Basic Strategy Question

[Angelo]

Personal Background:
I'm almost 50 years old with descent knowledge of Math - however I've never taken statistics. I JUST started playing blackjack and looking at the Basic Strategy.

I've found a few things in the basic strategy I don't agree with, but I thought I'd pick on this one scenario because to me - it is OBVIOUSLY wrong. Having said that - I also feel that if it was as obviously wrong as it appears to me - someone would have noticed and corrected it by now. So I do realize that while it seems obvious to me - the reality is probably just something I don't understand. So now for the scenario.

Question:
Why does basic strategy dictate that if PLAYER IS HOLDING A HARD 16 and DEALER IS SHOWING A 7 - PLAYER SHOULD HIT?

In my opinion - If you have a hard 16 - a hit will BUST you more than 60% of the time leaving you with only a 40% shot at winning the hand. Taking into account the dealers cards - you still cannot get over 40% chance of winning because of your busts. Therefore, why not just surrender and lose 50% of you money 100% of the time instead of 100% of your money way more than 60% of the time.

Thanx!

Angelo

 

[SuperTrump]

Personal Background:
I'm almost 50 years old with descent knowledge of Math - however I've never taken statistics. I JUST started playing blackjack and looking at the Basic Strategy.

I've found a few things in the basic strategy I don't agree with, but I thought I'd pick on this one scenario because to me - it is OBVIOUSLY wrong. Having said that - I also feel that if it was as obviously wrong as it appears to me - someone would have noticed and corrected it by now. So I do realize that while it seems obvious to me - the reality is probably just something I don't understand. So now for the scenario.

Question:
Why does basic strategy dictate that if PLAYER IS HOLDING A HARD 16 and DEALER IS SHOWING A 7 - PLAYER SHOULD HIT?

In my opinion - If you have a hard 16 - a hit will BUST you more than 60% of the time leaving you with only a 40% shot at winning the hand. Taking into account the dealers cards - you still cannot get over 40% chance of winning because of your busts. Therefore, why not just surrender and lose 50% of you money 100% of the time instead of 100% of your money way more than 60% of the time.

Thanx!

Angelo

Fred Renzie's explanation of 16 Vs 7:

"A dealer will break only 3% more often with a 7 up than with a playable 10 (26% of the time Vs 23%). And if you hit, your own chance to bust is exactly the same in both cases. The problem is, against the 10, if you manage to dodge the bullet by catching a small card like a deuce, you're still holding an underdog hand. But if you snag the same deuce against a 7 up, you've probably made yourself a winner. Yet you had the same chance to bust in both spots.

Hitting 16 against a 10 is a bit like drawing to an inside straight in poker; you almost need to catch the perfect card. 16 against a 7 however is more like drawing to a flush, since there are so many more cards that will win the hand for you"

-Blackjack Bluebook II-

 

[Unshake]

Oversimplified example but this is how I thought of it (I very well might be wrong):

Hitting you are going to get +1 unit 2/5 (40%) times and lose 1 unit 3/5 (60%) times (Assuming hitting to make a hand = a winner)

OR Surrendering you can lose .5 units every time.

After playing 10 hands using each strategy, strategy one gives you +4 + (-6) = -2, a loss of 2 units. After playing 10 hands losing .5 units each hand you would have lost 5 units. Either way you are in a negative expectation situation but you lose less hitting.

I guess I thought of your question as, is it better to lose .5 every time (-.5) or lose 1 60% of the time but win 1 40% of the time (-.2)

 

[davidpom]

The Nike slogan works well here: "Just do it..."

 

[Angelo]

Not that bright in math.

OK - so maybe I'm not that bright in math. Let me break this up a little.

First - If you hold a hard 16 and HIT - do you have a 8 out of 13 chance of busting (about 61.5%).

 

[Unshake]

OK - so maybe I'm not that bright in math. Let me break this up a little.

First - If you hold a hard 16 and HIT - do you have a 8 out of 13 chance of busting (about 61.5%).

Yes assuming the count is 0 8/13 will make you bust. Only A,2,3,4,5 will make a hand. I guess this might depend on your hand make up as well, but for discussions sake lets say the count is 0, since that is generally where basic strategy comes from.

 

[shadroch]

IF you stay on 16 vs 7, the only way you can win is if the dealer bust. In the long run ,he will bust roughly 26 percent of the time, meaning you will lose slightly less than 3 out of every 4 chances.
If you hit, you will bust 61% of the time, meaning you will survive 39% of your hands. By hitting, you've increased the possibility of winning or tieing by over 50%.
You are still a distinct underdog, but a 39% chance to win or tie beats a 23% of winning.
Surrendering, you have zero chance of winning and will lose half of every bet you make.Thats the same as winning 25% of you bets.

 

[TakingTheHouse]

I just think of it as you have to assume the dealer will have a 10 covered up. If they do, they have 17 and you lose. If they have anything lower, they can hit again and still have a good chance to wind up higher than a 16.

 

[Angelo]

I see a light.

First - UNSHAKE - I was looking at loosing 60% of the time with looking at winning 40% and the NET EFFECT. Thanx for that perspective. It makes a world of difference.

Second - Fred - You stated A dealer will break only 3% more often with a 7 up than with a playable 10 (26% of the time Vs 23%). I was looking at it assuming dealer 10 after the second card - hense a much larger chance of dealer bust.

I saw an obvious flaw in the basic theory. Now, as I suspected, I see the obvious flaw is in my thinking.

Thanx all of you who responded for turning on the beacon. Now, I see the light.