1. joeblackjack

    joeblackjack Member

    Let's say you are flat betting 1 unit on 6 spots on the following DD game: S17, DAS, DOA. What would the average profit per round be if you could successfully cut the cards such that an Ace is in the first 14 cards after 80% of all shuffles?

    I could be wrong - I haven't run the simulations - but I'm fairly certain that simply ensuring that at least 1 ace is in the first 2 cards of 1 of the 6 hands on the first round is enough to yield a significant overall advantage, even with flat betting throughout the remainder of the deck and accounting for the increased chance of dealer blackjacks.

    On a side note, is it realistic to believe that shuffle tracking is accurate enough to be able to track an Ace within 14 cards?
  2. HockeXpert

    HockeXpert Well-Known Member

    My guess is -EV. With only one ace guaranteed in the first 14 cards, you'll have 5 hands at a disadvantage IF the dealer doesn't get the A and 6 at a disadvantage if the dealer does get the A which will occur 1 in 7 times. You're also at -EV the other 20% of the time went you CAN'T cut the A where you want it.

    This would be true if you can guarantee YOU get an A in the first round. The shallower the pen, the better.

    Not that I am aware of.
  3. joeblackjack

    joeblackjack Member

    Thanks for the reply. As I said I could be wrong, but I believe the advantage yielded by having a guaranteed ace be a part of any of the initial hands is so large that it would more than offset -EV on the other hands played at a slight basic strategy disadvantage. I guess maybe I should run some simulations. I was just wondering if anyone had actually already run this type of simulation and could say for certain whether it would be positive or negative EV.
  4. HockeXpert

    HockeXpert Well-Known Member

    Don't get me wrong, I like the idea and I hear you about the power of a first card A (52%) but it's all the other bets that offset that +EV. You play all the other hands at a small HE but add the loss of the A and it's greater than you think. A better betting pattern like reducing your bet or # of bets in subsequent rounds (depending on count) may make this +EV.
  5. Sucker

    Sucker Well-Known Member

    Suppose you were to take 13 RANDOM cards, and add ONE ace to them. You shuffle, and deal out 6 hands (total of 7 hands, counting the dealer). Every player has approximately a 2% advantage.

    Of course it is!
  6. HockeXpert

    HockeXpert Well-Known Member

    On a modern DD shuffle? PM welcomed!:grin:
  7. joeblackjack

    joeblackjack Member

    Yeah I'm just trying to avoid any correlation of bet to count. So an ace gives a 52% advantage. Meanwhile, all the other hands are are played at less than a 0.5% disadvantage (the particular game I mentioned is actually less than 0.3% HE overall, not accounting for the removed ace). On a DD game with good penetration, you're looking at less than 30 total player hands being dealt on average. 30*0.5%= EV of -0.15 units, and the 1 hand played with a 52% advantage = EV of +0.52 units, for a positive EV of 0.37 units for every 30 units bet (e.g., per full shuffle played through). In practice the profit would likely be higher, since my guess is that on DD most houses actually only wind up dealing somewhere in the range of 25 player hands per double deck.

    Am I missing something?
  8. Gamblor

    Gamblor Well-Known Member

    I don't think your factoring in the possibility the dealer might catch the extra Ace, which is a powerful in their hands too. All 6 of hands is likely screwed.
  9. HockeXpert

    HockeXpert Well-Known Member

    You're assuming that you ARE going to get the ace. The dealer will get it 1 in 7 times putting all your hands at a large disadvantage (maybe 10% off the top of my head). Stick with Sucker's 2% advantage on each hand in the first round and see if it's worth it. It sounds like he has a better method to approach this without a sim. It looks like you can achieve a very small EV if the pen's shallow.
  10. joeblackjack

    joeblackjack Member

    Thanks for the advice...I'll investigate further.
  11. PonyPrincess

    PonyPrincess Active Member

    You have to factor in if the dealer gets the ace you're at a 36% disadvantage.

    But yeah you can figure out what the advantage is for that first 14 cards because it'd be in HiLo a +1 RC / (14/52) = ~+3.5 true count.
  12. Gamblor

    Gamblor Well-Known Member

    Right that's one good way to go about it, but keep in mind that since you know its definitely an ace (not a 20% its ace and 80% its 10), your EV is higher than a ~+3.5 TC.
  13. Ferretnparrot

    Ferretnparrot Well-Known Member

    +3.5 tc is a good approach to looking at it in the correct manor, but the HILO undervalues the EV of the ACE so the advantage is surely higher than what the ev of +3.5 woudl suggest.

    The ace IS NOT WORTH 52% in this manor!!! It is only worth that if you are to be dealt it as one of your initial cards. In this case, it is simply added to the shoe.

    The effect of removal for an ACE is 0.58% per deck so the effect of adding it to a quarter deck would be roughly 4 times that, say 2.3% change less the house edge and you have an approximation of your answer.

    14 cards is not a big window, but half a deck is, and you still have an edge with half a deck, and about break even with 1 deck so as long as you can cut the card ~INTO PLAY within a deck, you wont be losing money.

    I think its more feasable in shoe games, but if you can do it in DD, that awesome im sure youll score lots of bonus EV hands and good cover.

    while playing DD in this manor, a full table would require a lower bet spread for attacking this event since the number of -EV rounds dealt after is reduced.
  14. joeblackjack

    joeblackjack Member

    I'm aware that it's only worth that much if it's one of the first 2 cards...my original post was assuming that you were playing 6 spots, and thus an ace would appear in someone's first 2 cards (though possibly in the dealer's first two cards as well).

    You bring up an interesting point. If you were playing on a manually shuffled single deck game, it seems then that it would be relatively simple matter to track at least 1 ace and ensure that it is cut somewhere into the first 1/2 deck (assuming random distribution, obviously other aces would be in play most of the time as well, but by ensuring that 1 is always there, you would be at a minimum diluting the house edge). If this would yield an overall edge with flat or random betting, it seems like a far less detectable way to play with +EV than counting or virtually any other AP method. You could destroy places like Reno where single deck games are plentiful and not get barred.

    I also wonder if some places have at least entertained the possibility that merely cutting correctly could yield overall +EV. During a recent trip to Reno to practice counting, one of the places I went was the Grand Sierra. There, by default they do not cut the cards at all. Players can request a cut, but the dealer must call out "player cut" to the floorman. I have never seen a store that discouraged player cutting before, or that even allow games to be dealt without a cut.
  15. Gamblor

    Gamblor Well-Known Member

    LOL. I've never seen this either. Probably opening up other bigger problems by not cutting the deck.

    Definitely messes up the sacred flow of the cards!

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