Say,
I am betting on Red only and targeting two consecutive wins on that. What is the longest stretch of no such hits on any EC, this way?

If you learned even some BASIC programming you could run analyses and derive your own answers to these sorts of questions.

Tutorials are available online if you search for them.

By the way, I use JustBASIC.  But there are others.

Or -- shock...horror...-- pay one of the forum's programmers out of all your winnings, to do it for you.

cute question albahala

Quote from: Albalaha on December 23, 2013, 03:51:51 AM
Say,
I am betting on Red only and targeting two consecutive wins on that. What is the longest stretch of no such hits on any EC, this way?

Dear Sumit, the expected rate's Math is exactly as a single occurrence. My best guess is it's around the "less-than-twenty-times the cycle universe" rule*, which propagates throughout the game.

So with an universe of 2, you have 2 per cycle x 20 = at most 39 times ever recorded in live game conditions.

---

*: Please note in simulated TRNG anything goes; that "rule" (with quotes) goes for the regular player facing regular game speed in casino conditions.

20-30 is the longest

Quote from: Tomla on December 23, 2013, 07:04:51 AM
cute question albahala

You either do not know how to pronounce my username correctly or use incorrect one willingly.
Anyways, nice to see you back.


@Vic
            I have seen about 30 in zumma baccarat itself. Thanks anyways for your answer.

Quote from: Albalaha on December 23, 2013, 03:51:51 AM
Say,
I am betting on Red only and targeting two consecutive wins on that. What is the longest stretch of no such hits on any EC, this way?

81 spin

Quote from: Drazen on December 23, 2013, 09:45:06 AM
81 spin

In how big data you got this much extreme?

Many many spins

Well there is a formula with which you can calculate longest losing sequence for anything in the game if you know z-score.

As we know highest z-score for standard EC-s is about 5 or slightly over so it isn't hard to calculate how bad losing sequence it can get.

Cheers

Drazen

Here's a simple task by which you can determine just how long the red and black can go without hitting, or without hitting twice in a row.


Take any natural number (not zero) N  and divide it by 2.  Continue dividing by two until you reach zero.

Now, it's very very important that you count the number of steps that it took to reach zero.   The number of steps that it takes you to reach zero will match the maximum number of times that the red/black can go without hitting.  You can NOT use a calculator.  You must use pen and paper. 


By the way, in order to find the real maximum number of times, you must continue dividing by 2 until you actually reach zero.  If you still have a decimal with numbers remining, then you're not quite there, so continue dividing by 2.

Good Luck! 

Quote from: Xander on December 23, 2013, 07:00:20 PM
Here's a simple task by which you can determine just how long the red and black can go without hitting, or without hitting twice in a row.




Take any natural number (not zero) N  and divide it by 2.  Continue dividing by two until you reach zero.


Now, it's very very important that you count the number of steps that it took to reach zero.   The number of steps that it takes you to reach zero will match the maximum number of times that the red/black can go without hitting.  You can NOT use a calculator.  You must use pen and paper. 




By the way, in order to find the real maximum number of times, you must continue dividing by 2 until you actually reach zero.  If you still have a decimal with numbers remining, then you're not quite there, so continue dividing by 2.


Good Luck!

I beg to differ. This is a classic case of mistaking the map for the territory. Notice that Xander doesn't use the standard probability formula for calculating how many times R/B can go without hitting, but instead chooses to use a process which doesn't even give you the corresponding probability, so you can't see just how likely it is to get (say) 100 reds in a row.


The point he's trying to make is that because you will never actually reach zero by this method, so you will never come to the end of a losing streak, but never mind that the odds of even 50 reds in a row is so small as to be not worth worrying about.


You have to keep in mind that equations are just models and abstractions of reality - approximations which give useful results within certain parameters. To take an equation and push it to its limits, then claim that what the equation tells you is also an accurate representation of what happens in reality, can very often result in absurdity.


You see the same thing happening in quantum physics; just because the formula tells you that there is a non-zero probability that the computer you are now looking at COULD possibly vanish into thin air and be replaced by a bunch of bananas, doesn't mean that such a thing will ever happen, but because the almighty MATH says it might, we can't deny that it's possible. If you think about it, this is actually insane; it's giving the model priority over the reality that generated it.


Happy Christmas!

i confuse easily.....


isn't the question how long will an ec not hit twice


Does that mean


R B B B R B R B R B B R B B R B B B B B B B B R B R   

Yep, but Xander said:

QuoteHere's a simple task by which you can determine just how long the red and black can go without hitting, or without hitting twice in a row.

In other words, he's making no distinction between how long it can go before red hits either once, or twice in a row, or indeed anything else. The answer is the same in all cases: there is no limit (because the math says so).

Quote from: Turner on December 25, 2013, 01:50:04 PM
i confuse easily.....


isn't the question how long will an ec not hit twice


Does that mean


R B B B R B R B R B B R B B R B B B B B B B B R B R   

Exactly.

Quote from: Bayes on December 25, 2013, 02:27:07 PM
Yep, but Xander said:



In other words, he's making no distinction between how long it can go before red hits either once, or twice in a row, or indeed anything else. The answer is the same in all cases: there is no limit (because the math says so).

Confusion. What is your answer to my query, Bayes? I heard you saying that you do play such bets too.

You can arrive to 2 different results by following different paths.

I consider the following universal and more accurate way to estimate virtual limits.

If a single number could be absent for a maximum of 18 x 36 = 666 spins,
then all the rest derive from it;

666 : 2 numbers = 333 spins
666 : 3 numbers = 222 spins
666 : 4 numbers = 166 spins
666 : 5 numbers = 133 spins
666 : 6 numbers = 111 spins
666 : 7 numbers = 95 spins
666 : 8 numbers = 83 spins
666 : 9 numbers = 74 spins
666 : 10 numbers = 67 spins
666 : 11 numbers = 61 spins
666 : 12 numbers = 55 spins
666 : 13 numbers = 51 spins
666 : 14 numbers = 48 spins
666 : 15 numbers = 44 spins
666 : 16 numbers = 42 spins
666 : 17 numbers = 39 spins
666 : 18 numbers = 36 spins

Thus for EC's the virtual limit is 36 times to hit once, for twice in a row multiply 36 by 2 equals 72 times before you have your EC hit back to back.
If you wonder why 666 instead  of another arbitrary total the answer is:
0+1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+221+22+23+24+25+25+26+27+28+29+30+31+32+33+34+35+36 = 666

The other way is based on empirical observations since the recorded history of roulette, according to the most ever witnessed is 26 times to hit once, for two consecutive hits multiply 26 by 2 and we arrive to 52 bets/trials before success. 

However, if you adapt the mentality of 'how many times this could miss', then you are moving towards the wrong direction...