Thread Rating:

Zcore13
Zcore13
Joined: Nov 30, 2009
  • Threads: 39
  • Posts: 3537
June 6th, 2013 at 10:38:23 AM permalink
This is quite funny. But assuming it's true, please tell me about a wager in the casino that you can use this system on that is 50/50?

Also, if you can not find one, and you are saying you can use this system on a game that is close to 50/50 and overcome the small house edge, please tell me about the game that has no table maximum, so that when a bad streak does come, the table max doesn't ruin your progression.

Thanks,

ZCore13
I am an employee of a Casino. Former Table Games Director,, current Pit Supervisor. All the personal opinions I post are my own and do not represent the opinions of the Casino or Tribe that I work for.
Buzzard
Buzzard
Joined: Oct 28, 2012
  • Threads: 90
  • Posts: 6814
June 6th, 2013 at 11:40:40 AM permalink
Quote: 24Bingo

So somehow the progression stops working if someone else is opposite-betting you with a similar progression? How does that work?




It depends on who that other person is. Particularly if he is Statman oe EvenBob !
Shed not for her the bitter tear Nor give the heart to vain regret Tis but the casket that lies here, The gem that filled it Sparkles yet
24Bingo
24Bingo
Joined: Jul 4, 2012
  • Threads: 23
  • Posts: 1348
June 7th, 2013 at 3:06:38 AM permalink
Quote: Alembert

Notice, only very occasionally do the bets ever actually match value.



And that's the problem.

I see your system is a little different from mine, in that you start well above one unit, so that you can absorb quite a few wins in excess of losses. That's actually good, because under the assumption that your base bet, we'll call it x, is high enough that you never have to bet one unit (or even if you start betting the opposite in a similar positive progression), the math gets much simpler:

If your wins outstrip your losses, you will be ahead by (wins + (2*base bet - wins + losses - 1)*(wins-losses)/2).
If your losses outstrip your wins, you will be ahead if ((2*base bet + losses - wins + 1)*(losses-wins)/2) is less than your total number of losses, by the difference. You'll be behind by the difference if it's greater.
(And of course, if you have an equal number of wins and losses, that's what you'll be up by.)

Add it up and you'll find you break even.
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
Alembert
Alembert
Joined: Jan 14, 2013
  • Threads: 7
  • Posts: 57
June 7th, 2013 at 4:53:54 AM permalink
Quote: Zcore13

This is quite funny. But assuming it's true, please tell me about a wager in the casino that you can use this system on that is 50/50?

Also, if you can not find one, and you are saying you can use this system on a game that is close to 50/50 and overcome the small house edge, please tell me about the game that has no table maximum, so that when a bad streak does come, the table max doesn't ruin your progression.

Thanks,

ZCore13




Moving from theoretical to practical application, table Maximum is almost irrelevant. If you are running off a finite bankroll that approximates 2 standard deviations your bets will never approach any tables maximum. The highest bet I have ever had to place (on my way to a losing session) was 350.

I have tinkered with playing against Pai Gow Poker, Bacarat, Craps
KingFX
KingFX
Joined: Sep 7, 2013
  • Threads: 0
  • Posts: 3
September 7th, 2013 at 7:10:44 AM permalink
There is a one by Martin Blakey.


  • Jump to: