Tourney Teaser

Ken,

Thanks for this "teaser", because it will help some of the newer BJT players, like myself, see how you "pros" figure this out with the statistics, etc.

Personally this is what I would do and my reasoning:

Hard 9 - double my hand - this forces BR1 to double and hopefully bust
Hard 11 - double my hand - see above (I like this scenario best)
Hard 13 - double my hand - see above.

Hard 17 - this is the toughest one. If I stay and BR1 stays and dealer turns an 8, 9 or 10 I lose. But if dealer draws to 17 I push, BR1 loses and I win. If I double and bust then BR1 stands and laughs all the way to victory so in this case I would stand.

Now I'll wait for the right answers - please tell us students not just the "answers" but the rational also - PLEASE

 

I would do the following

1) Hard 9 - Hit to at least 17, because if you double and end up less than 17 the only way you would win is if the dealer busts. Hitting to 17 gives you a chance for a swing.

2) Hard 11 - this is a little closer, but I'd still do the same as hard 9.

3) Hard 13 - Hit to at least 17 to have a chance for a swing.

4) Hard 17 - Stand.

The only question in my mind is stopping at 17. You can get a swing with 17 but I would imagine the odds are a little better at 18 - 20 but I don't know those numbers.

 

Okay, let's try this...

Playing these hands within 30 seconds I would play them like this and why.

#1. 9, I would just hit to 17 or above, or bust trying.
Anything 16 or under and your a loser anyway.

#2. 11, same as above.

#3. 13, here is where I would actually DD. If I am going
to hit it anyway I might as well DD and hope I get
lucky and catch a 7 or 8 or I don't bust and the
dealer does.

#4. 17, I would stand and hope for a push.

(I actully posted "or dealers bust" also and then caught my mistake) sorry.

I maybe wrong on all three, but it is how I would play them.

 

Lets try this one

I’d like to start from the end
......1.BR2 has 17
Stand = 12.07%
DD = 17.30% (best play)
Hit to 18 = 16.20%
Hit to 19 = 15.44%
Now we know it is better to hit 17, but not 18 (we will need this later too).

........2.BR2 has 13
Stand = 0%
DD = 21.66%
Hit to 18 = 21.80% (best play)

.......3.BR2 has 11
Stand = 0%
DD = 42.79%
Hit to 18 = 43.05% (best play)

.......4.BR2 has 9
Stand = 0%
DD = 26.70%
Hit to 18 = 31.01% (best play)

These are the best outcomes for BR2, if he knows exactly that BR1 is going to double if he doubles, and always stand if he hit.

I think the only exception for BR1 will be when BR2 makes 20 by hitting. In this case BR1 has to hit to 17.

Very nice teaser Ken.

 

Double check

Ken, I would like you to check it because my numbers can be wrong too.
I will double check them today.

 

Errors Found

Thank you Ken for finding errors in my calculations.
Here are my new numbers

1. BR2 = 17

Stand = 12.07%
DD = 17.30% (best play)
Hit to 18 = 16.20%
Hit to 19 = 15.44%

2. BR2 = 13

Stand = 0%
DD = 22.22% (changed and best play)
Hit to 17 = 21.40%
Hit to 18 = 21.81%
Hit to 19 = 20.78%

3. BR2 = 11

Stand = 0%
DD = 42.79%
Hit to 17 = 42.58%
Hit to 18 = 43.05% (best play)
Hit to 19 = 41.87%

4. BR2 = 9

Stand = 0%
DD = 28.83% (changed)
Hit to 17 = 30.52%
Hit to 18 = 31.01% (best play)
Hit to 19 = 29.74%

 

Ok, here's where you "pros" need to fess up so us "new guys" can learn a little.

Where did you get those stats?

But more importantly I don't think you are considering everything. For example in the one scenario where you have the hard 11 and just want to hit up to 17 or bust out. So you don't double down. On this I totally disagree for this reason (ok I may be wrong but here's my logic)

When you double and DON'T bust it FORCES BR1 to double also.

WHY?

You've got 11 and hit a 6 to get 17. You've got 40 bet to BR1's 20. You've got 17 and BR1 has 16 with a dealer 10 up.

Ideally if I were BR1 I would surrender but that is not an option.

So when BR2 DOUBLES it FORCES BR1 to double to match the chip total. (AT LEAST THAT'S MY "LOGIC" in doing so many doubles)

Now what are the odds of BR1 hitting a winning hand on a 16?

See Arlaik, I understand you "odds", don't know where you got them, but I can live with that. I just think that since we are competing not agains the dealer but against BR1, that the odds for BR1 hitting, standing or DD should be included also. Then the best odds of the pairs would be the proper answer.

Once again I could be DEAD wrong. If I am that's ok, because I would rather learn this way then the way I did two weekends ago at the Nugget.

Please be patient with a new learner but let me know where I'm wrong.

 

You can figure them out fgk42!

Believe it or not all the information is here or on the sister site to figure these probabilites out. I asked similar questions a month or so ago and was very kindly told how to do it! I can't remember which posts now but if you review all the posts for the last 2 months in the strategy section I'm sure you'll find them. It's actually suprisingly easy(ish) if you know how! I am waiting to work on this problem myself but I have had no time to do so in the past day or so. I shall work on it over the next 24-48hrs and post my working out for scrutiny!

Cheers

Reachy

Ps. The information you need is on www.blackjackinfo.com; find the dealer outcome tables in the the "tournament" section. Also you need to find all the scenarios where you will beat BR1. For example if you hit to 18 you will win if the dealer hits to 17 or 18 because you either beat the dealer or push and BR1 loses. If the dealer hits anything greater than 18 or busts you lose. You also need to know infinite deck probabilites but I'm not sure where to locate those on the web. Calculate the probabliltes for all the different scenarios and the one with the highest %age is the best strategy

 

Dear fgk42.

When we do the calculations for specific case, we have to consider BR1's response to our action.

Ken clearly posted in his teaser that if BR2 doubles then BR1 doubles also, and
if BR2 hits then BR1 stands with his 16.
That is almost the best strategy for BR1 and that is included in my calculations.
To understand that, I can give you only one calculation (otherwise it is going to be few pages long calculations) when BR2 doubles his 11 and BR1 doubles hard 16 regardless of BR2's outcome.

BR2 doubles and advances when:

Dealer bust(22.98%) and BR1 bust (8/13) = 22.98*8/13 = 14.14%

Dealer 17(12.07%) BR2 win(7/13) BR1 Push/lose(9/13),
or BR2 push(1/13) BR1 lose(8/13) = 12.07*(7/13*9/13 + 1/13*8/13) = 5.07%

Dealer 18(12.07%) BR2 win(6/13) BR1 Push/lose(10/13),
or BR2 push(1/13) BR1 lose(9/13) = 12.07*(6/13*10/13+1/13*9/13) = 4.93%

Dealer 19(12.07%) BR2 win(5/13) BR1 Push/lose(11/13),
or BR2 push(1/13) BR1 lose(10/13)=12.07*(5/13*11/13+1/13*10/13)= 4.64%

Dealer 20(37.07%) BR2 win(4/13) BR1 Push/lose(12/13),
or BR2 push(1/13) BR1 lose(11/13)=37.07*(4/13*12/13+1/13*11/13)=12.95%

Dealer 21(3.74%) BR2 push(4/13) BR1 lose(12/13)=3.74*4/13*12/13 = 1.06%

Total we have 42.79% probability to advance by DD.

If we hit instead of doubling

Hit to 17 = 42.58% which is less than if we DD.

Hit to 18 (yes, you have to hit hard 17) = 43.05% better than 42.79% by DD

Hope this help.

 

Reachy,

Thanks for the link to Ken's site.

I understand the probability tables

BUT, and here's the important part, BUT

What about BR1's response to your bet?

Ok, I've got a hard 9 and hit to a 16-20. Well, BR1 doesn't have to DD because he/she had the chip lead.

IF I DD on ANYTHING AND DON'T BUST theny BR1 has to worry dont they?

I mean if I double on a 9 and pull an 8 for a 17 vs. BR1's 16 and dealer upcard of 10.

What does BR1 do?

Arlaik's number didn't figure THAT SCENARIO into consideration. Remember I'm not only playing against the dealer - my real enemy is BR1.

Or am I wrong?

 

Most important thing is to...

make him act on his 16 with a hit, then he will face a 60% chance of busting. So, if you double and card is face down he will be forced to act. If you hit to a standing hand he will be forced to act. The second you bust you are out. If dbl card is face down I would double 9-13 and hold 17. Force him to hit. I ran over a bunch of probabilities and they are all worthless unless you force him to act. I didnt hear any mention if the dbl card was face up or down. This is important.

Barney

 

double card

Funny but I ASSUMED that the double card would be face up.

What if it isn't?

I mean if BR2 doubles and the double care is face down doesn't that force BR1 to double regardless????????? Just due to margin of error and chip count????

I'm asking this because I DON'T KNOW. I'm NOT trying to be anything other than a student of the game.

 

So Ken,

Does that make a difference in the ODDS/PROBABILITY that arlick posted?

PS Thanks for posting this question BTW

 

Cards face up

Maybe I should study reading comprehension instead of Blackjack. LOL. Im sure araliks numbers are correct but I still couldnt hit or double the 17 because itself forces action by br1. The 13 might be a good double because it isnt a real bad hand to make a solid 19 and up total with one card, I say double. 9 and 11 I would go to "my" basic start against the dealer 10 and hit, and hit again if I catch craps.

17 stand
13 dbl
11 hit
9 hit

 

Here's my "basic" question Barney

If you're going to hit a 9 or 11 - you CANNOT BUST - why NOT double??????

This FORCES BR1 to double HIS bet and HE/SHE has a 16 versus a 10????

13 double - I'm with you.

11- I'm doubling ALL the time

9- doubling ONLY because I want BR1 to bust.

What makes it very interesting is the double care being up or down. On the upside BR1 knows what U have and can play according. Down and BR1 is blind.

Am I being THAT NAIVE ABOUT THIS???????

 

Hard 9, Hit or Double?

Ken please check my numbers here and tell me if you see anything wrong.
I would like to find it out myself.
BR2 = 9
BR1 = 16, Dealer = T

Here are the outcome probabilities

...................17...........18.........19..........20.........21
Hit to 17 ....12.00.......12.00.....35.08......12.00.....6.08
Hit to 18 .....----........12.92.....36.00......12.92.....7.00

Probability to advance by hitting to 18 is

P = D(17)*BR2(18-21)+D(18)*BR2(18-21)+D(19)*BR2(19-21)+
+D(20)*BR2(20-21)+D(21)*BR2(21)

P = 12.07*0.6884+12.07*0.6884+12.07*0.5592+37.07*0.1992+3.74*0.07=
= 31.01%.

Also what do you thing about BR1's best play when BR2 makes 20 by hitting?