Splitting 8-8

I know the conventional wisdom. Always split 8's. Yet instinctively and intuitively, I just don't see how splitting and getting two 18's vs. a dealer's 19 or 20 is a good deal.

I understand that splits/doubles present themselves as a result of initial split, yet I just have a very hard time wrapping my head around the advantages of risking so much more money, at least twice, and with doubles could be 3, 4, or even 5 x original bet, vs. just treating it as a 16 and taking either one win or one loss.

The videos posted, especially the Excel video, seems to imply that the calculations could be done rather efficiently by someone of Michael's ability.

Can someone please offer a numeric comparison between the odds of a simple hit vs. the split?

Thank you in advance.

Quote: bcmarshall

I know the conventional wisdom. Always split 8's. Yet instinctively and intuitively, I just don't see how splitting and getting two 18's vs. a dealer's 19 or 20 is a good deal.

Your instincts and intuition are correct. If the dealer is showing a 19 or 20, do not split 8s. Remember, most strategy charts say to split 8s only when the dealer is showing an 11 or less. There is some sarcasm in there, but it should show where your error is.

Quote: bcmarshall

Can someone please offer a numeric comparison between the odds of a simple hit vs. the split?

https://wizardofodds.com/games/blackjack/hand-calculator/

Value of standing or hitting on 8s against a 10 is -53%. Value of splitting is -47%.

Only split if you can’t surrender.

Quote: bcmarshall

I know the conventional wisdom. Always split 8's. Yet instinctively and intuitively, I just don't see how splitting and getting two 18's vs. a dealer's 19 or 20 is a good deal.

I understand that splits/doubles present themselves as a result of initial split, yet I just have a very hard time wrapping my head around the advantages of risking so much more money, at least twice, and with doubles could be 3, 4, or even 5 x original bet, vs. just treating it as a 16 and taking either one win or one loss.

The videos posted, especially the Excel video, seems to imply that the calculations could be done rather efficiently by someone of Michael's ability.

Can someone please offer a numeric comparison between the odds of a simple hit vs. the split?

Thank you in advance.

The Wizard has a blackjack hand calculator.
https://wizardofodds.com/games/blackjack/hand-calculator/

Just choose the rules, the number of decks, the dealer's up card, and your cards. Then click on calculate to compare the expected values for hit, split, etc.

Below is a picture of the calculator.

Quote: TomG

Your instincts and intuition are correct. If the dealer is showing a 19 or 20...

I’m confused.

Are you playing a version of BJ where you see both of the dealer’s cards before you make a decision?

Quote: TomG

Your instincts and intuition are correct. If the dealer is showing a 19 or 20, do not split 8s.


Value of standing or hitting on 8s against a 10 is -53%. Value of splitting is -47%.

First off, I assume you mean “dealer is showing 9 or 10”.

Your statements are contradictory. First you say don’t split 8s against a 9 or 10, then your figures show that, at -47%, splitting 8s against a 10 is the best play.

???

Quote: Ace2

First off, I assume you mean “dealer is showing 9 or 10”.

Your statements are contradictory. First you say don’t split 8s against a 9 or 10, then your figures show that, at -47%, splitting 8s against a 10 is the best play.

???

No, he means do not split 8s against a dealer 19 or 20.
Do split 8s against a dealer showing a 9 or 10 along with an unknown hole card.

I’m going to ask my question again:

Are you playing a version of BJ where you see both of the dealer’s cards before you make a decision?

Quote: DJTeddyBear

I’m going to ask my question again:

Are you playing a version of BJ where you see both of the dealer’s cards before you make a decision?

I'm guessing it was tongue in cheek suggesting that you would not split 8s if you 'accidentally' caught sight of the hole card and that you knew you were beaten..

Quote: DJTeddyBear

I’m going to ask my question again:

Are you playing a version of BJ where you see both of the dealer’s cards before you make a decision?

I’ll help. The OP asked why would you split 8’s to lose to a dealer 19 or 20.

Tom G’s excellent reply basically said you don’t know that the dealer has a 19 or 20, you only know the dealer has a 9 or 10.

a guess: OP learned to play BJ from dealers and other players so is going by the advice to 'assume the dealer has a 10'

think about this, sir. 8,8 is sixteen. Anything you can do to get rid of 16 is what you want to do

Quote: bcmarshall

I know the conventional wisdom. Always split 8's. Yet instinctively and intuitively, I just don't see how splitting and getting two 18's vs. a dealer's 19 or 20 is a good deal.

There is a false rule of thumb, that I've heard at least 100 times, that goes "Always assume the dealer has a 10 in the hole." I think the OP incorrectly believes that.

Quote: Wizard

There is a false rule of thumb, that I've heard at least 100 times, that goes "Always assume the dealer has a 10 in the hole." I think the OP incorrectly believes that.

I always assume the dealers has something around 8.5 in the hole.

A better assumption is that one of the dealers' three cards will be a ten. If the dealer is showing a non-ten, there is, I believe, a greater than 50-50 chance one of the next two cards will be a ten.
How useful that information is is debatable.

Quote: Wizard

There is a false rule of thumb, that I've heard at least 100 times, that goes "Always assume the dealer has a 10 in the hole." I think the OP incorrectly believes that.

And assume the next card out is going to be a ten.

I know this concept makes you want to pull your hair out, but trust me, with a few other rules added in, such as, you will always stay on a hard 17, you want the dealer to bust first actually helps a complete novice quickly learn a crude basic strategy. You start with that and then add some exceptions.

I can see the steam coming out of our blackjack gurus' heads, they are wondering why in God's name would anyone not just take the time to learn it right before they ever played a hand. Well, there have been quite a few juicy blackjack promotions that suddenly pop up, you know the promo won't last long. As an Advantage Player, you want as many people playing for you as possible, within reason, of course.

I have had really good success using my crash course blackjack training method that uses the "the dealer has a 10 and a 10 is coming" rule of thumb.

I know Barbra was skeptical as well, but she saw it in action.

Axel writes; "I have had really good success using my crash course blackjack training method that uses the "the dealer has a 10 and a 10 is coming" rule of thumb."

Axel. So this means you woild never hit 12-16. Always double 10,11. Always insure. No?

Quote: AxelWolf

I know this concept makes you want to pull your hair out, but trust me, with a few other rules added in, such as, you will always stay on a hard 17, you want the dealer to bust first actually helps a complete novice quickly learn a crude basic strategy. You start with that and then add some exceptions.

This would be worthy of a separate thread, and I'll split it off if you give a substantive reply.

That said, if anyone can give me a few rules of thumb that will result in a low overall error rate, I'm all ears. Please don't just say "Assume the dealer has a 10 in the hole." That wouldn't tell you how to play lots of hands. Does that rule imply assume EVERY future dealer card is a 10? For example, how does that rule advise playing 12 vs. 2? When you say a "ten is coming," does that indicate you would stand on 12 vs 2, because you would get that next ten if you hit? Also, do you still follow that rule when the dealer has an ace up and has already peeked for blackjack?

Mr Wizard. The assumption of a 10 coming or in the hole would only be correct 30.77% and wrong 69.23%. 😲 Still, I admit it seems far more often than not.🤔

The dealer has an unknown card. What are the chances that one of the next two cards is a ten?

Quote: billryan

The dealer has an unknown card. What are the chances that one of the next two cards is a ten?

There are 16 tens and 36 non tens in a single deck. So 30.77% from a fresh deck. Of course this differs depending cards played. But even at RC 20 on a 10 count system, the chance is still just 50-50%.

Am I doing this wrong?
The next card has a 4/13ths chance of being a Ten. As does the next one.
4/13 +4/13= 8/13.
At a zero count, aren't the chances much higher than 50-50?

Quote: billryan

Am I doing this wrong?
The next card has a 4/13ths chance of being a Ten. As does the next one.
4/13 +4/13= 8/13.
At a zero count, aren't the chances much higher than 50-50?

Yes your math is wrong. For infinite deck try:
1 - (1 - 4/13)*(1 - 4/13) for chance of neither next two cards a ten.

Quote: unJon

Yes your math is wrong. For infinite deck try:
1 - (1 - 4/13)*(1 - 4/13) for chance of neither next two cards a ten.

My math sucks. Can you walk me thru that in English, so I can understand it.

My original statement was one of the dealers' three cards will be a ten, not that one of his next two.
What are those chances. Thanks.

Quote: billryan

My math sucks. Can you walk me thru that in English, so I can understand it.

My original statement was one of the dealers' three cards will be a ten, not that one of his next two.
What are those chances. Thanks.

Let’s stick with infinite deck.

What is the probability that at least one of dealers three cards a ten?

That’s the same as 100% minus the probability that all three dealer cards are non-ten. The chance that any one card is non-ten is 9/13. So you have to hit a 9/13 three times in a row.

That’s 9/13 * 9/13 * 9/13 = 33.18%. So chance of at least 1 ten is 100% minus 33.18%.

Quote: unJon

Let’s stick with infinite deck.

What is the probability that at least one of dealers three cards a ten?

That’s the same as 100% minus the probability that all three dealer cards are non-ten. The chance that any one card is non-ten is 9/13. So you have to hit a 9/13 three times in a row.

That’s 9/13 * 9/13 * 9/13 = 33.18%. So chance of at least 1 ten is 100% minus 33.18%.

So the chances of the dealer having or getting a ten in his hand are 66+%?

Unjon writes:That’s 9/13 * 9/13 * 9/13 = 33.18%. So chance of at least 1 ten is 100% minus 33.18%.

Translation.. Are you saying there is a 66.82% chance a ten is either in the hole or the next card? Provided the dealers upcard is not a 10.

Quote: billryan

My math sucks. Can you walk me thru that in English, so I can understand it.

My original statement was one of the dealers' three cards will be a ten, not that one of his next two.
What are those chances. Thanks.

Eight Decks: 416 Cards

Player 8-8: (Two Non-Ten Cards)

We will say the dealer's upcard hasn't come out yet. There are (16 * 8 = 128) ten value cards remaining of 414 total remaining cards. 414-128 = 286

nCr(286,3)*nCr(128,0)/nCr(414,3) = 32.8611% (Rounded)---Zero Dealer Tens

nCr(286,2)*nCr(128,1)/nCr(414,3) = 44.4319% (Rounded)---One Dealer Ten

nCr(286,1)*nCr(128,2)/nCr(414,3) = 19.7995% (Rounded)---Two Dealer Tens

nCr(286,0)*nCr(128,3)/nCr(414,3) = 2.90761% (Rounded)---Three Dealer Tens, even though that can't happen.

However, by the time this information actually matters (when it comes time for the player to make a decision) the dealer's upcard either is already a ten or is not a ten. For that reason, you either remove the ten (or not a ten), as well as any other known player cards, and do the math accordingly.

For one example, let's say the dealer's card is a nine.

Eight Decks: 416 Cards

Player 8-8: (Two Non-Ten Cards)

Dealer Upcard: 9

nCr(285,2)*nCr(128,0)/nCr(413,2) = 0.4756811396600766 or 47.5681% that the next two cards are not tens. It will be slightly more likely neither are tens if a ten is the dealer's upcard.

Anyway, none of this is really pertinent to anything. If the dealer's upcard is a nine, then 8-A all end dealer actions and we do not even see the dealer draw any cards.

Let's go with the dealer showing nine and look at more precise numbers:

1.) The undercard is already a ten:

128/413 = 30.9927%

2.) The undercard is 8, 9, or A and the hand ends anyway (Two eights and a nine are gone already):

93/413 = 0.22518159806 or 22.5182%

3.) The hole card is 2, 3, 4, 5, 6 or 7 and the dealer takes (at least) one other card, which is a ten:

192/413 = 46.4891% (Overall)

Ten: (192/413 * 128/412) = 0.14443216812 or 14.4432%

Not Ten: (192/413 * 284/412) = 0.32045887303 or 32.0459%

Therefore, in an eight-deck game and with the dealer showing a nine, the probability of the dealer either having a ten undercard OR having an undercard that would require the dealer to draw AND drawing a ten on the next card is: .309927 + .144432 = .454359 or 45.4359% (Based on rounded numbers)

A dealer could also end up with a ten in a sequence such as 9 + 3 (Undercard) + 2 (Drawn Card) + 10...or any other number of ways that I am not going to go through.

I hope that helps. Either way, it's less than 50/50 mainly because the dealer could have (not a ten) and still not have to draw any cards.

Quote: moses

Unjon writes:That’s 9/13 * 9/13 * 9/13 = 33.18%. So chance of at least 1 ten is 100% minus 33.18%.

Translation.. Are you saying there is a 66.82% chance a ten is either in the hole or the next card? Provided the dealers upcard is not a 10.

If you are asking me, I'm not saying that.

If you assume the dealer has a ten in the hole, you will be right approx. 31% of the time.
If you assume the dealer will have at least one ten in his final hand, you'll be right at least 66% of the time.
Is either bit of the information useful in helping how to play hands?

Quote: billryan

If you are asking me, I'm not saying that.

If you assume the dealer has a ten in the hole, you will be right approx. 31% of the time.
If you assume the dealer will have at least one ten in his final hand, you'll be right at least 66% of the time.
Is either bit of the information useful in helping how to play hands?

I wouldn't think that either of those facts are useful. Those are a priori probabilistic facts to knowing the dealer's upcard; players make decisions based on what the dealer's upcard is.

Quote: moses

Axel writes; "I have had really good success using my crash course blackjack training method that uses the "the dealer has a 10 and a 10 is coming" rule of thumb."

Axel. So this means you woild never hit 12-16. Always double 10,11. Always insure. No?

NO, that doesn't mean that. That's just the basic premise of it, I obviously didn't lay everything out in detail, like their goal is to get 17 or better along with other assumptions and things to consider. It's not like I have an easy for beginners blackjack pamphlet all written up and this is a frequent thing that comes up.

You seem to have quoted the last part of my post and ignored this part where I said, "crude basic strategy. You start with that and then add some exceptions."

I'm not suggesting or promoting people do anything other than learn how to blackjack completely and correctly using all the proper tools available. I would never suggest anyone that learned blackjack in this way run out and play on their own.

Remember, this is for situations where the advantage is significant enough to overcome any mistakes or incorrect plays when they only a short timeframe to learn.

Let's be clear on "I have had really good success". Meaning, I have taken people who have never played a hand of blackjack and within a few hours, I was comfortable enough to have them play for me while betting significant amounts with positive expectation and results given the nature of the promotions.

On the types of promotions you'd be into, some of your people playing Blackjack terribly is probably good cover anyway.

Quote: billryan

If you are asking me, I'm not saying that.

If you assume the dealer has a ten in the hole, you will be right approx. 31% of the time.
If you assume the dealer will have at least one ten in his final hand, you'll be right at least 66% of the time.
Is either bit of the information useful in helping how to play hands?

Interesting concept. If we hit 12-16 and it's a 10? Hole card doesnt matter. Fat lady is already singing.

Quote: Mission146

On the types of promotions you'd be into, some of your people playing Blackjack terribly is probably good cover anyway.

On the flip side, you can't be looking too novice as you don't want pit wondering what's going on and why this guy/gal is sitting there betting thousands per hand and they don't even know what a chip is, but they seem to be winning.

My bad Axel. I should've included your words "crude basic strategy."

Quote: Wizard

Also, do you still follow that rule when the dealer has an ace up and has already peeked for blackjack?

There are and will have to be exceptions. Perhaps people misunderstand. It's not like I would just tell someone to "assume the dealer has a 10 and a 10 incoming"..... Okay, your ready, here's my money go start playing.

Quote: AxelWolf

There are and will have to be exceptions. Perhaps people misunderstand. It's not like I would just tell someone to "assume the dealer has a 10 and a 10 incoming"..... Okay, your ready, here's my money go start playing.

You give others your money to play for you?

Quote: moses

My bad Axel. I should've included your words "crude basic strategy."

I see what you did there, as predicted by me.

And yes, that's why I used those exact words, for guys like you who might take issue and so I could avoid having to explain every detail because I myself dont know all the details until I'm sitting there with someone and can assess how quickly they can learn and remember different things. There are some people who can look at a strategy card and have it down in no time, while there are other people who can't remember one simple rule no matter how many times you go over it.

Quote: moses

You give others your money to play for you?

I have, I do, I will. It really depends on the situation, the person, and other factors, but I guess the simple answer is, yes.

Of course, if someone contacts me asking for money to play for me, the chances are, I will politely tell them to go pound sand.

Quote: AxelWolf

On the flip side, you can't be looking too novice as you don't want pit wondering what's going on and why this guy/gal is sitting there betting thousands per hand and they don't even know what a chip is, but they seem to be winning.

Not so terribly that they would be hitting Pat 17's, or anything. By, "Terribly," I meant assuming a dealer ten.

Quote: AxelWolf

I have, I do, I will. It really depends on the situation, the person, and other factors, but I guess the simple answer is, yes.

Of course, if someone contacts me asking for money to play for me, the chances are, I will politely tell them to go pound sand.

Edit to add, there's almost zero chance I would ever invest in anyone str8 up card counting.

Quote: AxelWolf

I see what you did there, as predicted by me.

And yes, that's why I used those exact words, for guys like you who might take issue and so I could avoid having to explain every detail because I myself dont know all the details until I'm sitting there with someone and can assess how quickly they can learn and remember different things. There are some people who can look at a strategy card and have it down in no time, while there are other people who can't remember one simple rule no matter how many times you go over it.

My methods are not always conventional. It's based on what has been played and what remains to be played with a percentage thereof for making a final decision.

Quote: AxelWolf

I have, I do, I will. It really depends on the situation, the person, and other factors, but I guess the simple answer is, yes.

Of course, if someone contacts me asking for money to play for me, the chances are, I will politely tell them to go pound sand.

I have the "go pound sand" part down pat. lol

Quote: bcmarshall

I know the conventional wisdom. Always split 8's. Yet instinctively and intuitively, I just don't see how splitting and getting two 18's vs. a dealer's 19 or 20 is a good deal.

I understand that splits/doubles present themselves as a result of initial split, yet I just have a very hard time wrapping my head around the advantages of risking so much more money, at least twice, and with doubles could be 3, 4, or even 5 x original bet, vs. just treating it as a 16 and taking either one win or one loss.

The videos posted, especially the Excel video, seems to imply that the calculations could be done rather efficiently by someone of Michael's ability.

Can someone please offer a numeric comparison between the odds of a simple hit vs. the split?

Thank you in advance.

I should respond to this OP.

Honestly, you HOPE that you end up betting more than 2x your initial bet, because that is a very good sign for you. Part of what makes the EV in favor of splitting (rather than hitting) is the fact that it results in sometimes winning more on the hands that you do win.

The reason that this happens is because the splitting now sometimes leads to situations that are +EV. You should want to get into a situation where you have to double. You split 8's against a nine or ten and one of your eights draws a three, so now you can double against a dealer nine, which is a great situation for the player.

However, the calculator even says it's a split with no DAS (double after split), eight decks and only being allowed to split once. It doesn't really get any worse than that for 8-8 and is still better than surrendering and much better than just taking a card. The only way hitting becomes better is those same rules against a ten and the dealer does not check for Blackjack, so you can't eliminate the dealer from having an ace, and in fact, the other rules don't matter and you would still only take one card if the dealer doesn't check (unless you could surrender).

So, how can it be better than hitting with no DAS? Let's break this one down with minimal math:

Hit 16: You have eight ranks that bust you and the hand ends immediately. You have five ranks (A-5) where you make a standing hand.

Split 8's:

1.) For each eight, you have six ranks where you make a standing hand (9, 10, J, Q, K, A), so your prospects for at least making a hand that can win or push a playing dealer hand are immediately better.

2.) Each eight can draw a two or three, which even without DAS, gives the opportunity for a hit on those hands that literally can not bust and can lead to making hands that can win or push the dealer.

3.) With exception to another eight, all the other ranks (4-7) lead to you having the opportunity to hit to a different hand total (12-15) that is less likely to bust than your sixteen was.

One other thing you can use the calculator to do is look at a player Hard Eight v. Dealer Ten for a close enough comparison. If you double the negative EV on the Hard Eight hand's best decision (hitting, obviously) it still comes out to a lower Expected Loss than does hitting sixteen against a dealer ten. Actually, doubling the -EV on hitting the Hard Eight has a lower Expected Loss (slightly) than does surrendering a Hard 16, but again, that's because you cannot immediately bust hitting a Hard Eight and also that Hard Eight can turn into a good hand to draw to, such as eleven.

Why this comparison? Well, I think the ability to DAS makes the decision obvious, but even if you couldn't...two hands that are each Hard Eight have a lower combined expected loss than one hand that would see you either hit or stand on 8-8. If you can't DAS, then you're basically splitting 8's to make two separate hands that are both Hard Eight.

So, I hope this explanation helps without getting too far into the weeds. I also don't like Blackjack, so Blackjack discussion isn't really my strong point.

Mission 146: For someone who doesn't like blackjack, you certainly provided an excellent explanation. DAS vs No DAS is a huge factor for determining some splits IMO.

Quote: moses

Mission 146: For someone who doesn't like blackjack, you certainly provided an excellent explanation. DAS vs No DAS is a huge factor for determining some splits IMO.

Thanks for the compliment, Moses!

Isn't it exactly the same expectation of a player's hand? Player should have at least one ten 66% of the time?

Not that it changes anything. Just trying to understand the math.

Quote: Mission146

I should respond to this OP.

Honestly, you HOPE that you end up betting more than 2x your initial bet, because that is a very good sign for you. Part of what makes the EV in favor of splitting (rather than hitting) is the fact that it results in sometimes winning more on the hands that you do win.

The reason that this happens is because the splitting now sometimes leads to situations that are +EV. You should want to get into a situation where you have to double. You split 8's against a nine or ten and one of your eights draws a three, so now you can double against a dealer nine, which is a great situation for the player.

However, the calculator even says it's a split with no DAS (double after split), eight decks and only being allowed to split once. It doesn't really get any worse than that for 8-8 and is still better than surrendering and much better than just taking a card. The only way hitting becomes better is those same rules against a ten and the dealer does not check for Blackjack, so you can't eliminate the dealer from having an ace, and in fact, the other rules don't matter and you would still only take one card if the dealer doesn't check (unless you could surrender).

So, how can it be better than hitting with no DAS? Let's break this one down with minimal math:

Hit 16: You have eight ranks that bust you and the hand ends immediately. You have five ranks (A-5) where you make a standing hand.

Split 8's:

1.) For each eight, you have six ranks where you make a standing hand (9, 10, J, Q, K, A), so your prospects for at least making a hand that can win or push a playing dealer hand are immediately better.

2.) Each eight can draw a two or three, which even without DAS, gives the opportunity for a hit on those hands that literally can not bust and can lead to making hands that can win or push the dealer.

3.) With exception to another eight, all the other ranks (4-7) lead to you having the opportunity to hit to a different hand total (12-15) that is less likely to bust than your sixteen was.

One other thing you can use the calculator to do is look at a player Hard Eight v. Dealer Ten for a close enough comparison. If you double the negative EV on the Hard Eight hand's best decision (hitting, obviously) it still comes out to a lower Expected Loss than does hitting sixteen against a dealer ten. Actually, doubling the -EV on hitting the Hard Eight has a lower Expected Loss (slightly) than does surrendering a Hard 16, but again, that's because you cannot immediately bust hitting a Hard Eight and also that Hard Eight can turn into a good hand to draw to, such as eleven.

Why this comparison? Well, I think the ability to DAS makes the decision obvious, but even if you o hands that are each Hard Eight have a lower combined expected loss than one hand that would see you either hit or stand on 8-8. If you can't DAS, then you're basically splitting 8's to make two separate hands that are both Hard Eight.

So, I hope this explanation helps without getting too far into the weeds. I also don't like Blackjack, so Blackjack discussion isn't really my strong point.

Wow! I really didn't expect such a detailed and intelligent answer, but I am now convinced. The best play is to split 8-8 against all.

BTW, in case I wasn't clear in the original post, I never questioned the wisdom of splitting against 2 thru 8. It was only 9-A that puzzled me.

I'm new to the forum. Can anyone tell me how to get email notifications of replies to the thread?

Quote: bcmarshall

Isn't it exactly the same expectation of a player's hand? Player should have at least one ten 66% of the time?

Not that it changes anything. Just trying to understand the math.

I’m not sure it would work out in theory, because of the differences in player and dealer actions.

Stand on Hard 17+—BOTH

Dealer Tens—The dealer hits on hands that player sometimes does not. Specifically, the dealer must always hit on a hard 16, or lower hard total, so the dealer takes cards on hands where the player sometimes does not, depending on dealer up card.

Dealer Tens (Also)—Players may Double Down and dealers can not do that, which is another reason that the Dealer will draw more cards than a player playing optimally. In some situations, (EX: Player doubles on eleven against a dealer nine and draws a deuce) the player would hit for more cards, except he’s no longer permitted.

However, there are a few situations where the player gets cards that the dealer does not.

Player Splits: Obviously, the dealer can not split, so players draw cards the dealer cannot that way.

Player Hits: Players occasionally hit soft hand totals in which the dealer would be forced to stand—Dealer sometimes stands S17 (Depends on Rules), always S18.

So, the answer to the question depends on which side has the most expected cards in an average hand. I would guess dealer, but could be wrong.

Quote: bcmarshall

Wow! I really didn't expect such a detailed and intelligent answer, but I am now convinced. The best play is to split 8-8 against all.

BTW, in case I wasn't clear in the original post, I never questioned the wisdom of splitting against 2 thru 8. It was only 9-A that puzzled me.

I'm new to the forum. Can anyone tell me how to get email notifications of replies to the thread?

Thanks for the compliment!

I should caution you that my answers are ALWAYS detailed, but rarely intelligent, so please don’t think this is a standard for me and come to expect it in the future.

I think there’s a follow thread button? Welcome to WoV and thank you for starting a great thread with a very good question.

(Also, thank you for accepting answers that rely on math and logic. I wish I could say that for everyone who has ever visited here.)

EDIT: The button says, “Subscribe.”

BC Marshall writes: Wow! I really didn't expect such a detailed and intelligent answer, but I am now convinced. The best play is to split 8-8 against all.

Not so fast my friend. Suppose you have 88vs10 with a TC high enough to warrant a max bet. Hence, this means there are many 10's still in the deck. You get 18, 18. Dealer gets 20. You lose two max bets instead of one. The depth of an extra hand prompts the dealer to shuffle away a postive deck.

Those same two 10's would have been the first card of a two hand bet in a straight up game.