Multiple hand "single event" probability blackjack

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What is considered high? ≥50%?

In my example, a hand of a King and a four has a 56% chance to bust [see chart below] (note: this is the chance to lose immediately), however, even if you don't bust, the dealer could still beat the hand, so the chance to lose is even greater than 56%, more like 60-65%.

Basic Strategy has you hit that though, because you might get lucky, and if you do then over the long, long run, you would still come out ahead with your 35-40% chance to win, versus surrendering half the bet every time.

However, I am pointing out that I would find it more optimal to take the 50% gain, than to risk it as 'Basic Strategy' would tell you to do.



http://blackjackswitchonline.com/blackjack-odds.htm

As I said in the edit of my last post, this all depends on what exactly your goals in playing are, and your existing resources.

The theory seems too conditional to have any practical application.

What makes no sense (unless you are trying to give money to casinos) is that when playing two hands your choice of whether or not to play a hand would depend on what is in the other hand.

I imagine casinos would welcome anyone who did make choices like that.

Based on what ?

You walk into a casino .. you play 2 hands at once .. the first one is blackjack (so locked in profit), and the second is probabilistically a loser (you have not acted on it), so you can surrender the second, and walk out with guaranteed winning % on your money. That is the best statistical expectancy play (i.e., the best play to walk away with the most money for a single event) when the second hand is most likely a loser. It is because you know you won on the first hand that allows you take the half-bet surrender lose on the second, and still know you walk out a winner.

Basic blackjack strategy is for the long, long run, not a single event in blackjack. So, for a single event, popping in with a friend for a few hands, the probability theory is entirely different. You are not playing 1000 hands waiting for the probability to pan out. You want a mathematical strategy to maximize winning expectancy for just a few hands.

Playing poorly for a few hands does not create a good probability of walking out a winner.

I never suggested playing poorly.

You appear to be quoting off the basic strategy card the casino sells ? Those probabilities have nothing to do with maximizing the % winning expectancy in the short run. The card assumes you will play many hands, and what it tells you to do is based on that assumption.

If you only plan to play a few hands, you have blackjack on hand 1, and your second hand is a likely to bust hand say a '14', then you won't play enough hands for the probability to cycle around -- to make sense to risk a hit on 14, so surrender is the best play for someone who wants to maximize winning expectancy.

Define 'winning expectancy'.

If you are playing two hands and one of them is worth surrendering, the same hand would have been worth surrendering even if you were playing one hand. If you are playing one hand and a hand is worth surrendering, then it would also be worth surrendering if you were playing two hands.

Just because you might have a blackjack in one hand has no effect on what is the smart move on the other hand.

On the contrary, I showed the math below.

Assume: The second hand win probability is 39%.

Two possible outcomes:
a. a 39% chance to walk out of the casino with 250% winning of total bet on both hands (blackjack + second hand won = 250%)
b. a 61% chance to walk out of the casino with 50% winning of total bet on both hands (blackjack + second hand losses)
c. a 100% chance to walk out of the casino with 100% winning of total bet on both hands (blackjack + surrender =100% of total bet won)

Thus,
E(a) = .39 * 250% = 97.5% (you are expected to walk out with 97.5% gain of your total bet, blackjack + play 39% to win hand)
E(b) = .61 * 50% = 30.5% (you are expected to walk out with 30.5% gain of your total bet, blackjack + LOSE second hand)
E(c) = 1 * 100% = 100% (you can walk out with 100% gain on your total bet, blackjack + surrender)

100% > 97.5% >> 30.5%, thus the maximum winning expectancy is to surrender. (i.e., probability shows that you walk out with the maximum expected money if you surrender, though, not by much ) However, basic strategy has you play 25+% hands, so in those much worse cases, the expectancy would even stronger in favor of surrendering.

This is not what basic strategy would tell you to do, however, basic strategy fails to view both hands together to determine optimal winning expectancy. Note: This math is wrong for the assumption of playing many hands (i.e. "the long run") individually, because the expectancy would be calculated differently.

Last edited by LoveNotHate on 03 Aug 2014, 4:22 am, edited 3 times in total.

You cut off the other 61% from choice a. You can't do that and still calculate an accurate expected value.

If I understand this right, and correct me if I don't, this is the correct way to do it:

Two possible outcomes:
a. a 39% chance to walk out of the casino with 250% winning of total bet on both hands (blackjack + second hand won = 250%) and a 61% chance to walk out of the casino with 50% winning of total bet on both hands (blackjack + second hand lost = 50%)
b. a 100% chance to walk out of the casino with 100% winning of total bet on both hands (blackjack + surrender =100% of total bet won)

Thus,
E(a) = .39 * 250% + .61 * 50% = 128% (you are expected to walk out with 128% gain of your total bet, blackjack + play 39% to win hand)
E(b) = 1 * 100% = 100% (you can walk out with 100% gain on your total bet, blackjack + surrender)

39% to win means a 61% to lose. It should be obvious that when the second hand loses it cannot be the maximum expectancy.

Here is it is .. it is not relevant as you will see ...

E(c) = .61 * 50% = 30.5% (you are expected to walk out with 30.5% gain of your total bet, blackjack + LOSE second hand)

100% >> 30.5% .. I did not include it because it is obviously not maximum expectancy.

You don't maximize over individual paths, you maximize over expectation, which is a weighted average of all paths that arise from a particular choice. This is not a radical statement; it is the definition of expected value.

Yes, you could do that. That is how the "long run" expectancy is calculated to favor playing the second hand, and thanks for being so insightful. You are very intelligent for seeing that.

However, in this case, it is not weighted purposely, because I am playing for the "short run", and "averaging" only makes sense if you plan to play a lot of hands. The best decision in the short run is to compute the absolute expectancy for the three outcomes, and pick the one with the highest expectancy. I won't play enough hands for the "averaging" to work in my favor, so I suggest to compute the absolute expectancy.

Last edited by LoveNotHate on 03 Aug 2014, 4:45 am, edited 1 time in total.

Consider a choice between the following two options in a one-shot (not Blackjack) game:

a) 50% to win $1000 and 50% to win $999.
b) 50% to win $1000 and 50% to win nothing.
By your logic, there is no reason to have any preference between these choices.