Math Quizs

[Blue_Angel]

What is the probability for a streak of 6 to occur before a streak of 3 ?

What is the probability for a streak of any range to be exactly the same range on the next streak ?

[RouletteGhost]

Do you mean for example

B
B
B
B
B
R

Just occured. Now whats thr chances for B to streak 5 times again?

[Blue_Angel]

Do you mean for example

B
B
B
B
B
R

Just occured. Now whats thr chances for B to streak 5 times again?

Example:

B
B
B
B
B <------ 5 streak
R <------ end of streak for blacks
R <------ 2 streak
B <------ end of streak for reds, no bet
B <------ streak of 2, no bet
B <------ streak of 3, no bet
B <------ streak of 4, no bet
B <------ bet for black in order not to have same streak range as before
B <------ streak of 6, result +1
R <------ end of streak for blacks, no bet
R <------ streak of 2, bet for red in order not to have same streak range as before
B <------ end of streak for reds, result -1
B <------ streak of 2, no bet
B <------ streak of 3, no bet
R <------ end of streak for blacks, no bet
R <------ streak of 2, bet for red in order not to have same streak range as before
R <------ streak of 3, result +1

[Blue_Angel]

Another example coming from the 1st question;

I select 3 ECs, let's say that these are Red, Odd and High for the examples sake.
Now I'm going to bet all of them simultaneously and each time I'm losing, I double my next bet. (yes Marty!)

The bankroll should be sufficient for 6 steps of doubling up.
Here comes the tricky part, I'll bet this way until one of the selected EC's has 3 losses in a row, at that time first I'm going to finish the progression with a win on 4th, 5th or 6th step and after I'll exclude the specific EC and continue with the rest.
Continue till the same condition has been met to all EC's.

The point is that since a streak of 3 is more frequent than a streak of 6, it should happen before a streak of 6, thus we could use it as a signal to stop our progression while still in profit.
You could use 8 steps instead of 6 with stop signal on 4 losing streak, or 10 steps with stop on 5 losing streak, the principle remains the same regardless of the progression/signal range.

[NextYear]

What is the probability for a streak of 6 to occur before a streak of 3 ?

What is the probability for a streak of any range to be exactly the same range on the next streak ?

I don't think there is a rule for that.

[Blue_Angel]

I don't think there is a rule for that.

I think there is...

...a streak of 6 has half chance to occur before a streak of 5, a steak of 5 has half chance to occur before a streak of 4, a streak of 4 has half chance to occur before a streak of 3...

For any EC to repeat the same range of streak is:

streak probability minus all the rest streaks probabilities, continue deducting the same percentage with each successive streak of the same range and you would know what exactly is its probability in any case.

[NextYear]

I understand that, but there is no rule in which order they will appear...

[Blue_Angel]

I understand that, but there is no rule in which order they will appear...

It's one thing to say there's no guarantee and another to say that this event is more probable than the other.

The way I consider it is that the main problem which every Martingale has to deal sooner or later is that they don't know where to stop, they keep on and on till 1 rare losing streak wipes out their profit plus interest.

Another way to consider the classic Martingale is like a loan, you get some money till eventually you return all these money plus interest to the bank.

Don't get me wrong, I'm not saying that while you seat down at roulette table a streak of 6 or more could never happen, it could happen but having a condition in order to stop while ahead is improving the non stop doubling up in my opinion.

Not the best of approaches but just an idea for further discussion, sometimes it could lead to significant breakthroughs.
Consider each system idea as a lottery ticket, the more you buy the more your odds for a jackpot!

All of the important advances we are using today have begun by a single idea, an inspiration which eventually turn out to millions (if not more) of value.
Out of billions of worldwide population, don't you think that even 1 individual could have the HG?

Don't you think that this is a serious possibility?
Not all of them are gambling of course, that's why deduct 6 billion people out of 7 billion in total, 1 out of 1,000,000,000 has very good possibility but most of them are losers.

Casinos receive money from losers, then deduct 2.7% and transfer the rest to the very few.
Casinos will be always in business, always in profit even when HG is reality, because they arbitrage on total action. (arbitrage= house edge on total wagering)

[Blue_Angel]

I've borrowed a small part from Mad Professor (professional dice roller)

''Their logic is to “Ignore the rainstorm and just consider that it is single droplets of water that are flooding your basement. 
It doesn’t matter if it is pouring cats and dogs; a storm is only made up of individual drops of water. 
Therefore, storms only happened in the past-tense, and whatever is happening now can only be considered as individual raindrops. 
You have to overlook and disregard the totality of the deluge, no matter how much damage is being wrought”. 
Obviously, their logic is just plain FLAWED, and they don’t use the brains that God gave dogs, to have enough sense to come in out of the rain. ''

Devoted to all ''math experts'' like Turner.

[Bayes]

BA, I'm afraid you still don't get it.

link:s://:.youtube.com/watch?v=KOO5S4vxi0o

[FreeRoulette]

I think you are looking for a sequence probability.

The probability of 1 number on a double zero table is 1/38

So 2 numbers is
1/38 * 1/38

And n numbers is
(1/38)^n

So your original question of x sequence for y sequence is the exact same for any other sequence.

So, what is the probability of getting 3 zeros in a row followed by three 4s  is (1/38)^7

It doesn't matter what the numbers are, only the sequence length.

[FreeRoulette]

I think you are looking for a sequence probability.

The probability of 1 number on a double zero table is 1/38

So 2 numbers is
1/38 * 1/38

And n numbers is
(1/38)^n

So your original question of x sequence for y sequence is the exact same for any other sequence.

So, what is the probability of getting 3 zeros in a row followed by three 4s  is (1/38)^7

It doesn't matter what the numbers are, only the sequence length.

I wouldn't let me edit but it is (1/38)^6 not 7

[Blue_Angel]

I wouldn't let me edit but it is (1/38)^6 not 7

I assume you want to help me, thanks for your good intentions but I didn't ask for your ''help''.
Why don't you go to a maths forum to show them your brilliance, they could award you a Nobel prize for your novelties.

[Turner]

I assume you want to help me, thanks for your good intentions but I didn't ask for your ''help''.
Why don't you go to a maths forum to show them your brilliance, they could award you a Nobel prize for your novelties.

Nice..... 

[RouletteGhost]

Greeks.....pains in the ass