Lucky Charlie

T

tirle_bj

Member
Here's the situation:
Last Hand, only one player advances, Lucky Charlie (5 cards no bust) pays 3:1
Only two players left:
BR1 Bankroll $20,000, BR2 Bankroll $19,800
BR2 bets first - $1,000, BR1 bets second - $1,000
Cards: BR2 (A,2); BR1 (T,7), Dealer's Up Card 8
BR2 actually elected to hit and got 7 for soft 20
Question 1: Would it be better for BR2 to DD?
Question 2: How BR2 should act further (Stand or Hit trying to get Lucky Charlie)? And if hit, then when to stay on 4 cards?
 
My quick 30 seconds answers:
1: I think it would have been better for BR2 to double down given that BR1 had the hi and low. By doubling down, he gets the high against BR1 17, which is not bad place to be.
2. Tough one but I think BR2 should stand. He is chance of half-swing or full swing seem higher than getting 5 card charlie. He already has 10 or 20. If he were to hit, only ace will allow him to go for a 5 card charlie with no fear of busting.
 
I agree with The Professional on both answers. Especially for question 2, busting while going for the charlie would give BR1 a lock.
 
Well, I agree with the second answer, but disagree with the first.
If BR2 Stands on Soft 20, he has 68.59% chances to advance, comparing with 60.3% of Hitting to Hard 18.
Even Stand on Soft 19 is marginally better than Hit to Hard 18 (61.65% vs 61.11%).
Now we have a hint how to play optimally our Initial (A,2) in case of Hitting:
Hit to Soft 19, or Hard 18.
It is easy to estimate the chances to advance in case of DD:
24.47% (dealer busts) + 12.86% x 4/13 (d=17) + 35.93% x 4/13 (d=18) + 12.86% x 3/13 (d=19) + 6.94 x 2/13 (d=20) + 6.94 x 1/13 (d=21) = 44.05%
Actually BR2 has less chances since above mentioned number is when BR1 does not react (he needs to DD when BR2 totals stiff hand, 18 or better).
So by doubling down BR2 has less than 44.05% chances to advance.
I have done my calculations for BR2 hitting to Hard 18 or Soft 19 (taking advantage form Lucky Charlie Rule) and got 48.6%.
If interesting I can present them (too boring) :rolleyes:
 
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This is tangential to the main queston, but -
tirle_bj said:
It is easy to estimate the chances to advance in case of DD:
24.47% (dealer busts) + 12.86% x 4/13 (d=17) + 35.93% x 4/13 (d=18) + 12.86% x 3/13 (d=19) + 6.94 x 2/13 (d=20) + 6.94 x 1/13 (d=21) = 44.05%
Actually BR2 has less chances since above mentioned number is when BR2 does not react (he needs to DD when BR1 totals 18 or better).
Click to expand...
Looks like a typo in the last sentence - the emboldened references to BR2 and BR1 should be the other way round.

And shouldn't BR1 always react with a DD after a BR2 DD? (not just when BR2 totals 18 or better)
[Certainly for any BR2 total other than 17 a BR1 DD is a free hit. 17 presents the possibility of both players pushing if BR1 stands, but even then I think doubling will have the better chance of advancing.]

I did try to do the calculations to show the BR1 stand versus double probabilities of advancing when BR2 has 17, but I got myself into a terrible tangle, so I deleted the evidence.:)
 
OK!
If BR2 gets 6,7 or 8 on first Hit he Stands:
With total 21 he has 75.53%(d = 17,18,19,20,21), with total 20 - 68.59%(d=17,18,19,20), with total 19 - 61.65%(d=17,18,19)
So, for these cases we have 1/13 x (75.53 + 68.59 + 61.65) = 15.83%

Now we have to consider 9 or T on first hit (Hard 12 and 13):

On Hard 12 (1/13) if we draw 6,7,8,9 (1/13 each) we should Stand and our chances to advance are 1/13 x (75.53 + 68.59 + 61.65 + 48.79) PLUS
on A,2,3,4,5 (1/13 each) we should Hit again and our chances to advance are NO BUST (Lucky Charlie) 1/13 x (4/13 + 5/13 + 6/13 + 7/13 + 8/13)
That is 19.58% + 17.75% = 37.33% , so Probability to advance for this case is 1/13 x 37.33% = 2.87%

On Hard 13 (4/13) if we draw 5,6,7,8 (1/13 each) we should Stand and our chances to advance are same as above (19.58%) PLUS
on A,2,3,4 (1/13 each) we should Hit again and our chances to advance are 1/13 x (7/13 + 6/13 + 5/13 + 4/13) = 13.02%
That is 19.58% + 13.02% = 32.6%, so Probability to advance for this case is 4/13 x 32.6% = 10.03%

The last group is A,2,3,4,5 (1/13 each) on first Hit for Soft 14 through 18.
Here on second Hit (4 cards total) we can lock Lucky Charlie (5 cards no bust, pays 3:1) and automatically advance if we will total soft 21 or less.
Thus for Soft 14 (draw A) we have 7/13, for Soft 15 (draw 2) we have 6/13, for Soft 16 (draw 3) we have 5/13, for Soft 17 (draw 4) we have 4/13, and for Soft 18 (draw 5) we have 3/13 to advance with 100% of Probability.
So for A on first Hit we have 1/13 x (7/13 x 100% + 1/13 x 9/13 + 1/13 x 8/13 + 4/13 x 7/13) = 6.19%
Here the first addendum is to have after second Hit Soft 15 through Soft 21, second addendum is to get 8 on second Hit and then hit again and Not bust, third addendum is to get 9 and Not bust on 5th card, fourth addendum is to get T and Not bust on 5th card.
Same logic will apply to Soft 15 (draw 2), Soft 16 (draw 3), Soft 17 (draw 4), Soft 18 (draw 5)
We have accordingly:
Soft 15 1/13 x (6/13 x 100% + 1/13 x 9/13 + 1/13 x 8/13 + 1/13 x 7/13 + 4/13 x 6/13) = 5.74%
Soft 16 1/13 x (5/13 x 100% + 1/13 x 9/13 + 1/13 x 8/13 + 1/13 x 7/13 + 1/13 x 6/15 + 4/13 x 5/13) = 5.28%
Soft 17 1/13 x (4/13 x 100% + 1/13 x 9/13 + 1/13 x 8/13 + 1/13 x 7/13 + 1/13 x 6/13 + 1/13 x 5/13 + 4/13 x 4/13) = 4.68%
Soft 18 1/13 x (3/13 x 100% + 1/13 x 9/13 + 1/13 x 8/13 + 1/13 x 7/13 + 1/13 x 6/13 + 1/13 x 5/13 + 1/13 x 4/13 + 4/13 x 48.79%) = 4.70% (last addendum indicates Stand on four cards total 18)

Now we can Total: 15.83% (6,7,8 on first Hit) + 2.87% (9 on first Hit) + 10.03% (T on first Hit) + 6.19% (A) + 5.74% (2) + 5.28% (3) + 4.68% (4) + 4.7% (5) = 55.32%
Previously I got 48.6%, but actually that was an estimate (some shortcuts considering not optimal play)
As we can see Hit initial Soft 13 to Soft 19 or Hard 18 is way better (55.32% vs less than 44.05) than Double Down, having Lucky Charlie option handy.
 
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London Colin said:
This is tangential to the main queston, but -

Looks like a typo in the last sentence - the emboldened references to BR2 and BR1 should be the other way round.

And shouldn't BR1 always react with a DD after a BR2 DD? (not just when BR2 totals 18 or better)
[Certainly for any BR2 total other than 17 a BR1 DD is a free hit. 17 presents the possibility of both players pushing if BR1 stands, but even then I think doubling will have the better chance of advancing.]

I did try to do the calculations to show the BR1 stand versus double probabilities of advancing when BR2 has 17, but I got myself into a terrible tangle, so I deleted the evidence.:)
Click to expand...
Thank you London Colin, I corrected my typo. If BR2 has 17, then BR1 has 75.53% chances to advance (Dealer should make a hand), if BR1 doubles then he doesn't advance when Busts and then the Dealer busts or makes 17 (24.47% + 12.86% = 37.33%). So BR1's chances of NOT advancing are: 9/13 x 37.33% = 25.84%. So BR1's chances to advance by DD are 74.16% < 75.53 ! Also BR1 should DD when BR2 has a Stiff Hand (another typo :))
It is easy to prove, since BR1 gets extra cases when dealer busts and he doesn't.
So, BR1 doesn't DD only when BR2 has 17.
 
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