8 Card 21
  • Just played CVBJ and received the following hand:

    2,2,3,2,5,A,A,5....Froze Software...lol

    I never had or seen an 8 card 21.

    Enjoy the day,

    Jim
  • Can someone figure out the odds. (6D)

    I did a google and came across...

    www.talkaboutgambling.com

    Odds...I think...232692757835107.00 (They may of been talking about a higher number draw card 21, but I can't tell from the post.
  • First, let's take the really pathological case, max number of cards in a "sane" 21, as played by the dealer. That looks like this:

    2-2-2-2-2-2-A-A-A-A-5

    If you change the order of the 2's and A's, you reach a pat hand before reaching 21, so that order is required...

    takes at least 2 decks obviously, so assume 6 decks.

    first card has to be a 2. There are 24 2's in the shoe, out of 312 cards (ignoring burn cards and the cut card). So the probability of getting a 2 on the first card is 24/312. For the other 5 2's, you get 23/311, 22/310, 21/309, 20/308 and finally 19/307. What is the probability of drawing those 6 consecutive 2's? multiply those six fractions together. The probability of drawing each of the next 4 aces is 24/306, 23/305, 22/304, and 21/303. Finally the probability of getting the 5 is 24/302.

    If you multiply all 11 fractions, you get the probability of drawing that hand. It is vanishingly small, but still greater than zero. No idea whether this has ever happened in a real casino or not, but I'd guess it probably has, as many hands as have been played...

    Hope that helps... If you just remember the idea of multiplying probabilities, you can calculate the probability of drawing any particular hand you want. If order is immaterial, calculate the odds for (say) 10-6, and then 6-10 and add them.

    There are easier formulas for this (sample size N from population size M) that can be found in any good stat book... but the above is probably easier to understand (where the numbers come from) dealing with 6 decks of 52 cards...
  • In a six deck shoe game with 75% pen and assume a standard/average
    distribution of all cards: There are 233 cards available for the draw and
    18 2's and 18 A's. Initially the probability to draw an A or 2 is 18/233 and
    that assumes head to head play and you receive the very first card.

    One burn card (234 minus 1 = 233), further the burn cd is not an A or 2
    75% pen = 1.5 dks =(6)A's and (6)2's...unavailable
  • In both of the above examples, what is the probability to draw the 2nd
    2? In my example you may think that it is 17/232, but it is not. I got the first 2 and lets say the dealer draws a 10. So the probability that I will
    draw the second 2 is 17/231 and not 17/232. The next card that comes out
    is the dealer down card...Guess what... I no longer understand the value
    of the "configuration space" because the dealer hole card is unknown
    and as a result I can say nothing about probabilities thereafter.

    The same is true for steels example. The probability to receive a 2nd 2 is
    not 23/311....it is 23/310 provided that the dealer does not have a 2
    showing. After the dealer gets his hole card, nothing can be said
    about probabilities thereafter.................
  • much of that really can't be directly handled. If there are 6 players, then the probability of that second 2 is lower if another player gets a 2 first, if no 2's are dealt my probability for a second 2 is higher because there are fewer cards left.

    My example was really taking 6 decks, shuffling, and then dealing yourself one hand with no dealer hand or anything else. I think that is the easiest way to understand things. In a hand-held game you really don't know much at all about the cards that have been dealt except for your two and the dealer's up card, so any analysis there has to make lots of assumptions anyway...

    But the basic idea is to take the population (312 cards less any number of cards you know are already missing) and the remaining cards you are interested in (number of 2's still left) and the probability for drawing a single 2 is (number of 2's left) / (number of cards left). Probability of drawing a 2 and then another 2, is the probability of drawing the first 2, multiplied by the probability of drawing the second.

    that gets you into the right "neighborhood" without pulling your hair out. :)
  • Odds of a dealer >=8 card 21 depends on the number of players. Odds if the dealer completes his hand are about 980,000:1.
  • I should have split the 22 vs 7. I don't care about the lost fraction in this play, I make it all the time. My reasoning is why fight 2 bad hands when i can get tired with one.

    QFIT...Three players

    I sure would appreciate knowing the odds on an 8 card 21 under the additional information. Appreciate help with number.

    Best, Jim
  • it's a near-impossible question to answer. Unless you recall exactly what cards had already been dealt precisely. The probability for drawing a specific card is (number of that card unplayed) / (number of cards left in deck). To compute the probability of a specific hand, it is just the probability of drawing each required card, multiplied together..

    The probability of _any_ 8-card 21 is a different animal. There you need to compute the probability for each possible 8 card 21, and add them together. Any good probability and stat book will include formulas to calculate the probability of sampling from a population with replacement (infinite deck idea) or without replacement (finite deck idea) among other things...

    As Norm said, it is very small. I have seen some large count 21 hands by dealers, whether I have seen an exact 8 card or not I really don't remember...
  • What steel has said is the reason I tried to apply the hand to a real
    play situation....head to head. If you add more players the problem
    is even more difficult. If the dealer played face up you could figure the
    thing without much trouble, but that is a lot of add/subtract to try and keep
    the ratios known at all times.

    Back in December, I won 9 hands in a row playing two hands and It was
    more than I had ever experienced, but it don't compare to this example.
  • Would the odds of any 8 card 21 be that difficult to figure out? I was simply curious on any 8 card 21. I know it's not common, but just don't how uncomon.

    Thanks for all the information and discussion.

    Best, Jim
  • There is a simple way to figure the pobability of an event when you
    can't determine or fix the configuration space exactly. The short name
    for it is "trials". For Example: You set up a BJ game...say you and the
    dealer and you want to know the probability of your example or any
    other example. How to? Set your computer up to run a few 100 trillion
    hands and count the number of times the event occurs. If it occurs once
    ever 10 million hands, that is your probability, etc, etc
  • The odds of an 8 card or greater dealer 21 with three players and a hand of 22v7 with 6 decks are about 1,100,000:1.
  • Thank You, QFIT...Jim
  • Jim... you may want to un-check the "Randomize player speeds option" this will prevent the occasional "Lock-up". Confounding the House Players sometimes causes that.

    QFIT... there wouldn't be any cheatin goin on would there be? Ya know the left hand Player slippin one under the table?

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