Stand a pair of 7s against a 10? Really?

Single Deck, H17, NDAS, No Surrender.

Stand a pair of sevens against a dealer 10?

Only 2 sevens. Hit any other 14.

For sure it will draw a comment, especially sitting third base.

But that aside, it must be because you already have two of the four 7s that can make your 21.

Just tell me it's not a typo, and, if the math is not too complex, why those two sevens mean so much, although it's the SECOND 7 that matters, since you hit 7-5, 7-6 7-8 7-9 against that 10, but not 7-7.

That single stand amongst all those hits against a dealer 10 sticks out like a sore thumb in the chart.

And if we're into the obscure parts of the basic strategy canon, why not stand those 7s against an Ace?

Maybe the simulations tell us so, but beyond it being a matter of faith, why?

Because there are only two 7's that'll get you to 21 (by hitting only once).

A dealer with a 10 up is likely to have a 20.

The '7' is a magic card in the 7,7 hand because it gets you to 21. It doesn't get you to 21 when holding 7,5, 7,6, 7,8, or 7,9.

Dealer with an Ace up is not super likely to have a 20. You already know the dealer does NOT have a ten in the hole, so he either has A-9 in the hole. You're in a better position with dealer having an ACE up (after checking for BJ and not having it) than you are with a Ten up.


You also have to take into account the 7 as being the dealer's hole card OR a hit card. If dealer has 2, 3, or 4 in the hole, then having more 7's in the deck is bad, because he can draw to 19, 20, or 21 respectively. If he has a 5 or 6 in the hole, more 7's in the deck is better.

In a fresh single deck, when you have two sevens and the dealer shows a ten, the dealers hole card will be a ten fifteen out of fortyeight times. Less than one third of the time, so to say it's likely is incorrect.
If you hit the two sevens, what can you expect? Of the remaining cards, you can get 15 tens, four nines and four eights that will break you. That's almost half the cards. An ace, or deuce doesn't really help you. So what will help? There are four threes, four fours, four fives,four sixes, and two sevens. 18 cards that help you. So the chances of drawing a card that substantially improves your hand is well under fifty percent, thus staying is the best move. You will lose most of the time, but less often if you stay than if you hit.
Hope that helps.

Quote: billryan

In a fresh single deck, when you have two sevens and the dealer shows a ten, the dealers hole card will be a ten fifteen out of fortyeight times. Less than one third of the time, so to say it's likely is incorrect.
If you hit the two sevens, what can you expect? Of the remaining cards, you can get 15 tens, four nines and four eights that will break you. That's almost half the cards. An ace, or deuce doesn't really help you. So what will help? There are four threes, four fours, four fives,four sixes, and two sevens. 18 cards that help you. So the chances of drawing a card that substantially improves your hand is well under fifty percent, thus staying is the best move. You will lose most of the time, but less often if you stay than if you hit.
Hope that helps.

I was just about to post same thought. Absolutely, less likely,I agree and couldn't have said it better. ( only with the WIZ"s approval of course )

Quote: billryan

Hope that helps.

In two ways. It's the right answer and it explains it in examples that I can "see" with a deck of cards in front of me.

There are 31 cards that will hurt or do no good, and only 18 that will help (I get to 49 because I assume not knowing the burn card in the fresh single deck). Only 37% of the time do I improve my hand. Duh...

Gotta learn to think these things through a little bit more. Then mental exercise of thinking more about it before I fire off a thread here will help keep the synapses firing.

Thanks for a great answer.

I didn't read all the posts in this thread so pardon me if it's already been mentioned, but hitting a two-card 14 until you have a pat hand will bust 56% of the time. And you would surrender 7,7 vs 10-up in SD. (Not that you'll ever find surrender in SD.)

Quote: racquet

In two ways. It's the right answer and it explains it in examples that I can "see" with a deck of cards in front of me.

There are 31 cards that will hurt or do no good, and only 18 that will help (I get to 49 because I assume not knowing the burn card in the fresh single deck). Only 37% of the time do I improve my hand. Duh...

Gotta learn to think these things through a little bit more. Then mental exercise of thinking more about it before I fire off a thread here will help keep the synapses firing.

Thanks for a great answer.

Actually, if you think like that you will make a lot of wrong decisions in BJ. You MUST consider what the Dealer is likely to make.

For example, with 7-7 vs TEN in a single deck game, the dealer will make the following hands with these probabilities:

17: 8.7% (its a low frequency because you have two 7s)
18: 13.0 %
19: 12.9 %
20: 37.1 %
21: 3.9 %
Bust (22 or more) 24.5%

So, by standing, you only win 24.5% of the time and you lose 75.5%. We say that the STAND option has an Expected Value of EV = -0.51 (which is 0.755 - 0.245). The higher your EV, the better -so an EV = - 0.51 is pretty bad.

Now look at what happens if you hit and draw a 3. You get a 17, which will win 24.3% of the time (when the dealer busts), pushes 8.5%% of the time and loses 67.2% of the time. Your Expected Value with a 7-7-3 hand is EV = -0.43. So drawing a 3 helps, but it only helps a very tiny amount.

If you were to draw a 7 and make a 21, your 21 will beat the dealer 96.1% of the time and you will push only when the dealer gets a 21 -3.9% of the time. With a 7-7-7 against the dealer, your expected value is EV = +0.94. That is a huge improvement over -0.51. What you can see is that, when the dealer holds a 10, the ability to get a 21 against the dealer is very important. The hand 7-7 vs 10 is a freak hand because the presence of two sevens in your hand greatly discourages you from HITTING your hand.

In single deck BJ, the Expected Value of your options when holding 7-7 vs 10 is:

STAND: EV = - 0.510
HIT: EV = -0.515
SPLIT: EV = -0.624
DOUBLE EV= -1.035

So all, you options are bad, you will, on average, lose more than 50% of your original bet no matter what you do. But the STAND option has the highest expected outcome, so it is the best way to go..

I've had this exact scenario once. A friend and I were paying the "blackjack pays 2:1" coupons at Casino Royale. I was at first base and he was on my left in the second spot. The table was full so it was one round then shuffle. I got the 7, 7 vs 10 and tucked the cards. My friend had 10, 3 and hit. Low and behold he draws a 7! Go figure that the first time this shows up, the 7 is there for me. Luckily the dealer busted. :-) It took nearly an hour for me to get the snapper, but it was a good session netting about $300 flat being quarters on this abysmal 6:5 game.

Quote: Ibeatyouraces

I've had this exact scenario once. A friend and I were paying the "blackjack pays 2:1" coupons at Casino Royale. I was at first base and he was on my left in the second spot. The table was full so it was one round then shuffle. I got the 7, 7 vs 10 and tucked the cards. My friend had 10, 3 and hit. Low and behold he draws a 7! Go figure that the first time this shows up, the 7 is there for me. Luckily the dealer busted. :-) It took nearly an hour for me to get the snapper, but it was a good session netting about $300 flat being quarters on this abysmal 6:5 game.

I played those BJ pays 2:1 coupons at Casino Royale for quarters for years.