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Thread: negative progression

  1. #1


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    Smile negative progression

    i know im gonna get killed for this because it sounds like a martingale system but bear with me. ok if i walk into a casino everyday to just win $20 and wait for the dealer to go twice without busting and then i jump in with a $20 negative progression until i win one time and then immediately leave the casino after the first win. fyi table max is $2500 so that would enable me to go to 7 rounds. Mathematically could i go 365 days without losing? so probability of losing being less than 1 and 365? i know it can happen at any time but this would be nine straight hands of the dealer not busting and 7 straight hands of my losing. my math puts it at about one every 400 days. is my math correct? thanks. interested in your expert opinions.

  2. #2


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    I got once every 170.35 days. First, you should know that dealer not busting the prior 2 hands is irrelevant. Your chance of losing the hand is 48%. Your chance of losing 7 consecutive hands (will discount pushes for the sake of this example) is 0.00587, which is likely to happen 2.14 per year

  3. #3


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    I'm not getting into a negative progression discussion but let's take another look at those calcs. If for every 100 hands you expect to lose 48 and win 43 then the chance of losing a hand is 48/(43+48)=53% rounded to whole numbers. So the chance of losing seven hands in a row is (0.53)^7=1/85.
    Last edited by davethebuilder; 08-25-2017 at 11:33 PM.
    Casino Enemy No.1

  4. #4


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    thank you for the info.

  5. #5


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    thank you for the info. hey banker i got a question. why is the fact that the dealer not busting the previous two hands is irrelevant? the chances of the dealer busting once every nine hands instead of once every seven hands would be higher, right? thanks man.

  6. #6


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    I have a question to, why are you insulting Sam Snead?

  7. #7


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    Quote Originally Posted by samsnead View Post
    i know im gonna get killed for this because it sounds like a martingale system but bear with me. ok if i walk into a casino everyday to just win $20 and wait for the dealer to go twice without busting and then i jump in with a $20 negative progression until i win one time and then immediately leave the casino after the first win. fyi table max is $2500 so that would enable me to go to 7 rounds. Mathematically could i go 365 days without losing? so probability of losing being less than 1 and 365? i know it can happen at any time but this would be nine straight hands of the dealer not busting and 7 straight hands of my losing. my math puts it at about one every 400 days. is my math correct? thanks. interested in your expert opinions.
    So does this mean you will not split hands or double at anytime? If you do, then how do you count it? Rounds or bets? If I get it correctly, you place a $20 bet, you lose, now you place a $40 bet, get a pair of Aces so you split, placing 2 bets of $40 each (so your second round is now $80), you lose. You then place a bet of $160 in your third round and you get a pair of 8's against a dealer 7. So you split, you now have two bets of $160 each out. Dealer gives you a 3 on the first 8 and you double (so third bet of $160 on table), you get a 9 for a 20. The other 8 gets you a 2, you double (another $160), get a 10, now you have 4 bets of $160 each on your 3rd round. Dealer has a 4 under his 7, gets a 10 and you have lost $640 in your 3rd round. You have to place a bet of $1280 on your 4th round. You lose that and you hit the max bet before you can bet on the 5th round.

    Its not such an unlikely scenario.

  8. #8


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    Dear Sam Snead
    For better luck, have your wife kiss your balls first - old golf joke.

    I normally don't get involved in dumb questions like this - are you out of your fucking mind. With a bet if $20, with successive doubling, you have 7 chances to win 1 hand for a profit if $20, 8 chances to break even since you are now constrained by table max, on an occurrence which happens more often than you think.

    Schmuck, what if you lose the 8th bet trying to break even. Why don't you add up the total sum value of your 7 doubled bets made, in order to make $20.

    Now, I realize I have a problem expressing my thoughts in a clear and concise manner. I trust you can forgive the puffery if my comments, and conclude that the only way for you to not make, but average your $20 a day is through some counting endeavour.

    Welcome to the site.

  9. #9


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    Quote Originally Posted by ZeeBabar View Post
    So does this mean you will not split hands or double at anytime? If you do, then how do you count it? Rounds or bets? If I get it correctly, you place a $20 bet, you lose, now you place a $40 bet, get a pair of Aces so you split, placing 2 bets of $40 each (so your second round is now $80), you lose. You then place a bet of $160 in your third round and you get a pair of 8's against a dealer 7. So you split, you now have two bets of $160 each out. Dealer gives you a 3 on the first 8 and you double (so third bet of $160 on table), you get a 9 for a 20. The other 8 gets you a 2, you double (another $160), get a 10, now you have 4 bets of $160 each on your 3rd round. Dealer has a 4 under his 7, gets a 10 and you have lost $640 in your 3rd round. You have to place a bet of $1280 on your 4th round. You lose that and you hit the max bet before you can bet on the 5th round.

    Its not such an unlikely scenario.
    Zee
    One if your better posts. You deserve a reward. Treat yourself to a rub and tug joint.

  10. #10


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    Quote Originally Posted by samsnead View Post
    thank you for the info. hey banker i got a question. why is the fact that the dealer not busting the previous two hands is irrelevant? the chances of the dealer busting once every nine hands instead of once every seven hands would be higher, right? thanks man.
    The reason it is irrelevant is because we assume the outcome of every hand as an independent event. In other words, we assume the dealer is busting because of random event. If, on the other hand, you think the dealer is busting because of specific composition of the deck, then it would be relevant. As Davethebuilder pointed out above, my calculation was not accurate because it ignored the push hands. So, we try it again assuming 48% loss, 44% win and 8% push. This gives us an effective loss of 51.84%. Then the chance of losing 7 hands in row would be pretty much 1%, so it would happen every 100 days according to your scheme (3.65 times a year). This loss will be about (3.65 X 2540) = $9,271. Your wins will be (365-3.65) X 20= $7,227. This will result in a yearly loss of $2,044.

  11. #11


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    Quote Originally Posted by samsnead View Post
    i know im gonna get killed for this because it sounds like a martingale system but bear with me. ok if i walk into a casino everyday to just win $20 and wait for the dealer to go twice without busting and then i jump in with a $20 negative progression until i win one time and then immediately leave the casino after the first win. fyi table max is $2500 so that would enable me to go to 7 rounds. Mathematically could i go 365 days without losing? so probability of losing being less than 1 and 365? i know it can happen at any time but this would be nine straight hands of the dealer not busting and 7 straight hands of my losing. my math puts it at about one every 400 days. is my math correct? thanks. interested in your expert opinions.
    This is not a winning strategy. It never will be. Please feel free to move on from this -EV mindset. I suggest you start studying now if you want to become a long-term winner in the game of blackjack.

  12. #12


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    Quote Originally Posted by samsnead View Post
    i know im gonna get killed for this because it sounds like a martingale system but bear with me. ok if i walk into a casino everyday to just win $20 and wait for the dealer to go twice without busting and then i jump in with a $20 negative progression until i win one time and then immediately leave the casino after the first win. fyi table max is $2500 so that would enable me to go to 7 rounds. Mathematically could i go 365 days without losing? so probability of losing being less than 1 and 365? i know it can happen at any time but this would be nine straight hands of the dealer not busting and 7 straight hands of my losing. my math puts it at about one every 400 days. is my math correct? thanks. interested in your expert opinions.
    Read up on the Oscars Grind here..Maybe it will mak things a little clearer for you...

    http://www.blackjackforumonline.com/...ack_System.htm
    http://bjstrat.net/cgi-bin/cdca.cgi

  13. #13
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    Quote Originally Posted by Freightman View Post
    Treat yourself to a rub and tug joint.
    Zee has admitted that he would need a "Little Blue Pill."
    My suspicion is that he is too cheap to foot the cost. But if
    we each contribute a nickel or a dime for pharmaceuticals ...


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