The ticking clock ...

I found myself in yet another decision-making quandary, with the online, 20-second timer ticking.

BR2: 2596. Bet 1000
BR1 (me): 2599. Bet 1000

Dealer: 2

BR2: T,T. Split to T,2 and T,T
BR1: A,9. ????

I won't say what I did, but I've had the chance to consult the tables and discover that my action was in fact correct by quite a margin (assuming I've got my arithmetic right). However, I really wasn't sure at the time.

I'm curious to know if the right play is as unclear to others as it was to me, when given only a short time to think about it.

 

Off the top of my head, if you do nothing you lose only if dealer busts about 35% of the time. So about 65% win.

If you double for 1000 then you advance for sure if dealer busts or if you get 10 or Ace. That is about 73% at a minimum so you should double.

Larry

 

For me the answer to that is "yes". I did quickly narrow it down to standing or doubling, however I reasoned that BR2 stood a good chance of pushing overall with his stiff and his 20 and so my initial feeling was to stand on the soft 20. With a little more thought (probably more than 20 seconds), I realized that it came down to whether the dealer would bust but still decided that standing was better.

I too was surprised by the wide margin in favour of doubling over standing. A great example of the power of correlation!

 

I spotted that too. However, the difference is about 13%, not 5%. So the shortcut doesn't give us an answer, as it only comes to around 60%.

 

Thanks for all the replies so far. It's good to know that these things can be as surprising to others as they are to me.

After the full post-mortem, my final analysis was -

Ways to win the table:

Standing:
Dealer does not bust [64.7%]


Doubling:
I win the hand [55.1%]
I push the hand [7.8%]
I lose to a dealer 21 [11.9% * 12/13 = 11.0%]

Grand Total: 73.9%

At the table I was very tempted by the same logic as Leftnut, comparing the one way that standing could lose (dealer bust) to the myriad of ways that doubling could lose. I saw it as a probable coin-flip, and in the end went for the double. (And happily I drew an ace.)

When it comes to employing a shortcut at the table, I would still need to have commited the various tables of probabilities to memory, which sadly I haven't. But in principle just adding the dealer's probability of making 21 to my probabilty of winning shows that doubling is better than standing.

 

Your method and arithmetic are sound. A quick sim confirms both results.

Showing, once again, since we can't have our tables and computers with us during play, that experience and a clear head are our most valuable assets at the table!

 

London,

I don't get to play in a lot of tournaments, but it appears you and the robot are having a fun time.

Your scenario got me to thinking. I have not calculated this stuff, but I believe, in all similar scenarios where only the bust will give your opponent 2 wins, that you are better off doubling vrs standing as long as a 10 count will make you advance.

The soft hands seem obvious to me. Per your example, your advancement rate is 64.7. Standing on a soft 20 or BJ gives the same advancement rate, where as, doubling either will put you at the 73% rate. I believe, without calculating, that if he would have had 18 instead of 20 and you have soft 18, 19, 20, or BJ and double that you would advance more than standing.

I also believe the scenario might be true for when you have 7, 8, 9, 10 and 11 and you double vrs hit.

Just a possible shortcut.

Larry

 

It was a human opponent in this case; at blackjack21 rather than at GameAccount where the bots reside.

What do you mean by 'a 10 count'?

Did you mean to say a '10 card'? That does give you a 30% chance on its own, before you add in any other possibilities.

I guess it's a generalization of your earlier point. If doubling will advance you if the dealer busts or you are dealt a 10, then that gives you a minimum of -
35.3% + (4/13)*64.7% = 55.2%

So unless standing is significantly higher than this, the unknowns to be added to the above figure are likely to take doubling over the top.

That's definitely true if I have 20 or BJ. It may well be true for the other cases too, but I don't really have an intuitive feel for these things.

I've been thinking that it might be useful to compile a list of which DDs have a combined 'win or push' probability higher than the 64.7% probability of the dealer making a hand. That might be a useful starting point for guesstimating the right action in some situations.

 

Yes, by 10 count I mean 10, J, Q or K.

However I have been thinking about the case when opponent has 18 and a stiff. You should actually double anything that will not bust. If you don't double opponent will have a minimum of 35.3% advance rate (and higher against dealer 3, 4, 5 and 6). If you double even a 5, the only thing opponent has a chance to win with is dealer 17 or 18 which is only about 27.5%.

 

If you meant that your opponent already doubled down his hand to a total of 18, (or did split and his highest hand got 18) then means he is now a big favourite to win the game. You have to add these percentage when dealer bust and when dealer draw to 17 or 18 because both of these situations will work against you. So this make our opponent ~ 62.8% (when dealer shows 2) up to 69.3% (when dealer shows 6) favourite to win the game. Therefore yes, you have to double anything up to hard 16 (you will bust 62%) and when dealer upcard is a 6 you will have to double even in hard 17 (busting 69%) !!! And this is because a bust will not damage your winning chances, but will improve your winning chances between 35% up to 42% for when you will not be busting. (ie: you will still win when dealer draw to 19,20,21 even if you bust)

I would add to this scenario when your opponent double down to 19, (or did split and his highest hand got 19) in this case you will even have to double in a hard 18 when dealer shows a 4,5 or 6.

If I am wrong, please someone to correct me, I may be deadly wrong.

 

If you permit me, I would change the dates a bit. (just as an exercise for myself)

BR2: 2596. Bet 1000
BR1 (me): 2599. Bet 1000

Dealer: T

BR2: T,T. Split to T,T and T,T
BR1: A,9. ????

If you stand you will win only if the dealer has (or or draw to) 20 (about 33.5% of the time) or if dealer draw to 21 (add another 3.5%) So this would be a total of ~ 37% winning percentage. (still a big underdog)

If you double you will get T about 30% of the time, and Ace another 8% of the time. So, this double in itself add up to a bit more than 38% of time when you will win, which is slighty more winning percentage than in the case you would have stand on that soft 20. But here we also have to add the percentage when dealer bust which is about 21%. If we add up those percentages now we are 59% favourite to win.

Of course we can double down and receive anything else but not 10 or Ace more than 60% of the time, and dealer have a T face down about once in 3 similar situations, but we are still a good favourite by doubling down and big underdog if we just stand and hope for the dealer push. Please correct me if Im wrong.

 

You are right that doubling is once again a lot better than standing, for the reasons you state. But you've made a slight error in your calculation; the same error that Larry made earlier in the thread :-

You've counted twice the cases where the dealer busts and you draw a T or A.

(Also the dealer bust percentage is about 23%, rather than 21%, according to Ken's tables - http://www.blackjackinfo.com/bjtourn-dealercharts.php#T4DS17)

Rather than add that on, you need to take 8/13 of it: i.e. those occasions when the dealer busts and you don't draw a T or A.

So doubling is worth about 30% + 8% + (8/13)*23% = 52%

 

Many, many thanks ! Your formula really helps me to understand where my mistake was, as I was not aware of the mistake until now and otherway it would remain a repetitive mistake which at times would cost me winning percents which are sooo precious.

As for 23% instead of 21% .. I was using some other chats.. and I think I may be better if I will update/change the charts I have with the ones from Ken

 

Chances for dd opponents has TT, TT

In addition to advancing when your double card is a Ten or an Ace giving you 30 or 21, and when the dealer busts with you getting cards other then Ten or Ace, you need to add chances when your total double is 19 and the dealer has 17 and 18, as well your total double is 18 and the dealer’s 17. Also, you advance with the dealer’s 21 no matter what you get.
Chances of getting a Ten or an Ace can be simply written as 5/13th.
Double card anything but Ten or Ace times dealer’s bust is 8/13 x 23%
Since dealer with Ten up (no bj) finishes with 17, 18, and 19 about 12.1% each, double getting an eight is one chance of thirteen and getting a nine is two chances (one beats d’s 17 and one wins with d’s 18). This can be written as 3/13 x 12.1%
And finally we need to add all d’s 21 except the already counted when we’ve got 20 or 21, which is 8/13 x 3.8%
Total chance of winning with our opponent when we double A9 is
5/13 + 8/13 x 23% + 3/13 x 12.1% +8/13 x 3.8% = 57.7%

Other way to obtain your chances is to use percentage number of winning double A9 vs. dealer’s Ten (Ken has this table available somewhere on his site) plus chances of the dealer ending with 21 and when we push with 20.
42.5% +3.8% + 4/13 x 36.8 = 57.6%

S. Yama

 

Oops. Good point. That's very similar to the method I used in the original case.

(For what it's worth, it also occurs to me that the table giving a dealer bust rate of 21% is probably showing the percentages before the dealer checks for blackjack.)

 

When is a proper case for double in hard 20 instead of split it into two hands ? I assume it is true only where can not be splitted because of the remaining chips sum being lower than the bet or when it is due to rules like ``unlike ten value cards can not be split``