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Lets assume a very basic set of rules and table for them, these rules are unlikely to be seen in any casino and the reason is clear, there is only a 0.04% edge in favour of the casino, this could be beaten without ever counting cards simply by applying basic strategy and not even knowing the exceptions to it. Here are the rules.

1). We always start with a full deck of cards.

a). Assume this for 2 reasons, first the count (for card counters) is irrelevant as there is not enough cards out to establish a pattern (such patterns can affect the way you play using basic strategy and therefore the stats of the game).

b). Second it means we are always working from the same base numbers outlined below

2). We have 1 full deck of 52 cards

3). 4 of these cards are gone out of the deck leaving 48 (these 4 cards are the first 4 used in play above)

4). One of the 4 cards above in point 3 are unknown (because it is the dealers face down card)

5). This leaves only 3 cards whose value are known to us (from which we make all our decisions)

6). After the game is finished, all the cards are replaced in the deck and the deck is reshuffled to begin again (this brings us back to point 1 to repeat the process for every game).

I cannot include screenshots because I am new to this forum. But I wanted to post one of a basic strategy table with the rules specified.

If the dealer up card is 6, and your two cards show a total of 14 (perhaps 8 and 6, or 9 and 4, it does not matter the combination), you should stand (take no more cards).

However if your hand was an A3 (soft 14), you should double down.

Finally if your hand was a 77, (double 7), you should split.

The questions I have are, what are these decisions based on.

The reasons for questioning these tables are because almost every website or book quotes such tables without explaining the maths behind them. And when the maths is outlined, it’s usually done in either a convoluted and inefficient way, or over simplified to the point that it does not explain properly the other decisions in the table.

Essentially what I am asking is how do you calculate your next move given a set of rules and a specific hand. Meaning how exactly are the three examples above worked out. Is there a specific formula that can be used so that I may program it and build my own tables.

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    $\begingroup$ I think a lot of it is simulation based rather than with a mathematical proof. $\endgroup$
    – picakhu
    Aug 3, 2011 at 17:48
  • $\begingroup$ Does anyone know if there is a computer algorithm which claculates the EV for each situation, because for a math project i need to know how to do this but i'm stuck :(? $\endgroup$
    – Kees Til
    Jun 11, 2014 at 19:24

2 Answers 2

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Taking your example: If the dealer up card is 6, and your two cards show a total of 14 (perhaps 8 and 6, or 9 and 4, it does not matter the combination), you should stand (take no more cards).

The basic idea is to see if your probability of winning the hand is increased by drawing over standing. If you stand, you only win if the dealer busts. You can calculate this either exactly or by simulation. If you hit, you lose if you go over 21. If you don't go over 21, you might get a higher total than the dealer. Again, you can calculate this either exactly or by simulation. The easier way is to simulate lots of hands. Note that it does matter what your 14 is-if it is two 7's, you have a higher chance of busting than if it is 8+6, 9+5, or 10+4, as there is one less card left that you can draw without busting.

I would not say the basic strategy is useless, as playing that way reduces your losing percentage substantially from two alternatives: either never bust (stop when you are at 12) or play like the dealer. It is not sufficient to get above even, but one part of a long term winning strategy will include losing as little as possible when you can't win.

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  • $\begingroup$ I can appreicate the need for simulation (to test a theory or formula). But perhaps I am missing something. How is it faster to simulate as opposed to just calculate? As you suggested we only have 8+6, 9+5, 10+4, 7+7, Ace(A)+3, also can have A+A+2, and A+A+A+A. I dont think there are may other combinations (might be wrong). With todays computers even a basic low spec laptop could calculate mathematically every probabbality for this given set of game rules and card combinations (there is only 7 combos) in a second or less. Is it possible to show how I can calculate them. I would appreicate it. $\endgroup$
    – Francis Rodgers
    Aug 3, 2011 at 19:37
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    $\begingroup$ For a full calculation you need to consider all the possibilities, and making sure you have them all and keeping track is a lot of work. So you have seven ways to make 14 to play against (in this case) a 6 showing. Assuming you stand, you have to calculate the chance dealer will bust, which means pairing all types of card with the 6 (13 cases) and we already have 7*13=91 things to check. The number of possibilities explodes. $\endgroup$
    – Ross Millikan
    Aug 3, 2011 at 19:51
  • $\begingroup$ Oh, ok now I get it. You have to take the prob of your hand busting, against the prob of the dealer busting. Ok that does make a difference as you pointed out. But I still think it is doable for a given hand combination in less than possibly 2 or 3 seconds even on an 4 or 5 year old computer. If I just had the steps to take for 1 example I would be very happy. Thanks for everything so far. $\endgroup$
    – Francis Rodgers
    Aug 3, 2011 at 21:03
  • $\begingroup$ So for 8+6 vs 6+10, if you stand, he busts drawing 6 or more, which is 28/48 probability. The deck has 32 cards that are 6 or more, and 4 are accounted for. It gets harder at 8+6 vs 6+9, as he busts drawing 7 or more (26/48), or drawing A+6 or more (4/48 * 28/47). As the dealer total gets lower, the possibilities expand, as he can draw a small card and need to keep drawing. $\endgroup$
    – Ross Millikan
    Aug 3, 2011 at 21:14
  • $\begingroup$ Thanks for your help here. Just to clarify above you say 8+6, since this = 14 I assume this is the player hand. Then we have 6+10. I assume this is the dealers hand. If so we dont know the down card is a 10. Are my assumptions correct so far, and am I correct in assuming that 6+10 is only one possibible hand for the dealer. And that we would have to do another set of possibilities for the dealer with say 6+9, and another with 6+8 and so on. Then calculate the rest as you did for every set of dealer hands before going on to the next player hand. Thats a lot of calculations! Thanks so far. $\endgroup$
    – Francis Rodgers
    Aug 4, 2011 at 0:01
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Ed Thorp was the first to devise a system for beating casino blackjack, by simulating millions of deals on an IBM computer and turning the results into a card counting system. But if you always start with a full deck the casino has the edge. Basic strategy just gives the player (almost) an even chance.

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