What is it? - page 14
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You don't have to move anywhere. It is simply a series of Bernoulli tests, with their inherent laws. Yes, with probability p=0.5 a result of 600 over 400 is indeed unlikely, but not at all from the series of impossible. But if a series of 10000 trials results in 6000 on 4000, this is where you have to think hard because it will be almost 100% non-random deviation from the expectation (although the success rate will be the same, 60%).
You don't have to move anywhere. It is simply a series of Bernoulli tests, with their inherent laws. Yes, with probability p=0.5 a result of 600 over 400 is indeed unlikely, but not at all from the series of impossible. But if the series of 10000 tests contains 6000 for 4000, this is where you need to think seriously because it will be almost 100% non-random deviation from the expectation (although the success rate will be the same, 60%).
6000 vs 4000 at 10000 is understandable. We won't go beyond normality.
Once again the same question, but put it in a different way.
We create a new object - a system of events (e.g. roulette). There are no zeros. Red/Black - 50/50. We have done 1000 trials. Event A1 happened (one event) in which Red fell out 600 times, Black fell out 400 times. Correspondingly, there is extremely small, but admissible P(A1) for example = 0.0001.
That's it, we have forgotten about this thousand tests. We start with a clean slate.
Question: With next 1000 trials (in the same system) the probability of which event is more - A3={Red falls out 600 times, Black falls out 400 times} or A4={Red falls out 400 times, Black falls out 600 times}
Or P(A4)=P(A3) ? How to calculate it according to Mr. Bernoulli's scheme?
The probabilities are equal because the probabilities of the elementary outcomes (red/black) are 0.5. Let me find the formulas. Here:
Theclassical formula for the probability of k successful outcomes in a series ofn trials in the Bernoulli scheme is as follows (the probability of success is p) :
In your case it is simpler, because p=q=0.5.
But usually people are not interested in the probability of the outcome {600, 400}, but, say, the probability that the next series of trials will fall at least 600 on red. The corresponding sum will be obtained.
Получится соответствующая сумма.
... which, by the way, is convenient to calculate approximately, using tables of Gaussian distribution - it approximates very well Bernoulli at large n
or rather not Bernoulli, but binomial
The probabilities are equal because the probabilities of the elementary outcomes (red/black) are 0.5. Let me find the formulas. Here you go:
OK. P(A4)=P(A3). And the theorem is just right. And tables are sometimes necessary. But...
Try to understand me, put yourself in my shoes. Otherwise you won't be able to explain anything. Try to forget TheorWer, which you (this is a reference to all) studied perfectly (or not quite) in your time.
So, again. Create a new object - an event system (e.g. roulette). There are no zeros. Red/Black - 50/50. Made 1000 trials. Event A1 has occurred (one event) in which Red has fallen 600 times, Black has fallen 400 times. Correspondingly there is extremely small, but acceptable P(A1) for example = 0.0001, i.e. is in the region of third sigma (in our case already further).
Now (if you will) calculate probabilities and get that P(A3) ={the next series of 1000 trials will fall not less than 600 for red} equals P(A4)={the next series of 1000 trials will fall not less than 600 for black}.
I.e. we get equal probabilities that the other theorem works or does not work
Because at event A4 quantity Red = quantity Black (deviation 0 RMS), and at event A3 quantity Red = 1200, quantity Black = 800 at n = 2000. I.e. SV deviated by 9 RMS.
Contradiction however .....
............
ps I am writing at work, so there may be some inaccuracies, but the point is correct.
There are many paradoxes in terver. Your paradox looks quite plausible. True, the deviation is not 9, but only 4.5 brackets, but that's not the point.
Let's clear up the confusion in the notation of events.
A1 = {600K, 400Ch in series 1}
A2 = {600K, 400F in series 2}
B2 = {400K, 600F in series 2}
A3 = A1 && A2 = {(600K, 400F in series 1) AND (600K, 400F in series 2)}
A4 = A1 && B2 = {(600K, 400F in series 1) AND (400K, 600F in series 2)}
Yes, the probabilities of A2 and B2 are equal. But where did you get that the probabilities of A3 and A4 are equal?
In short, I don't know how to reassure you yet. If you're that bothered, try reading some of the classics, say Feller. There's also a classic book on the paradoxes of terver, but I can't remember the author.
Т.е. мы получаем равные вероятности того, что работает или не работает другая теорема
так как при событии A4 кол-во Красное = кол-ву Черное (отклонение 0 СКО), а при событии A3 кол-во Красное = 1200, кол-во Черное = 800 при n = 2000. Т.е СВ отклонилась на 9 СКО.
Противоречие однако ....
You have calculated the RMS incorrectly, for this process it is proportional to n. After the second series of tests, the relative deviation from the expectation has decreased.
There are many paradoxes in terver. Your paradox looks quite plausible. True, the deviation is not 9, but only 4.5 brackets, but that's not the point.
Let's clear up the confusion in the notation of events.
A1 = {600K, 400Ch in series 1}
A2 = {600K, 400F in series 2}
B2 = {400K, 600F in series 2}
A3 = A1 && A2 = {(600K, 400F in series 1) AND (600K, 400F in series 2)}
A4 = A1 && B2 = {(600K, 400F in series 1) AND (400K, 600F in series 2)}
Yes, the probabilities of A2 and B2 are equal. But where did you get that the probabilities of A3 and A4 are equal?
In short, I don't know how to reassure you yet. If you're that bothered, try reading some of the classics, say, Feller. There's also a classic book about the paradoxes of terver, but I can't remember the author.
Thank you at least. Although it's not a fact either, because by events A3 and A4 I meant
Does it hurt? I don't know. I used to get meetings with TV professors, heads of departments at reputable universities, and what? Either they told me that I did not understand anything, or (those who tried to get into it) they just threw their hands up.
Probably, many people think that the situation we are discussing is an off-topic. But it is not.
The situations are essentially the same. The money is "in the plus", and pips (for the topicstarter), or the balance of the Player with a negative Calculated Expectation (in my case) is "minus".
Where does the profit come from? We must find an answer. Otherwise why did we come here?
Did you miss my post or didn't you understand it? :)
The fact that after the second series of tests the deviation in units of RMS (or rather RMS expectation) for A3 has decreased (relative to A1) and means that very "aspiration". Notice, it decreased even with a very unlikely and unfavourable outcome of the second series. Better calculate the probability ratio to increase and decrease the relative deviation from MO in the second series.
There are many paradoxes in terver. Your paradox looks quite plausible. True, the deviation is not 9, but only 4.5 brackets, but that's not the point.
Indeed, not the point and decided not to elaborate. But since a second rejoinder has appeared that my calculation is wrong, let's check our Chimes.
My calculation was as follows. (Marking of events according to Mathemat)
.......
After 1st series n = 1000 A1 = {600K, 400Ch in series 1} MO=500 Disp= 1000*0.5*0.5 RMS=15.8 3*SCO = 47.43 Deviation(A1)=(600-500)/15.8=6.32
After 2nd series n = 2000 A3 = A1 && A2 = {(600K, 400Ch in series 1) AND (600K, 400Ch in series 2)} ...........................................................................................
..................................................................................... MO=1000 Disp= 2000*0.5*0.5 RMS=22.36 3*SCO = 67.08 Deviation(A3)=(1200-1000)/22.36=8.94