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.