EV vs Probability in Splitting 10s

EVs are good for long run expectations, but for one hand, probabilities make more sense. For example, a hat with nine 'Win $1' coins and one "Pay $10" coin. The probability of winning once is very high (90%), but in the long run, the EV will dominate and you will lose overall.

I think a similar thing happens with splitting 10s when the dealer has a bust card like a 6. Without splitting, there are three (not equally likely) outcomes (Win, Lose, Push) which may be straightforward to count and calculate probabilities. With splitting, however, there are five (not equally likely) outcomes: Big Win (WW), Small win (WD or DW), Break Even (DD or WL or LW), Small Loss (DL or LD) or Big Loss (LL). I think we can also assume a player who splits knows it's risky, and is happy to consider a Break Even as a Win. Let's call it "not lose".

Counting all those possibilities to determine P(not lose) vs P(Lose) is a challenge, but I'd love to see that analysis. I'm sure P(not lose) is greater than P(lose), but I get brain-twisted when I try to think it through.

Additionally, I'd love to compare P(Big Win) for splitting, vs P(Win) for not splitting. We already know how the EVs match up, but it's that chance to win both hands that lures people in.

The next question (assuming I am correct), is how many hands is it before the long term EV outweighs the short term Probabilities?

Game rules (assumptions): 6-deck, split to two hands and stand on 17 (other rules not important in working out EV for the scenario(s) below).

Here are the simplified values*** for comparison below:

simplified values***: rounded the value to the nearest whole % and doesn't take the chance of "pushing with the dealer" into consideration.

Stand on 20 (T+T) vs Dealer 6:

Win 1 Unit 85% of the time

Lose 1 unit 15% of the time

For an EV of: 70%


Split Ten's into two hands vs Dealer 6:

Win 2 units: 41% of the time

Draw (win 1 unit and lose 1 unit): 46% of the time

Lose 2 units: 13% of the time

For an EV of : 56% of the INITIAL Bet (or 28% EV on each individual split hand).

Note: A SINGLE split hand with a ten vs a 6 wins about 64% of the time and loses about 36% of the time.

Also Note: You would only consider splitting tens, if you were playing a BJ variation were you could see the dealer's first two cards or possibly against a flashing dealer^^^.

flashing dealer^^^: Is when the dealer unknowingly/unintentionally shows/exposes their 2nd card in a regular BJ game before they are meant to (if you split tens in a 'flashing dealer' type of scenario, make sure to cover the play by pretending to be drunk or a
beginner).

IMPORTANT: if you had $100 initial bet, then you are about $14 in EV better off NOT to split Tens against a dealer 6 up card (if I wasn't using simplified values, the real figure would be closer to about $13.60 EV better off).


Sorry I didn't answer it the exact way you wanted, but this was just a simple explanation (someone else may come along and give you a more precise one).

Hope this was helpful

Update:

Quote: riverman

...Additionally, I'd love to compare P(Big Win) for splitting, vs P(Win) for not splitting. We already know how the EVs match up, but it's that chance to win both hands that lures people in....

Here are the "Win" and "Big Win" values against a dealer 6 (from figures I mentioned above):

About 85% chance of a WIN for not splitting.

About 41% chance of a BIG WIN for splitting.

Thanks for the reply ksdjdj. Those are the traditional calculations, but you have set up one of my most burning questions, thanks.

In a non-split 10-10 hand, what is P(draw)? P(win) is 0.85, and P(loss) is calculated as 1-P(win), so we are counting a draw as a loss. That's fine for figuring out EV, since it adds a zero to the total losses, but for probabilities, it's an actual outcome and needs to be counted. P(loss) is less than 15%.

P(draw) plays a crucial role in counting the probabilities of split hands. Those numbers dont address the percentage of time you win or lose 1 unit (which requires a draw).

The chance of a draw/pushing when standing on a player 20 vs a dealer 6 is about: 10.15...%*** (6 Decks — Dealer Stands on Soft 17)

The chance of losing when standing on a player 20 vs a dealer 6 is about: 9.72...%***.

Therefore, the chance of winning must be: about 80.13%, which is 100% - (chance of pushing + chance of losing).

***: below is a link to show you the dealer probabilities (compared to dealer's up card):

https://wizardofodds.com/games/blackjack/appendix/2b/

Also, I chose to use "simple figures" in my original reply, since the loss in EV of splitting 10s is so big for the basic strategy player.


Hope this helps, and sorry I couldn't answer all the questions you had fully, but there are a lot of people better than me at this kind of thing on this site, and they may answer your questions "more fully".

Quote: riverman

...Big Win (WW), Small win (WD or DW), Break Even (DD or WL or LW), Small Loss (DL or LD) or Big Loss (LL).,,,

For splitting 10's, let us assume that you got a total of 17 and 19 for that split.


A single handed 17 v s 6 l/d/w chances are as follows:

Player loses: about 41.1%

Player pushes: about 16.6%

Players wins: about 42.3%

A single handed 19 vs 6 l/d/w chances are as follows:

Player loses: about 19.9%

Player pushes: about 10.6%

Players wins: about 69.5%

and the chance of a dealer getting an 18 is about 10.6% (important for DD)

From the above, the rough chances of getting a WW, WD, DD(Overall push), DL and LL are as follows:

WW: about 42.3% of the time

WD: about 16.6% of the time

DD: about 10.6% of the time

DL: about 10.6% of the time

LL: about 19.9% of the time

This adds up to an EV of about 50.8%^*^ of your initial bet for the above example .

^*^: the real EV for the above scenario is about 50.2% (I used very rough figures to get 50.8% EV)

Note: the above DOES NOT change the average EV of splitting 10s vs 6, in other words it will always be about 56% EV of the INITIAL BET for the player who decides to split 10's vs 6 every time.

Also, someone with plenty of free time could work out all the combinations for splitting 10's vs 6 (if they want), but with the link below and more than a few hours(or days) you could do it yourself (re-posted the link below, to make it a bit easier for you to find):

https://wizardofodds.com/games/blackjack/appendix/2b/

I used 6 deck and dealer stands on 17, for all the figures in this post.

Lastly, I forgot to mention another situation where splitting 10s against a 6 can be the best strategy (in one of my earlier replies) and that is when you are going to receive the next card out of the shoe and KNOW that the card is either a 9, 10 or Ace.

Thank you, ksdjdj. One last question. In a split hand, you didn't include WL as one of the draw options. Can you find that info also and provide a more complete table of probabilities for all outcomes (WW, WD, WL, DD, DL, LL)?

And if you are willing, I'd love to know how you are calculating these. ;-)

I didn't include WL because DD and WL mean almost the same thing (in my previous post I should have used WL for that example).

The reason is:

A potential DD (push) can only happen if you get two hands with the same total after splitting (that are between 17 and 21).

A potential WL (push) can only happen if you get two hands that are different by at least two numbers, for example the player getting a 17 and a 19, or if you get a 12 to 16 and an 18 .

There is also a chance of a DD/WL(push) never occurring on two hands, and that is when there is only one gap between the two hands, for example, the player getting a 17 and 18.


Got to go now, I will see if I can try to answer the other questions you had later (but I don't think I will be able to answer it, sorry).