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I was watching Beating Blackjack with Andy Bloch where he runs through the basic strategy charts that outline the best strategy with playing the game. Later he also talks about the methodologies to count, but that is not relevant to my question.

He states in that video that the basic strategy is an outcome of computer simulations, and that suggests to me that its a rather weak method to go about doing it. I have a two-fold question, one is when is a computer solution considered good enough to be "true", and the second, surely, this is a much easier problem that does not need computer simulations to solve, are there any known methods to derive the basic strategy.

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None of the answers here really tackle the question that I am actually asking, which is that how come there is no simple rigorous method to prove the optimal strategy. For example, assuming that no form of card counting has occurred, is is still true that if you have a 10+5 or 9+6 or 8+7 vs a 6 you stand? (that is what basic strategy says) If the answer is yes, there should be some reason why.. –  picakhu Jan 15 '11 at 5:37
    
I added a concrete example to my answer based on your comment, I hope you meant something like this. –  daroczig Jan 16 '11 at 13:00
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up vote 3 down vote accepted

I'm not sure I understand your first question. You don't have a "computer model" for blackjack. The rules of the game are given. You can model the dealing of the cards as if they were uniformly random draws from a collection of objects (distinct objects if you are using a single deck of cards) without replacement.

Since it is a game with randomness built in, it is a good candidate for Monte Carlo simulation. To my knowledge, it is not very simple to derive basic strategy using just pen and paper because you would have to deal with a number of different cases and keeping track of them all can be tedious. I'm not aware of any elegant solutions to this problem.

If you go further and look at card counting strategies, things become even more complicated and the methods and strategies appear to be more and more heuristic.

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Thanks for your answer. I used the term model to mean 'modelling'. Meaning using a computer to obtain a monte carlo solution. Perhaps my wording can be improved. About my first question, I will edit it. –  picakhu Jan 14 '11 at 9:31
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The sort of question you would do with Monte Carlo is to decide "If the dealer shows a 6, what value should I draw to?" You would do this by simulating many deals with the dealer showing a 6, try out various strategies, and see which has the highest expectation. Under certain statistical assumptions, you can state that a particular strategy wins a% of the money at stake with a standard deviation of b%. Raising the number of trials will decrease b%, roughly by the square root of the number of trials. So if one strategy is clearly better than another, it won't take too many tries to know that. If they are close, it will take a lot. If they are really close, you will never know, but then it doesn't matter much.

As svenkatr states, the details of Blackjack make it difficult to do an analytic solution, but you could in particular cases. Going back to dealer shows a 6, you could say suppose I draw to 15. After tedious counting, you could exactly calculate your winning percentage. After more counting, you could compare that with the winning percentage of draw to 14, and you would have a rigorous answer.

Added: counting just changes the frequency distribution of cards that are dealt. The same approach works-you just ask questions like "If the proportion of cards that count 10 drops from 4/13 to 2/13, how does that change the results?" and adjust your strategy accordingly. Generally if the high count cards are depleted you draw more because you bust less. More high count cards are good for the player, because s/he can stop and let the dealer bust.

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The scenario that you described is a single deck game where the dealer must stand on Soft 17, which is a disadvantage to the casino. Every casino table has its own specific rules. On some tables, the dealer will hit Soft 17, others will stand.

In a single deck game, there are also other factors which can have an effect on the overall game odds. One of these is deck penetration, another is the number of "rounds" played, which is just the number of times a new hand is played from the same deck before it is shuffled.

I noticed that there is no consideration of the cases where the dealer ties the player (but this is only applicable when the player is hitting). Also, the dealer can bust on 4 or more card. Example: dealer has 6-2 and then draws 4,10 to form 6-2-4-10 = 22.

I have a blackjack Monte Carlo simulation program which runs on my Apple notebook. I checked the data for the scenario described above.

In my simulation, I have used deck penetration of 65% and limited the number of rounds to 6. The results show that in the case where the player draws 10-5 and dealer draws 6, and the player stands, the player will win 42.1858%, lose 57.8142% and push 0%. The average loss by standing is 15.6283%. If the player hits however, the player will win 28.5349%, lose 67.1360% and push 4.3291%. The average loss by hitting is 38.6011%.

This data was extracted from the following table for the case that Dealer Stands of Soft 17:

The columns represent the following in this order: (1) Player Card (2) Player Card (3) Dealer Card (4) Strategy (Stand or Hit) (5) Frequency (Trial size during simulation) (6) Wins (7) Losses (8) Pushes (player tied dealer) (9) Net Gain/Loss (10) Win probability (11) Loss probability (12) Push probability

  • 10 5 6 S 2985424 1259426 1725998 0 -0.156283 0.421858 0.578142 0
  • 10 5 6 H 2983974 851474 2003320 129180 -0.386011 0.285349 0.671360 0.043291

  • 9 6 6 S 562585 230554 332031 0 -0.180376 0.409812 0.590188 0

  • 9 6 6 H 562917 157084 380592 25241 -0.397053 0.279054 0.676107 0.044840

  • 8 7 6 S 748253 305639 442614 0 -0.183060 0.408470 0.591530 0

  • 8 7 6 H 748480 222004 492233 34243 -0.361037 0.296606 0.657643 0.045750

Notice that the trial size in both of the 10-5-6-x cases is higher than the other four cases. This is simply due to the higher number of 10 valued cards in the deck, i.e. each value is the set {10,J,Q,K} is equal to 10. Also note that the sample sizes the two 10-5-6-x cases are not identical to each other. This is due to the random selection of the players strategy during simulation where the probability of the player standing is assumed to be 1/2 and probability of the player hitting is also assumed to be 1/2.

The calculated 63.26 percent chance of busting by hitting (which was posted earlier), is pretty close to the simulation loss result of 67.1360%.

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I do not think that the basic strategy would be an outcome of any Monte Carlo simulation, as blackjack (and other games in casinos) had been played a lot before enough computer capacities were available for such computations. And these games are made up cleverly (just see Pascal's roulette), there is no need to compute the house's edge, as it has been well calculated for ages.

With probability and game theory any outcome can be expressed, and the probabilities (and by this: wins and losses) could be computed, so the basic strategy could be defined. E.g.: you have 15 and the dealer an A, the probability you will go bust with another card is quite high: $\frac{29}{49}$ (around 59 percent) to risk. This could be even worse with a 5 in hand (~61 %). The odds of the dealer has another card worth 10 is quite high: $\frac{16}{49}$ and other cards also help him (7, 8, 9 and also smaller ones with smaller probability). And so on.

It is true that with modern computers a simulation could be implemented a lot faster then doing all computations for probabilities, that is why I think these techniques are so popular nowadays.

Another issue is that the outcome of a Monte Carlo method could hold an acceptable accuracy with enough repetitions, so human driven (planned) computations just do not worth the time or energy to deal with.


After editing, I try to answer for the concrete question, which says: "assuming that no form of card counting has occurred, is is still true that if you have a 10+5 or 9+6 or 8+7 vs a 6 you stand?" (OP)

Let's say you have a 10+5 (hard total: 15) and you can see a 6 at the dealer. The basic strategy says you have to stand. Why?

You know nothing about the other card of the dealer, but you are quite sure about he will ask for another card. This is not 100 percent true, as he may have an ace ($p=\frac{4}{49}$), and could stand, if you have no more than 17 and did not bust.

What if you stand? You will loose if the dealer have an ace ($p=\frac{4}{49}$) or have a smaller card, hit and the overall sum will be between 17 and 21. It could be done with a sum of the 2 unknown card being between 11 and 15. It could be any of the followings: 2-9, 2-10, 2-J, 2-Q, 2-K, 2-A, 3-8, 3-9, 3-10, 3-J, 3-Q, 3-K, 3-A, 4-7, 4-8, 4-9, 4-10, 4-J, 4-Q, 4-K, 4-A, 5-6, 5-7, 5-8, 5-9, 5-10, 5-J, 5-Q, 5-K, 6-5, 6-6, 6-7, 6-8, 6-9, 7-4, 7-5, 7-6, 7-7, 7-8, 8-3, 8-4, 8-5, 8-6, 8-7, 9-2, 9-3, 9-4, 9-5, 9-6, 10-2, 10-3, 10-4, 10-5, J-2, J-3, J-4, J-5, Q-2, Q-3, Q-4, Q-5, K-2, K-3, K-4, K-5, A-2, A-3 and A-4. Note: you have 4 pieces of all cards in the deck except for the 5, 6 and 10, as you have seen 1-1 in your or the dealer's hand. That will also affect the computed probability, but I will not deal with this small problem here. So the prob. that you you will loose against the dealer's 3 card is: 68 hands will be bad for you from the possible 169 ($=13^2$) combinations. This and the above soft 17 gives a $p=\frac{4}{49}+\frac{68}{169}=0.4839995$.This means you have around 51.6 percent chance to win!

If you do not stand, you will have a quite high risk of busting: $p=\frac{8*4-1}{49}=\frac{31}{49}$, which is a lot higher chance (around 63.26 percent) to loose. And I did not compute the chance of you would not bust but the dealer would hit a higher sum than you. That's why you should stand.

I might have maiden some mistakes in the above example, but I hope the theory can be seen. And of course I do think the above computation could have been done in a lot simpler way using combinatorics, but that was done such a simple way only using the classical theorem of probability.

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In your example, you are using one deck.. –  picakhu Jan 16 '11 at 15:57
    
@picakhu: yes, I do, as blackjack can be played with only one deck also. The number of decks in play only affects the probability of getting a card what have been seen before (what I left out of my example, as I stated), which is not a very big deal in blackjack - not like in poker. Also, 4+ decks in play serves the purpose of making more difficult counting the cards (so: cheat), and also the ease: as the dealer do not have to shuffle the deck every turn. And of course also increases the house edge. In your example, the prob. is so high, that more decks do not make tangible difference. –  daroczig Jan 16 '11 at 17:19
    
"he may have an ace, and could stand, if you have no more than 17 and did not bust" - are you suggesting that the dealer's action depends on whether I have more than 17 or not? The dealer should act the same whatever value I stand at. –  Chris Moore Jul 5 '12 at 10:02
    
Well, if the player has 10+5 and the dealer has a 6 and an ace, I think the dealer could stand. –  daroczig Jul 5 '12 at 22:31
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