it gets.
Have you tried running for chief of Michurinians? You'd have a good chance with the real Michurins.
... junkie.
As a junior and inquisitive naturalist, that's insulting!
;)
No, it's not. I'm stumped. How do I get my point across? Read again about the original problem https://www.mql5.com/ru/forum/122871/page14#254008
and her interpretation of the slingshot https://www.mql5.com/ru/forum/122871/page16#255508
read it.
And where did you get this from:
II) With large number of trials n, number of events A will tend to n*P(A) -- I understand and accept.
There is no such a thing. The number of events A can deviate any far from n*P(A). Look up laws of arcinus. http://polbu.ru/safonov_dealing/ch61_all.html
Кол-во событий А может как угодно далеко отклоняться от n*P(A).
О))))
О))))
Yeah, whatever. You increase the number of throws, the possible deviation increases. In the limit, infinity can deflect infinitely far ;)
Of course, I do not think that with 10 coin tosses (heads=+1,tails=-1) the cumulative sum will go more than 10 away from the origin O))))