Basic Strategy Algorithm

What all is considered in the development of Basic Strategy for stiff hands?

It must be more than simply calculating favorable odds of the player not busting vs. odds of the dealer busting. Otherwise, you'd be better off hitting vs. all dealer up-cards from 12 to 14...

If you stand on a stiff hand, your only chance of winning is if the dealer busts. So...

Odds of player not busting on 12 = about 70% vs. Highest Dealer bust card (42%). If that's the only consideration, you should hit on all 12s.
Odds of player not busting on 13 = about 60% vs. Highest Dealer bust card (42%). if that's the only consideration, you should hit on all 13s.
Odds of player not busting on 14 = about 44% vs. Highest Dealer bust card (42%). If that's the only consideration, you should hit on all 14s.

This definitely does not follow Basic Strategy, so there must be something more to it... Either that, or BS is essentially aptly named ;)

Can anyone shed some more light on this please?

Thanks!

the wizard's appendix 1 might be what you are looking for

https://wizardofodds.com/games/blackjack/appendix/1/

Quote: BlkJkPlyr

What all is considered in the development of Basic Strategy for stiff hands?

It must be more than simply calculating favorable odds of the player not busting vs. odds of the dealer busting. Otherwise, you'd be better off hitting vs. all dealer up-cards from 12 to 14...

If you stand on a stiff hand, your only chance of winning is if the dealer busts. So...

Odds of player not busting on 12 = about 70% vs. Highest Dealer bust card (42%). If that's the only consideration, you should hit on all 12s.

You start by figuring out whether or not to hit a hard 21 (hint: don't), then determine the expected value of standing on hard 21 (which is the probability of dealer having 17-20 + probability of dealer busting).
Next, you figure out what the expected value of hitting a hard 20 is (it's 1/13 x the EV of a hard 21 + 12/13 x the expected value of a player bust (which is -1)) and compare that to the EV of standing on a hard 20 (which is the probability of dealer having 17-19 - probability of dealer having 21 + the probability of dealer busting).
Keep going down the numbers.
When you get to 12, the EV of a hit is (1/13 x the EV of stiff 13 + 1/13 x the EV of stiff 14 + ... + 1/13 x the EV of stiff 21 + 4/13 x the EV of a player bust), and the EV of a stand is (probability of dealer busting - probability of dealer having 17-21).

Note that if you hit a stiff 12, four of the 13 ranks (A, 2, 3, 4) put you in a position where you are no better off than you were, since you still lose unless the dealer busts, and you are now more likely to bust if you hit again.

Thanks odiousgambit! Great reference!

Thanks DonGuy!

Well, that answers my question. There's definitely more to it. haha. Will just take some practice getting familiar. Will continue to use BS, but was just wanting to solidify in my mind the logic behind it. Thanks!

Quote: BlkJkPlyr

there must be something more to it...

Don't overlook the fact that the dealer wins if both the player and the dealer bust.