Question about basic strategy in Blackjack

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.


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.


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%.