Question

D

DCB

Member
Hi Ken/All,

I just downloaded Ken's e-book "How to Win More Blackjack Tournaments." I have a question about the scenario at the end of chapter 10 on page 36:

If BR2 doubles for less for a total of 190K, then doesn't that mean that BR1 should stand with 9/9 instead of doubling for less? I understand that we don't know what BR2 has since the double down is blind, but we have to assume that the dealer likely has 18, and there is a chance that BR2 may have drawn an A or 2 for a total of 17 or 18. I don't understand why doubling for less is the best play for BR1. Any help would be greatly appreciated!
 
Until Ken sees the question, may be I can make a guess. I do not know the whole scenario, since I do not have that book at the moment but it might be the idea is to force BR1 to split. So if you are BR2 and BR1 got you locked out which means he wins if you both win and he wins if you both lose your bets and you play first, this strategy might help. You double for less. He has 9/9 and cannot double for less on 18, so he will have to split. To split you have to match the original bet, as you cannot split for less. This gives you the upper hand if both of you lose, which is an improvement on the original scenario.
 
That's right, Professional. Doubling for less, but enough to take the high, forces the opponent to put more at risk by splitting. You then win with the opponent's L/L, and if you win your hand you beat the opponent's W/L. I've used this strategy, first put forward by Ken, several times with good results.
 
Correct, but Ken is saying that after BR2 doubles for less, BR1 should double for less as well (instead of splitting or standing), with 9/9 here. That's what I am confused about.

I don't want to post the entire hand because it's Ken's book and I don't want to infringe on the copyright.
 
DCB said:
Correct, but Ken is saying that after BR2 doubles for less, BR1 should double for less as well (instead of splitting or standing), with 9/9 here. That's what I am confused about.
Click to expand...
Ken describes it as a 'free' double. That is -

Splitting would be an error, since it would give up the low (which was the whole point of BR2's sneaky tactic of doubling for less, rather than for the full amount).

Standing would achieve nothing, since, win or lose, your fate would depend solely on the outcome of BR2's hand.

So you double for less. If you bust you are no worse off than if you had stood. If you don't bust there is a chance that both you and BR2 win your hands and you prevent BR2 from overtaking you.


Actually, it does seem to me that it is really an 'almost free' double -

BR1 busting does open up two possible ways for BR2 to advance that standing would prevent. As you say, BR2 could have made 17 or 18 and could push with the dealer making 17 or 18.

We have to weigh the downside of those two possible outcomes against the upside of the additional ways for BR1 to advance that come from drawing an A,2, or 3, and both BR1 and BR2 beating the dealer (or pushing with a total of 19,20 or 21).
 
Thank you. I am new at this, so please bear with me.

"If you bust you are no worse off than if you had stood." I am having trouble wrapping my head around this statement because you don't know what BR2 pulled. If you have 18, and the dealer likely has 18, and BR2 doubles for less and pulls A,2, or busts with any card 6 or higher, you win by standing with your 9/9. Those are pretty good odds. BR2 has to pull a 3,4,5 AND has to beat the dealer if you stand pat on 18. But if you double for less (and likely bust), then BR2 can win with A,2,3,4,5.

It would be AWESOME if someone was able to quantify this scenario with probabilities.
 
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You lose very little by doubling for less. There are just two scenarios -
BR2 makes 18 AND Dealer makes 18 AND BR1 busts . [ 1/13 * 36% * 10/13 = 2.13% ]
BR2 makes 17 AND Dealer makes 17 AND BR1 busts. [ 1/13 * 12.89% * 10/13 = 0.76%]
(Using https://www.blackjackinfo.com/dealer-outcome-probabilities/#T6DH17 to get dealer probabilities.)

For any other combination of BR2 and dealer hand, with BR1 busting, the result will be the same as for standing - BR1 only advances if BR2 loses to the dealer.

So you give up a total of 2.9% by risking busting.

That's the debit side of the balance sheet. The credit side would be a lot more complicated to figure out, and I think should come to a much higher figure.
 
London Colin said:
That's the debit side of the balance sheet. The credit side would be a lot more complicated to figure out, and I think should come to a much higher figure.
Click to expand...

After a certain amount of head scratching, I think this may be it -
Code: 
Dlr Bust & BR1 19-21 & BR2 17-21  [ 24.37% * 3/13 * 5/13 = 2.16% ]
Dlr 17 & BR1 19-21 &  BR2 18-21   [ 12.89% * 3/13 * 4/13 = 0.92% ]
Dlr 18 & BR1 19-21 & BR2 19-21    [ 36%    * 3/13 * 3/13 = 1.92% ]
Dlr 19 & BR1 19 & BR2 19          [ 12.87% * 1/13 * 1/13 = 0.08% ]
Dlr 19 & BR1 20-21 &BR2 19-21     [ 12.87% * 2/13 * 3/13 = 0.46% ]
Dlr 20 & BR1 20 & BR2 20          [ 6.92%  * 1/13 * 1/13 = 0.04% ]
Dlr 20 & BR1 21 & BR2 20-21       [ 6.92%  * 1/13 * 2/13 = 0.08% ]
Dlr 21 & BR1 21 & BR2 21          [ 6.95%  * 1/13 * 1/13 = 0.04% ]
-----------------------------------------------------------------
Which adds up to:                                          5.69%
-----------------------------------------------------------------

Which would mean a net gain for doubling for less of 5.69% - 2.89% = 2.8%
 
I think you've got it there. Much appreciated! I am new to this, and it seems to be very hard to internalize the "free double" rules.
 
As usual, nice work by the regulars here answering this question in my absence.
I've been having fun in New York City for a few days. (Jazz at Smalls Live was the highlight!)

I should not have described this as a true "free double" in my article, but London Colin has demonstrated that it still is an effective play.

By the way, for anyone who doesn't have the book, here is the excerpt that describes the situation:

Here's the example (Note: BR1 and BR2 designates the table players with the
highest and second highest bankroll respectively):

BR2 has $262,000, bets $100,000 and is dealt a hard 16.
BR1 has $265,000, bets $100,000 and is dealt a pair of 9s.

The dealer shows an 8 up.

BR2 predictably doubled down, but the double-down amount is critical. He chose
to double for $90,000. (Actually, any double amount between $4000 and $96,000
accomplishes the same thing.) The key here is to double for enough to pass a straight
$100K win by BR1, but little enough to not allow BR1 to keep more unbet chips if he
splits.

In this particular example, the rules were those of the new Ultimate Blackjack
Tour, where all double down cards were dealt face down. Therefore, BR1 didn't know
whether BR2 had busted or made a hand.​

The article also mentions that this situation was discussed here at BJT. Here is the original thread:
https://www.blackjacktournaments.com/threads/ubt-tournament-teaser.1711/

I haven't re-read it yet, but it's nice to look back over these ancient threads of wisdom. :-)

 
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