Not the Grail, just a regular one - Bablokos!!! - page 225

 
b2v2:

Trouble with theorist:)
After any combinations on random numbers there will be an equal probability (about 25% of the time) of TT, TH, HT, HH.
A coin cannot have a memory.


What does theorist say about this?)

I.e., of course, the bottom combination will meet earlier... But if part of that combination has already come, there remains, as you correctly said, 25% for each deuce. But earlier we figured out which FULL combination will come EARLY...

 
qimer:


And what does the theorist say about it?)

I mean, of course, the bottom combination will occur earlier... But if part of that combination has already arrived, there remains, as you correctly said, 25% for each deuce. But earlier we figured out which FULL combination will come EARLY...

The coin gives a continuous sequence of HTTHHHHHHHTTTHTHTHTHTHTHH, etc., so it is in this sequence that the probability of HHTT appearing is greater than the probability of HHNH appearing. Why? Because, often the appearance of combinations of HHTT is "killed" by combinations (HHTT, HHTT, HHTT), when the first two reports ( in 75%), and when "killing", often falls "more likely" combination. So at the expense of this "kill", the probability of the "more likely"increases . There is nothing supernatural, you just have to understand...

I hope I've made myself clear.

Try to calculate the probability of the combinations falling out if the combinations are not looked for in a continuous sequence, but count the combination in four, i.e. HTTH, HHTH, THHH, TTTH, TTHH

 
Talex:

Try to calculate the probability of combinations if the combinations are not searched in a continuous sequence, but count the combination in four, i.e. HTTH, HHTH, THHH, TTTH, TTHH


So we're not looking for fours, we're looking for twos in a continuous sequence ... and the HH pattern will be followed by 60% of the TH and TT patterns... Wrong again?

Even if you divide the patterns into individual dropdowns it's still the stuff (supposedly) supposed to work, but it doesn't... I'm not arguing here, but I don't understand why))

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I figured out what the mistake was

 
I checked the filter according to the algorithm described by User in this thread, and it didn't work for me personally.

I also checked it, using this scheme:

1 We have a sequence of coin tosses, for example:
0 1 1 0 1 1 1 0 0 0 0 0

2. Convert it to spins of three, we get:
3 3 4 0

3. By inequalities we get:
|x3 - x2| - |x2 - x1| = |4 - 3| - |3 - 3| = 1 > 0
|x4 - x3| - |x3 - x2| = |0 - 4| - |4 - 3| = 3 > 0,
Therefore the next increment must be less than zero, and this can be at x5 = (0 ... 3)

(4) Since in (2) we assigned numbers to combinations at random, we repeat the previous items
n!-time, hence 8! = 40320 times and find all falls where the conditions of two
in a row with one sign. For this example we obtain the following vector in which all discontinuities are reduced to
to an initial permutation:

0 - 18720
1 - 12720
2 - 12720
3 - 13440
4 - 0
5 - 12720
6 - 12720
7 - 12720

5. Move back to the next coin flip. For each number in the vector:
Number of outcomes with next zero (all combinations where first 0): N0 = 18720 (000) + 12720 (001) + 12720 (010) + 13440 (011) = 57600
Number of outcomes with the following one (all combinations where the first one is 1): N1 = 0 (100) + 12720 (101) + 12720 (110) + 12720 (111) = 38160

6. Hence probability 0:
P0 = 57600 / (57600 + 38160) = 0.601
Probability 1:
P1 = 38160 / (57600 + 38160) = 0.399

7. Conducting a test, the betting is done only with probability 0 or 1 greater than 60%:
Total: 14,774 - guessed hits

14,420 no hits, which is the same 50% / 50%.

I also tried another way - the result is the same.

That is, the probability of about 75% - gained through betting on all numbers at once 100%, in other cases, it is proportional to the number of numbers, and the average

We get 75%.

 

I think it's time to end the Penny game. It's long been understood.
Joker has already said that it is effectively unused.

You can still read a bunch of literature for educational purposes (including wikipedia links).
The games are fundamentally different:

1. Flip 2 times and look - 25% HH, TT, TN, NT are equally likely. Game ends after two rolls always.
2. Bet on HH and HT and play to a particular combination. The game may take longer. Pay attention in the Java game to the parameter - average duration of the game. And maximum duration is also an interesting parameter.

You can play the second game only by screwing analogue of martini to take your two coins in such series TTTTTTT.... TH.
Well martinis can be screwed to a lot of things, but the disadvantages are known.

The moduli of the increments are the same.

I'd be happy if someone with an IQ equal to Perelman could convince me otherwise.

 
b2v2:

I think it's time to call it a day with the Penny game. It's been clear for a long time.
Joker has already said that it is effectively unused.

It was just interesting to try it - what if.
b2v2:

Pay attention in the Java game to the parameter - the average duration of the game. And the maximum duration is also an interesting parameter.

There's a bug in the game - with HH vs. TH - you get an average series length of 2, apparently something to do with rounding.

Although the correct series length would be:

HH - 0.25 * 2 = 0.5

TH - 0.25 * 2 = 0.5

HTH - 0.125 * 3 = 0.375

TTH - 0.125 * 3 = 0.375

HTTH - 0.0625 * 4 = 0.25

TTTH - 0.0625 * 4 = 0.25

HTTTH - 0.03125 * 5 = 0.15625

TTTTH - 0.03125 * 5 = 0.15625

And so the sum of the series tends to 3. Other than that, I agree.
 
Achernar:

Checked it too, with this pattern: ...


Yes, I also checked it like this. Checked it with real data. The response was 50/50 with >=67% probability (according to the filter). Hence, all this talk about the reality of the filter is bullshit. Penny's game should really be finished.

 

The only thing perhaps (in terms of incremental and penny modules) may be due to the non-randomness of the quotes.
The distribution on the forex is not binomial slightly. Who knows, you can run a tic-tac-toe indicator (red/green) and count the number of series.
The probability of reversal/continuation is not always 50/50.

 
b2v2:

A coin can't have a memory.


Exactly. Imagine a coin flipped forty times in a row with heads. Which is more likely in the 41st flip: heads or tails?

Most people would say tails, mathematicians would say it's the same. In fact, the odds of it being heads are greater.



 
A normal mathematician also realises that on an eagle roll 40 times a coin is almost guaranteed with two eagles:)