# UBT Tournament Teaser

Discussion in 'Blackjack Tournament Strategy' started by Joep2, May 17, 2006.

1. ### Joep2Banned User

We all know that in the UBT format all double downs are dealt face down

So here is a hand for you to play

You are Br 1 acting last against Br 2

This is an elimination hand and only 2 players left and all secret bets have been used

Min bet 1,000

Max bet 100,000

Surrender is available

Br 2 has 262,000 bets 100,000

Br 1 has 265,000 bets 100,000

Br 2 is dealt 10-6 and doubles down for 90,000 (face down)

You as Br 1 are dealt 9-9

Dealers up card is 8

What should you do as Br 1 with your 9-9

Part 2 of this question

Is your decision changed at all with a different dealers up card

Joep2

My initial reaction was to double the (9,9) for 90K.
A quick check of a few numbers shows that is a better choice than standing. (It's close, less than 2% better.) Splitting would be tougher to calculate, but I'm guessing giving up the low is too costly, and would produce a lower win rate than either standing or doubling.

I don't think the upcard matters. I double behind BR2 no matter what the dealer is showing.

3. ### ReachyNew Member

I like what BR2 has done. If BR1 splits as BS says he should then BR2 has the low (must remember that!). From what I can remember the most likely dealer outcome is 18 about 35%(?) with about a 24% (?) chance of busting and 12% chance of hitting 17? Anyway if BR2 wins BR1 can only beat them with a split W W; a BR2 push will be beaten by a BR1 split W W, single bet win, split W L or push and a BR2 loss can only beat a BR1 split L L.

BR2 outcomes: Lose 72K, push 262K, Win 452K
BR1 outcomes: Split lose 65K, Single bet Lose 165K Split, Surrender 215K W/L/Push 265K, Single Bet win 365K, Split W/W 465K

If BR1 stands he will push 35% of the time vs a 24% chance of BR2 winning (See Ken Charts BJI! )

If Br1 splits I estimate that he will L L 23% of the time, L W 45% of the time and WW 21% of the time. (There are the push potentials but I'm ignoring them!)

After all that I have decided that the best decision is to split. My rationale is BR2 is likely to lose or push 75% of the time. To beat a win BR1 has to split and WW, to beat a push he can either push, WL a split or win a single bet. In this scenario BR1 will push 35% of the time if he stands or 45% of the time if he splits. BR1 will L L 23% of the time vs 70% for BR2. In summary splitting increases the chances of a push and give you the opportunity to beat an (unlikely) BR2 win.

There. What do you think?

Part 2:

The dealers up card has a relatively small affect on BR2 chances of winning a DD. 26.5% chance of winning vs Dealer 6 compared to 20% vs Dealer 10. The most important effect is on BR1s outcomes. I would suggest following BS so stand vs 10, A or 7 and split the rest.

Cheers

Reachy

4. ### ReachyNew Member

hadn't thought of the DD angle....

Should have looked into DD options. Damn!!!

Mind you if you DD you only win 20% of the time vs 24% for BR2 so you're a dog. If you stand you beat the dealer 37% of the time. Ahhhhh! Now I getting it. 20% of 190K is 38K whereas 37% of 100K is, well, 37K. The penny is starting to drop...

Mind you before you know the dealers upcard (Heisenbergs Uncertainty Principle) if my rudamentary knowledge of probability is correct the odds of winning are 21% (i.e. 0.46 x 0.46). That would give an outcome of 42K. For Splitting to be less favourable than DD the probabilty would have to decrease to 19% which I guess with an upcard of 8 might be there or there abouts. Whatever you do it's a very close decision

Regards

Reachy

Last edited: May 18, 2006
5. ### maxwellMember

9 9

I Would Split Those Puppys With Only Two Hands To Go Hoping For A Double Down On Split:d You Guys Are Something Else I Hope I Do Not Have To Play Any Of You All- Reachy You Fit In Well In This Group Good Luck To All

Actually maxwell, there are not two hands left. This is the final hand.

7. ### ReachyNew Member

Are my numbers any good Ken?

Cheers Reachy

8. ### maxwellMember

Ken

My Eyes Are No Good Ok Last Hand I Would Still Split And Go For The Win

I haven't calculated any of the split numbers, so I can't say there. Your other numbers look OK.

However, I do have one concern about your post. Where you say...
Optimizing the hand EV is an idea that isn't really pertinent to most final hand decisions. Consider this example...

BR2 has \$1000, bets \$500 and doubles all-in. He ends up with a stiff vs a dealer 8.
BR1 has \$1800 with a bet of \$250 working. He has a hard 12.

In this case, considering the EV of BR1's hand would lead you away from the obviously correct play. BR1 is a lock if he stands, although the hand EV is much lower by standing than hitting.

Sorry to sidetrack the thread with a completely unrelated example.

I really ought to go back and figure the split probabilities on the original question I guess. Maybe later today.

10. ### ReachyNew Member

Thanks Ken.

So how did you get to your figure for DD being about 2% better than standing? I assumed you'd used EV calculations because they seemd to fit my calculations.

Cheers

Reachy

I used a spreadsheet to figure each of the possible outcomes for BR1, BR2 and the dealer. With just standing vs doubling for BR1, it's not too tough. Splitting would be a much more involved calculation.

12. ### ReachyNew Member

I really want to know how you figured it out. Can you tell me what calculations you did?

I've tried to figure them out myself and I just can't get the 2% differential between DD and standing.

My estimate is that if you DD you will beat the dealer 22.3% of the time. Am I anywhere near?

My calculations:

Dealer % BR1 win %
17 0.1289 18,19,20, 21 0.308 0.0397012
18 0.36 19, 20, 21 0.231 0.08316
19 0.1287 20, 21 0.154 0.0198198
20 0.0692 21 0.077 0.0053284
21 0.0695 0 0 0
Bust 0.2437 18,19,20,21 0.308 0.0750596
0.223069

Dealer will hit 17 12% of the time; hitting 18, 19, 20 or 21 will win it for you. That will happen 30.8% of the time therefore you chance of winning that scenario is 3.97% or 12.89% x 30.8%. Am I on the right track? I then did the same thing for all the other DD outcomes, added them all up and came out with 22.3%. Is this correct? If so I will continue with my calculations to figure out the probabilty of winning when standing. Then get back to you again!

Cheers Guys

Reachy

How to calculate the probabilities...

How often will BR2 advance if BR1 stands? We'll look at each dealer outcome and consider ways for BR2 to win, assuming BR1 stood with 18.

For dealer results, I'll use the 6D S17 chart here: http://www.blackjackinfo.com/bjtourn-dealercharts.php
For player results, I'll use infinite deck approximations.

If the dealer busts (24.37%), BR2 advances if he did not bust. We can ignore BR1's result. So, that's 24.37% X (5/13) = 9.3731

Dealer makes 17 (12.89%), BR2 must have 18-21. 12.89% X 4/13 = 3.9662

Dealer makes 18 (36.00%), BR2 must have 19-21. 36.00% X 3/13 = 8.3077

Dealer makes 19 (12.87%), BR2 advances with 19-21 (push OK). 12.87% X 3/13 = 2.9700

Dealer makes 20 (6.92%), BR2 advances with 20-21. 6.92% X 2/13 = 1.0646

Dealer makes 21 (6.95%), BR2 advances with 21. 6.95% X 1/13 = 0.5346

Add all those up, and you get p(BR2 advances when BR1 stands) = 26.2162%

----------------------------------------------------------------------
Reuse the dealer outcome percentages from above.

Dealer Busts, BR2 must not bust and BR1 must bust.
24.37% X (5/13) X (10/13) = 7.2101

Dealer 17, BR2 can push or win while BR1 busts.
12.89% X (5/13) X (10/13) = 3.8136

Dealer 18, BR2 can push or win while BR1 busts.
36% X (4/13) X (10/13) = 8.5207

Dealer 19, BR2 can push or win while BR1 busts.
12.87% X (3/13) X (10/13) = 2.2846
Dealer 19, BR2 can win while BR1 pushes.
12.87% X (2/13) X (1/13) = 0.1523

Dealer 20, BR2 can push or win while BR1 busts.
6.92% X (2/13) X (10/13) = 0.8189
Dealer 20, BR2 can win while BR1 pushes.
6.92% X (1/13) X (1/13) = 0.0409

Dealer 21, BR2 can push while BR1 loses.
6.95% X (1/13) X (12/13) = 0.4935

-----------------------------------------------------------

For BR1, this answer shows that doubling is better than standing by 2.88%. (Remember, the probabilities above are for BR2 succeeding, so BR1 wants to see lower percentages.)

As you can imagine, this gets tedious if you want to figure the results when BR1 splits. Especially since it's not immediately clear what the optimal strategy post-split would be. I don't think the usual must-win-both-splits strategy is best, as a push may be useful for BR1.

I don't want to tackle that one with a spreadsheet, but I may be able to apply another software tool to the task.

14. ### ReachyNew Member

Thanks Ken

This is brilliant Ken. I really appreciate you taking your time to do that for me.

Fantastic!

Cheers

Reachy

15. ### tirle_bjMember

split ~ 63.3%

You got it right, Ken!
If BR1 does NOT bust on both hands (hits 'til 12), his chances to advance are about 63%.

Thanks tirle! You saved me all that effort. Glad to know my instinct was on track this time.

17. ### WaltNew Member

Confused!

I had an answer before reading all those calculations. But I'll stick with it. I'm doubling for the same 90k and retaining the low, regardless of dealer up card.

That's the correct answer, based on all those calculations, both mine and tirle's.

19. ### ReachyNew Member

What would BR1 done if acted first?

Would BR1 still DD if they acted first? Would this affect what BR2 does at all? My guess is "Yes" and "No" but I'm rarely right ;-P

cheers

Reachy

20. ### S. YamaActive Member

Reachy, for the first time I have a feeling you’re playing with us.
Are you asking if BR1 acting first, should really double down for at least 97,500 on his two nines with bet of 100K out brl 265K, and BR2 betting 100K out of 262K, and that’s on the last hand?
It would reverse the chances, and now BR2 would have almost eighty percent chance to advance.
If BR1 did double and busted out then BR2 would be guaranteed to win. If BR1 got to twenty or twenty-one, then BR2 would have to hit to one point more. Not seeing the double down card would require BR2 to hit to 20.

By the way, when I was checking the numbers yeasterday using spreadsheet I mistakenly typed in 19, instead 18 (two nines) for BR1.
It is still a better play to double down the nineteen in this situation vs. BR2 doubling his 16.
In all cases, for infinite deck, as well as for one to six decks including effects of removal, it is better play by a quarter to one half percent.

S. Y.