A probability theory problem - page 2
You are missing trading opportunities:
- Free trading apps
- Over 8,000 signals for copying
- Economic news for exploring financial markets
Registration
Log in
You agree to website policy and terms of use
If you do not have an account, please register
What are the dependent events: there are three balls in a bag, two of them red, one blue. The probability of getting the blue ball out on the first try = 1/3, the probability of getting the red ball out = 2/3. Let's say the red one is taken out, and there are two balls left. Now the probability (already conditional probability UW) to pull both red and blue balls = 1/2.
Ooooh, what a topic... As an old card player, it's a sin for me not to put in a word.
Let's rephrase the question: there is a bag, it contains three balls, the balls may be red or blue, but how many are in the bag is unknown (we don't know what candles the market has in stock). We have already drawn two balls, both red. The question is, which ball is left in the bag, or rather, what are the chances of it being blue/red?What are dependent events: there are three balls in a bag, two of them red, one blue. The probability of getting the blue ball out on the first try = 1/3, the probability of getting the red ball out = 2/3. Let's say the red one is taken out, and there are two balls left. Now the probability (already conditional probability UW) of pulling both the red and blue ball = 1/2.
Ooooooh, what a topic... As an old card player it's a sin for me not to insert my word.
Let's rephrase the question: there is a bag, it contains three balls, the balls may be red or blue, but how many are in the bag is unknown (we don't know what candles the market has in store). We have already drawn two balls, both red. The question is, which ball is left in the bag, or rather, what are the chances of it being blue/red?timbo, why are you twisting things around?
Originally I wrote that it's just an example of dependent events and it's completely inapplicable to financial markets.
The author's problem statement exactly referred to dependent events.
In financial markets we deal with independent or weakly dependent events.
We have already drawn two balls, both are red. Attention question - which ball is left in the bag, or rather what are the chances of it being blue/red?
And the answer to your question is "almost 0.5."
Why almost? Because the eventsare"almost independent" and after 5 or 9 white candles the probability of the 6th or 10th white candle will still be a little bit lower than 0.5
And the answer to your question is "almost 0.5".
I'm not twisting the subject, I'm developing it. I'm just taking your example and giving the next one.
By the way, the answer is wrong.
And the answer to your question is "almost 0.5".
I'm not twisting the subject, I'm developing it. I was just taking your example and giving the next one.
By the way, the answer is wrong.
OK, give me the right one and argue it.
ZS I would put it this way: the more consecutive white/black candles you have, the lower (less than 0.5) the probability of the next white/black one. But I don't see how that probability can be expressed in numbers without statistical research.
OK, give the correct one and argue it.
I would put it this way: the more white/black candles in a row, the lower (less than 0.5) the probability of the next white/black one. But I don't see how that probability can be expressed in numbers without statistical research.
You have to look at the problem globally, not locally. The right question is not which ball is next, but which ones are even in the bag. If two red ones have already been drawn, then originally there were either three red ones in the bag, or two red ones and one blue one. Now, let's estimate the probability of pulling two red balls in each of the scenarios.
If there were three reds, the probability of getting two reds in a row is 1, and if there was one blue, the chances of getting two reds in a row is only 1/3. The odds of the sampler (two balls) are the same as the odds of the whole set (three balls), i.e. the odds of there being a red ball are three times higher than the odds of a blue ball.
This is the classical problem about the CV (conditional probability) of dependent events from the classical theorist.
Unfortunately, it is of no practical use in the financial markets.
This is the classical problem about the CV (conditional probability) of dependent events from the classical theorist.
Unfortunately, it is of no practical use in the financial markets.
Why not? "A trend is more likely to continue than to change direction" is classic.
If you count the candles over a long enough period, you get about evenly split. And in general it seems that the chances of an up/down candle are 50/50. However, somewhere there is a thicker candle up and somewhere there is a thicker candle down. So, we should find this dense candle and open in its direction. I.e. "follow the trend" is another platitude.
It's getting closer to the case, but I don't think that's where we need to dig. I am currently processing the data and if there is anything of interest I will post it here.