Martingale: the maximum possible chain of continuous losses/profits - page 9

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goldtrader:

Print out every trader with the 48th Bold and hang it above your monitor!

And in general it feels that the subject of martin and lock is as eternal as the market itself.


Internet generation - they want freebies:)

I got sick of martin a long time ago (of course the last one was a drain). (Of course I lost the last one.) Locke, thank goodness, I have avoided it)))

 
maxfade:

There is a number in the region of 13-14-15 continuous occurrence of one of the events, let 15 be an example, let the events be 0 and 1, and let them be equal probability (if not, then why this whole circus)

The probabilities of 11111111111111111 (15 units) and 11111111111111110 (14 units + 1 zero) are equal (=1/(2^15)) (as well as any other combination of 2^15 pieces)

After we "observe" 14 units in a row (oh, miracle! - probability = 1/(2^14) ), what is the probability of "1" in the 15th event?



My question was different...

Your reckoning is as old as time.

 
 
Mischek:


:)))
 
Mischek:


but if he'd killed her 20 years ago... :)
 
sever30:


My question was different...

what page is the question on?
 
maxfade:
what page is the question on?

first post
 
sever30:


My question was different...

Your little story is as old as time.

You're right. According to this counting, every time you enter a forest, you can meet a bear. They use martin secretly, but they do, and they use it on the pumas.
 

if you run this test many times, the maximum possible series length will slowly tend towards 1000000

 
Tantrik:
You are right. According to this saying, every time you enter a forest, you may encounter a bear. They use martin in secret, but they do, and they use it on the pumas.

And some bank did, just recently lost everything almost to zero, it was on the news.
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