Out of thin air!

[GLC]

I've been thinking about my betting against a certain dozen for 6 times in a row.  That's 1-1; 3-3; 9-9; 27-27; 81-81; 243-243  = 844 units if you lose.


Here's my thought for a hit-n-run.


Randomly pick a dozen or a column.  Let's say you chose the 2nd dozen.  Start betting that the 2nd dozen won't spin 6 times in a row.


What are the odds that you can get unlucky enough to pick a dozen that is going to hit for the next 6 spins? 


I have a booklet with 20,000 certified spins on a single zero wheel recorded continuously  in a marathon roulette challenge at Casino de Macao.


I have been randomly picking a spot to start betting and I haven't even come close.  Four in a row so far and 1 was a zero.  I've done this 400 times.


I know that it's possible.  But there's no way to determine the odds because how do you calculate the odds of picking the dozen that's going to hit in the next 6 spins?


I know that if you bet just 1 of the dozens continuously, you will lose because each dozen will hit 6 or more times in a row periodically.  But what are the odds that you will pick the exact time when that dozen is going to hit 6 times in a row?


It only has to happen more often than every 844 times.  I think some of us could go our whole lives and never pick a bad time to start betting.


Any thoughts?


GLC

[RouletteExplorer]

I know that if you bet just 1 of the dozens continuously, you will lose because each dozen will hit 6 or more times in a row periodically.  But what are the odds that you will pick the exact time when that dozen is going to hit 6 times in a row?


GLC the odds are exactly the same as if you would bet just 1 of the dozens continuously.

This idea is so effective like when the ostrich puts her head in the ground when she is in danger and thinks that if she can t see the enemy that is gonna kill her,then the enemy will not be able to see her too.

Nothing changes by picking a dozen randomly.

[vladir]

He's rigth... odds are the same. BUT... what if you wait for a dozen to repeat 6 times... THEN you play this for some games (20-50), well... I think there you migth have some winnings... A dozen will hardly repeat 6 times for two times in so litle games, I guess...

[GLC]

He's rigth... odds are the same. BUT... what if you wait for a dozen to repeat 6 times... THEN you play this for some games (20-50), well... I think there you migth have some winnings... A dozen will hardly repeat 6 times for two times in so litle games, I guess...

Help me understand this.

If I wait for a dozen to hit 6 times in a row and then I bet for it to not hit for 6 times in a row, how is this different than just randomly picking a point to start betting that the same dozen won't hit 6 times in a row.

If I start betting continuously against the 1 dozen hitting 6 times in a row.  I know that I will lose because dozens hit 6 times in a row fairly often.  But, if I pick random points to bet 6 times in a row against the  dozen hitting, I can see myself not picking the exact times the 1 dozen hits 6 times in a row.

[Bayes]

But, if I pick random points to bet 6 times in a row against the  dozen hitting, I can see myself not picking the exact times the 1 dozen hits 6 times in a row.

Hi George,

you're right about playing continuously - sooner or later you'll hit 6 in a row. My beef with the alternative, random entry kind of approach is that it doesn't solve the problem if you continue to play that way. Advocates of hit & run think that they're avoiding the bad sessions somehow, but how?  Imagine a continuous sequence of spins within which your system is highly likely to bomb out (say 1000 spins). The "solution" of the hit & run player is to only play, say 10 spins at a time, with random entry. Sooner or later, during one of those 10 spin sessions, you will hit a losing run. The question is - has the probability of your finding this losing sequence decreased merely by the fact that you're limiting your play in any one session? I you think it has (and I don't), can you attempt to explain why? If you run a bot for 1000 spins playing continuously and find that the probability of losing is say 90%, by what mechanism is this probability lowered by playing 10 spins every day for 100 days instead?

I agree that waiting for a loss won't make any difference.

[Bayes]

And we can even increase it more. If the standard deviation for a dozen that has just repeated 6 or more times is high for, lets say, the last 50 spins - when I say high, I mean really high, or by other words this dozen has came out a lot more then normal - 25 times or more in a total of 50 spins), then it's even more hard that a new set of 6 spins on the same dozen will hit in the next spins. That would make the standard deviation to go out of the norm - yes since this game is random, it can still happen, but maybe not in our lifetime...)

I agree. 

But having said that, I wouldn't under any circumstances use the triple up progression. If the last 50 spins were in 3+ standard deviation with respect to the dozen in question, I would play the next 30 - 50 spins using a mild progression  on the other 2.

[Bayes]

Hi vladir,

I meant to quote your post but instead modified it by mistake. Sorry mate. 

[RouletteExplorer]

you're right about playing continuously - sooner or later you'll hit 6 in a row. My beef with the alternative, random entry kind of approach is that it doesn't solve the problem if you continue to play that way. Advocates of hit & run think that they're avoiding the bad sessions somehow, but how?  Imagine a continuous sequence of spins within which your system is highly likely to bomb out (say 1000 spins). The "solution" of the hit & run player is to only play, say 10 spins at a time, with random entry. Sooner or later, during one of those 10 spin sessions, you will hit a losing run. The question is - has the probability of your finding this losing sequence decreased merely by the fact that you're limiting your play in any one session? I you think it has (and I don't), can you attempt to explain why? If you run a bot for 1000 spins playing continuously and find that the probability of losing is say 90%, by what mechanism is this probability lowered by playing 10 spins every day for 100 days instead?


Exactly.
The odds are always the same
I agree that waiting for a loss won't make any difference.

[vladir]

Hi vladir,

I meant to quote your post but instead modified it by mistake. Sorry mate. 

Can't you undo that? I don't feel like writting all again ^^