I have a theory based on anecdotal evidence that when the count get lower than -2, a push 22 is about to happen. I start betting one unit on the 22 side bet (which pays 7x, 20x or 50x if the dealer gets 22, depending on color) and keep adding one unit whenever the count gets when lower. I've had a few instances of a 50x payout on a $25 bet and almost never got a 22 when the count was high.
Is this dumb luck, or are the odds of a push 22 truly better when a large number of Aces and a low number of 5s have been pulled in the last few rounds?
Thanks
Quote: jdmartineven though it's a continuous shuffling machine
I have a theory based on anecdotal evidence that when the count get lower than -2,
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might be dumb luck i dont think counting works on continuous shuffle
Quote: jdmartinThanks! Assuming I can find a table where it's not a continuous shuffling machine, could the Aces count help me predict the push-22?
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just an fyi i dont really know too much about counting but i was always under the impression that it was dependent on deck penetration
i have heard people talking about counting with a CSM but i never got into the conversation if it was true or not
It's not complete yet so don't use this to gamble your own money, but I did get interesting results. I implemented the strategy outlined by the Wizard for Free Bet, but only the real money part, with no bankroll limit. Basically I just wanted to see what's the status of aces vs fives whenever the dealer hits 22.
Parameters:
- 8 decks, moving to new deck after 85%
- reset counts after 8 decks
- 2 players using the strategy described above
- dealer hit soft 17
I ran 1,000,000 rounds. Whenever the dealer got 22, the count was pushed to an array. At the end of the million rounds, the array was analyzed in pandas. And I ran this whole analysis 100,000 times to make sure there's no weird state leftover.
I got consistent results out of this half-baked simulator: about 55% of push-22 happened when the count was 0, about 30% when the count was either 1 or -1. As one would suspect it's an almost normal distribution.
This may appear obvious in retrospect since the further from zero the count gets, the more likely either side is going to win. Or it could be statistical white noise and I'm basically reading tea leaves.
In the coming days I will make a new iteration of this simulator, taking into account the entire Wizard strategy and a number of scenarios (single player, one or many players not following the correct play, etc). I am recording my progress on video and I'll share the YouTube url when I'm done.
I'm not really a math person so I'll have a friend review my conclusions as well. I don't plan on sharing the python code directly because I don't want to support it but it will be on display on the video (screen cast).
Again this could be entirely garbage but I'm confident enough that from now on I'm going to gamble a dollar on push-22 when the count is 0.
If I have any success with this or if it's fun enough, I'll tackle other side bet scenarios in various table games, at the risk of being shamed by the Wizard. I understand the house edge can get ridiculous (ex; king's bounty is like 25%) but if I can identify situations where the side bet is slightly less ridiculous I'll have a way to get my fix of high-volatility gambling while not feeling like a complete fool.
To be continued!
My instinct is that you will not have a winning system here, but who knows, could squeeze out an edge. Also, for your original question... if I understand it correctly then you're both just trying to play the hand the best you can, even though you're in charge of different hand signals to the dealer, correct? If that's the case the fact that there's 2 of you means absolutely nothing to the math of the game, past as someone mentioned if your "teammate" didn't want to double for the full amount and you did... but that's you just scooping up his lost EV, not actually making +EV from the deck/game/casino.
Lastly, CSM's are countable, when the dealer doesn't immediately put the cards back in the tray. Tho even then there's like a 15 card buffer before they get mixed back in, but if you're playing with more than 4 decks (and even then) your count frequencies are just god awful. It's the same as playing a 6D shoe and someone putting the cut card 15 cards in, then shuffling every time they hit the 16th card. Basically renders counting useless, so I wouldn't trust any short term results from that game and would be very willing to bet it's -EV even with what little counting you can do. You have to find a lazy dealer that let's 2 or 3 decks go, at least. Then you have to blast a massive spread and fade insane variance (kinda like playing online live blackjack).
Good luck to you on your adventure!
Over my head.Quote: RomesIf you're a programmer this should be simple; find the effect of removal on each card. I.E. Remove all of the 9's, then run the game with perfect basic strategy and rules and see how many 22's happen after 1,000,000 shoes. Then do the same for every other number, and for a complete deck to compare the changes to find the EoR. Then you'll find the cards that correspond the most with the outcome you're looking for and you can create a specialized count for the push 22 on it's own. Then program your count and find at which + count it becomes positive to place a wager (might be too high to consider real... or never).
My instinct is that you will not have a winning system here, but who knows, could squeeze out an edge. Also, for your original question... if I understand it correctly then you're both just trying to play the hand the best you can, even though you're in charge of different hand signals to the dealer, correct? If that's the case the fact that there's 2 of you means absolutely nothing to the math of the game, past as someone mentioned if your "teammate" didn't want to double for the full amount and you did... but that's you just scooping up his lost EV, not actually making +EV from the deck/game/casino.
Lastly, CSM's are countable, when the dealer doesn't immediately put the cards back in the tray. Tho even then there's like a 15 card buffer before they get mixed back in, but if you're playing with more than 4 decks (and even then) your count frequencies are just god awful. It's the same as playing a 6D shoe and someone putting the cut card 15 cards in, then shuffling every time they hit the 16th card. Basically renders counting useless, so I wouldn't trust any short term results from that game and would be very willing to bet it's -EV even with what little counting you can do. You have to find a lazy dealer that let's 2 or 3 decks go, at least. Then you have to blast a massive spread and fade insane variance (kinda like playing online live blackjack).
Good luck to you on your adventure!
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Hi Romes.
Quote: jdmartinI was curious about this problem so I decided to do an analysis using Python (pandas, Jupyter, etc.).
It's not complete yet so don't use this to gamble your own money, but I did get interesting results. I implemented the strategy outlined by the Wizard for Free Bet, but only the real money part, with no bankroll limit. Basically I just wanted to see what's the status of aces vs fives whenever the dealer hits 22.
Parameters:
- 8 decks, moving to new deck after 85%
- reset counts after 8 decks
- 2 players using the strategy described above
- dealer hit soft 17
I ran 1,000,000 rounds. Whenever the dealer got 22, the count was pushed to an array. At the end of the million rounds, the array was analyzed in pandas. And I ran this whole analysis 100,000 times to make sure there's no weird state leftover.
I got consistent results out of this half-baked simulator: about 55% of push-22 happened when the count was 0, about 30% when the count was either 1 or -1. As one would suspect it's an almost normal distribution.
This may appear obvious in retrospect since the further from zero the count gets, the more likely either side is going to win. Or it could be statistical white noise and I'm basically reading tea leaves.
In the coming days I will make a new iteration of this simulator, taking into account the entire Wizard strategy and a number of scenarios (single player, one or many players not following the correct play, etc). I am recording my progress on video and I'll share the YouTube url when I'm done.
I'm not really a math person so I'll have a friend review my conclusions as well. I don't plan on sharing the python code directly because I don't want to support it but it will be on display on the video (screen cast).
Again this could be entirely garbage but I'm confident enough that from now on I'm going to gamble a dollar on push-22 when the count is 0.
If I have any success with this or if it's fun enough, I'll tackle other side bet scenarios in various table games, at the risk of being shamed by the Wizard. I understand the house edge can get ridiculous (ex; king's bounty is like 25%) but if I can identify situations where the side bet is slightly less ridiculous I'll have a way to get my fix of high-volatility gambling while not feeling like a complete fool.
To be continued!
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This is very interesting. Remember of course most 22's happen at a count of 0 because most of the time the count is 0 anyways. Better would be to compare the hit rate of true -1 to the hit rate of 0 etc. It only makes logical sense that a Hi-Lo counter would get more 22's at a negative count with a lack of 10's giving more combinations. As for Aces and Fives, I'm assuming the count is high when more 5's have been dealt than A's? I would say more aces lead to more soft hands which leads to less busts and therefore less 22's. Even If you dod gain an edge, these games are so volatile that you could possibly never notice a difference if you played your whole life. I ran a sidebet on this game called "pot of gold" and saw that 9's 10's and A's are bad for the player. With Hi-Lo just about any negative count made the side bet profitable. I wouldn't dare try and make money off this due to the volatility though. Good luck on your math quest
Quote: richodude
This is very interesting. Remember of course most 22's happen at a count of 0 because most of the time the count is 0 anyways.
What makes you think that most of the time the count is 0? I would say just the opposite, most of the time the count is not 0.