# Tourney Teaser

Discussion in 'Blackjack Tournament Strategy' started by KenSmith, Jun 30, 2006.

For all you number-crunchers, here's another set of teasers. I'll warn you ahead of time... The strategy decisions in these questions are all very close, generally 0.3% between the best choice and the second best choice.

Rules: Infinite decks. No surrender. All cards dealt face up. One advances.

Only two players remain, and it's the final hand. You're in trouble, because you are on the button and you trail with a bankroll of 100 compared to the leader's 105.

You bet \$20, he bets \$20. He is dealt a 16, and the dealer has a ten up. She peeks, and there's no blackjack. How do you play the hand if...

a) You have a hard 9.
b) You have a hard 11.
c) You have a hard 13.
d) You have a hard 17.

You..........\$100.........\$20..............Hard 9 or 11 or 13 or 17
BR1..........\$105..........\$20.............Hard 16

Dealer Ten up, no blackjack.

Remember, in each case, you're acting first, before the leader who has a hard 16. Your opponent will play perfectly behind you, but I'll give you his optimal strategy right now:

If you double, he's going to double regardless of what your hand total is.
If you don't double, he's going to stand on his 16, again no matter what your cards are.

2. ### fgk42New Member

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.

3. ### mrbillNew Member

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.

4. ### TXtourplayerExecutive Member

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.

Last edited: Jul 1, 2006
5. ### arlalikMember

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.

Last edited: Jul 1, 2006

Hmm, my numbers are noticeably different than arlalik's. I'll have to double-check my results, which means it will likely be after I return home from Vegas after the weekend.

7. ### arlalikMember

Double check

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

8. ### arlalikMember

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%

9. ### fgk42New Member

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.

10. ### ReachyNew Member

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

11. ### arlalikMember

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.

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.

12. ### fgk42New Member

Reachy,

Thanks for the link to Ken's site.

I understand the probability tables

BUT, and here's the important part, BUT

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?

BR2 cannot stand at 16, else he has no way to win. So, he must hit to at least 17, or better yet, 18.

Then, regardless of where BR2 stands (17-21), BR1's optimal play is to stand with 16 and hope the dealer busts. If BR2 has stood with less than 21, BR1 can also advance if the dealer beats them both.

However, in the case of hard 9, BR2 is better off doubling anyway.

BR1's optimal play if you double and do not bust is to double as well. Again, no matter what card you drew on your double, he should double.

So, in each of these cases, we've assumed that BR1 will do whatever is best for him to do. Of course in the real world, players don't always do what's best. But that makes the math all messy!

Arlalik, my numbers now agree with you in all but one case...

For BR2 with hard 9, I get BR2 advancing 28.6% by hitting to 18, which means doubling is better than hit-to-18. Since all my other numbers now agree, I suspect there's an error lurking in your calculations for that one.

(I used the same piece of software for all of these, so my answers are more likely to either be all wrong or all right!)

Assuming my number for that is accurate, it leads to an interesting observation on this hand...

Hard 9, double.
Hard 11, hit to 18.
Hard 13, double.
Hard 17, double.

Hmmm, you double in every case EXCEPT when you have the strongest double-down hand. Weird, huh. I think I understand why, and I'll have more to say after I'm back home. For now, I'm back out to play while in Vegas.

Last edited: Jul 2, 2006
14. ### Barney StoneNew Member

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

Actually,the initial message mentioned that all cards are face up. And, you're right, it matters.

16. ### fgk42New Member

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.

17. ### fgk42New Member

So Ken,

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

PS Thanks for posting this question BTW

18. ### Barney StoneNew Member

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

19. ### fgk42New Member

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.

20. ### arlalikMember

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?