Late Surrender Basic Strategy: Why the Correct Strategy Doesn’t Make Sense
By Arnold Snyder
(From Casino Player, February 1994)
© 1994 Arnold Snyder
Question from a Reader: The widely published and disseminated blackjack basic strategy for late surrender seems to me to be wrong. The strategy that is accepted is to surrender a hard 15 vs. a 10, and hard 16 (excluding 8, 8) to a 9, 10 or Ace. Although I am not using high-level blackjack mathematics or computers for my reasoning, it seems to me that the correct strategy would be to surrender every hard 15 or hard 16 (except 8, 8) any time a dealer shows 7, 8, 9, 10 or Ace.
My reasoning is that any time you take a hit on 15 or 16, over half of the cards in a full deck will bust you. If you have a 15, drawing a 7, 8, 9, 10, Jack, Queen or King will bust you. That equates to 7 out of the 13 possible cards to draw. If a blackjack game consists of four decks, and the player holds an 8 and a 7 versus a dealer 7, and those are the only cards that have been played, then 109 of the 205 unseen cards will bust him.
If a player is over 50% likely to lose all his bet, then it seems advantageous to surrender one half of his bet all the time. If my reasoning is flawed, I would greatly appreciate an explanation as to why.
Answer: This is a wonderful question because your argument is so persuasive, that to anyone — other than a mathematician — it makes perfect sense. First of all, you are 100% correct in most of your analysis. In a four-deck game, there are a total of 208 cards (52 x 4). If you remove a player hand consisting of a Seven and an Eight versus a dealer upcard of Seven, you will have 205 unseen cards remaining.
Of those 205 remaining cards, 109 cards (all remaining Sevens and Eights, plus the Nines, Tens, Jacks, Queens and Kings) will bust the player’s hard 15. One hundred nine cards represent 53% of the 205 possible hits. Furthermore, you would also expect to lose at least some of the 47% of the hands which you did not bust with one hit, depending on the ultimate totals of your own hand and the dealer’s hand.
So, since you know you’re going to lose more than 50 percent of your 15s played out against a dealer Seven—no doubt about it!— why don’t the “experts” tell you to surrender this hand as basic strategy, and hold your losses to an even 50 percent and no more?
This makes perfect sense, right?
WRONG!
This is what happens when amateurs try to do a dangerous stunt like statistical analysis. Statisticians are the Evel Knievels of the math world — trained professionals who dare to perform their feats of mental wizardry without any safety nets. But please, don’t try it yourself at home. You’re liable to start devising your own “basic strategy,” and the next thing you know, you’ll be panhandling for pocket change, wondering where your savings went!
Let’s simplify this problem. Forget about decks of cards. Instead, let’s use marbles in a vase.
Put 100 marbles in an opaque vase — 47 white marbles and 53 black marbles. You have to reach in and draw out one marble. If you draw a white one you win $1, and if you draw a black one you lose $1.
It is obvious from the start that you are going to lose 53 times out of 100 draws.
Therefore, if I offer you a surrender option, whereby you may simply give me 50¢ per draw, rather than risking a dollar to draw a marble, would you surrender?
No!
You’re forgetting that when you don’t lose, you win. In our marble example, you will expect to lose $53 on every 100 draws, but you offset this loss by winning $47 out of every 100 draws. So, your net result after 100 draws will be a loss of only $6. If you surrender 50¢ on all 100 draws, you will lose $50 instead of losing only $6! So, just because you know you’re going to lose more than 50 percent on a specified hand does not make it a surrender decision. No way!
What percentage of your hands do you have to expect to lose before you would be better off surrendering half your bet?
Consider the marbles…
If I had 60 black marbles (losses) and 40 white marbles (wins), what would be my net result from 100 draws?
-60 + 40 = -$20.
Still not enough black marbles to make surrendering a wise decision. What about 70 black marbles and 30 white ones?
-70 + 30 = -$40.
Still not enough black marbles. How about 75 black marbles and 25 white ones?
-75 + 25 = -$50.
Aha!
This is our break-even point, where 100 draws would result in the same expectation (-$50) as 100 surrender decisions.
Since 75 is exactly 3 x 25, then you know that surrendering half your bet is the optimal strategy decision only if you will lose more than three times as many hands as you will win!
With cards, instead of marbles, the math is not so simple because some hands will push, and we don’t know the precise win/loss percentages for Seven, Eight versus Seven without a more detailed analysis showing all of the possible player totals versus all of the possible dealer totals. But the simple fact remains that unless you expect to lose more than three times more often than you expect to win — don’t surrender.
If you want to work out all of the possible outcomes yourself, use a computer to cycle through all of the possible hands, because there are many thousands of possibilities. You could spend months trying to figure out what to do with 15 versus 7.
Instead, trust the dozens of mathematicians and computer programmers who have all come to the same conclusion — don’t surrender that hand. It’s true you will lose more than 50% of these hands. But you will not lose more than three times as often as you will win. And that’s how you determine your surrender basic strategy. ♠