R.R. offers a new teaser

T

tirle_bj

Member
Here is the Case:
In the last hand R.R. got (7,7) vs 9 when in order to advance he should've win one bet. He had enough funds to split, double, etc.

First let's consider the case when he doesn't split. Then according to "must win" strategy he has to hit 'til 16.

Here for simplisity we consider infinitive deck.
P16=P17=P18=P19=P20=P21=1/13 + 1/169 = 8.28%
P win = D bust x (6 x 8.28) + D17 x (4 x 8.28) + D18 x (3 x 8.28) + D19 x
(2 x 8.28) + D20 x (1 x 8.28) = 25.11%

In other words by not splitting the probability for R.R. to advance is 25.11%

Let's consider now the case when R.R. SPLITS.
Here's the information we will need to find out the optimal play.

Probabilities to win by doubling the first hand:

(7,A) vs 9 = 38.39%
(7,2) vs 9 = 35.69%
(7,3) vs 9 = 49.26%
(7,4) vs 9 = 50.66%
(7,5) vs 9 = 28.59%
(7,6) vs 9 = 26.83%
(7,7) vs 9 = 25.08% (*)
(7,8) vs 9 = 23.32%
(7,9) vs 9 = 21.56%
(7,T) vs 9 = 19.81%
P(stand on stiff) vs 9 = 22.84%

(*) this case needs to be considered separately, because that's what happened with R.R. and there is another option to resplit.

As we can see doubling on 16 and 17 total is worse than just stand.
Let's find out how to play hard 15, 14, 13, 12 on the first hand.

Double on Hard 15: P adv > 23.32% because if we make 17 thru 20 then second hand must be hit 'til one point over to win 1 bet, because only in case of pushing first hand there might be the difference (if we lose double on first hand - we're hopeless, if we win double on first hand - we're done).

Stand on Hard 15: P adv = 1/13 x (50.66 + 49.2 + 35.69 + 38.39 + 28.59 + 26.83 + 25.08 + 23.32) + 5/13 x 22.84 = 31.15% This formula represents doubling on the second hand everything but total 16 or 17.

Very important point: there is NO "free hit" on the first hand available in this cases since we might lose the option to STAND on the second hand with total of 16 or 17 in case of busting on the first hand and will be forced to double instead.

Double on Hard 14: P adv > 25.08%
Stand on Hard 14 : P adv = 30.15% (same as above)

Double on Hard 13: P adv > 26.83%
Stand on Hard 13 : P adv = 30.15%

Double on Hard 12: P adv > 28.59 = 30.17%
Stand on Hard 12: P adv = 30.15%

As you can see Hard 12 is turning point (thanks to extra chances mentioned above). However it may be ignored in a live money Tourneys because of really marginal difference.

So, finally we have 1/13 (38.39 + 35.69 + 49.26 + 50.66) + 9/13 x 30.15 = 34.26% This formula represents double on the first hand with (7,A), (7,2),
(7,3), (7,4) and stand with "stiff" hands.
Actually we have even more than 34.26% because in case of doubling on the first hand and pushing with the dealer we may still win on the second hand.

Here we have the Final Playing Strategy:
1) split (7,7)
2) Double first hand with A, 2, 3, 4 otherwise stand
3) If we stand on the first hand then we should double on the second everything but hard 16 or 17. If we double on the first hand then with 17 thru 20 we have to hit second hand 'til one point over, otherwise no difference how to play (just playout).

In case of third 7 coming after split there is no need to resplit the hand since even if you win double, it may not be enough in case of losing other two hands.

Final Probabilities,as shown above are:
>34.26% for split vs 25.11% vs not split (hit 'til 16)

P.S. Per our conversation with S. Yama this case became an interesting point to understand how to win at least one hand with splitting option involved.
 
Nice work!

This is definitely an interesting avenue of research. I'll have to come back to work through the details as I'm pressed for time now. I'm hoping there are some simple rules that can be formulated for these types of decisions.
 
Need more details on the hand?

tirle bj does R.R. have to win and others lose? Dose R.R. just need to win? Can anyone DD to beat R.R.? What order is R.R. betting in? A true determination cannot be made without knowing more details.

Just from what you have on your post I would either stand or hit, I wouldn't think of spliting unless I was over a max. bet down (and you said all R.R. needed was to win). 14 against the dealers 9 sucks to begin with so why split and get two seventeens against the 9? Your hoping for a dealers bust either way if you need a win.

I would take my chances with one hand, more then likely I would stand since you cannot bust your hand and have a chance. If I really believed the small cards were coming I might hit, it depends on the flow of the game. Bottom line is your needing the dealer to bust anyway or a third 7.
 
An explanation of the situation

TXtourplayer said:
tirle bj does R.R. have to win and others lose? Does R.R. just need to win? Can anyone DD to beat R.R.? What order is R.R. betting in? A true determination cannot be made without knowing more details.
Click to expand...
tirle_bj is examining what happens when RR needs to win at least one bet. To be clear, consider this simple scenario: RR and only one other player left, but that player has just collected on a blackjack on the final hand, and RR must win at least one bet to beat his blackjack. Assume RR has plenty of money to split, resplit, double, etc as needed.

How does RR maximize his chance of netting at least one bet win here?

It's immediately apparent that standing on any stiff pair is a poor decision. Why stand on (7,7) vs 9 when you can split and you might draw a 4 then a ten on each? Worst case, if they turn into stiffs again just stand on them then, and you're no worse off than you were to begin with.
 
Tirle,

Excellent job analyzing this very complex situation. It is an unusual and interesting hand.
I am wondering about “no free hit” comment. Are you saying that one ought to stand on twelve (and for sure thirteen) on the first hand, to avoid busts, so one could stand on 16 and 17 two-cards on the second hand to benefit from the dealer’s bust?
What we are giving up in EV is 4/13x5/13x28.88% = 2.7%
Did you calculate all the benefits of winning first hand and pushing doubled second hand, as well as modifying play when we get good first hand, nineteen to twenty-one?
For example, if you hit twelve (or thirteen), and you get nineteen or better on your first hand then hitting some second hands to seventeen increases you total chances of winning at least one hand. Perhaps, having a good first hand, “custom” plays, like doubling totals of 10 and 11 and hitting some other hands, produce even better results.

Also, it would be interesting to know exactly how much worse (expressed in %) is hitting instead doubling on the total of nine on the first hand.

S. Yama.
 
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