Insurance Paid 2.5:1 - Spansish 21

#1
Hi,

It's me again. The Spanish 21 offered in my local casino pays insurance 2.5:1, instead of 2:1 as in regular Blackjack.

I checked and the EV is -12.5%, which is worse than regular Blackjack. I'm thinking of using 10-Count (T(JQK)=-2, Ace~9=+1) as mentioned in Stanford Wong's book, Professional Blackjack.

I'm starting count at -64 as 8 decks and all the 10s are removed. If the running count goes above -16, I should have an edge on insurance bet (+6 per deck, +48 for 8 decks. Or start at 0, +48 to have an edge). It would be appreciated if someone could check it and let me know if I do the math correctly, thanks.

Chisanchen
 

daddybo

Well-Known Member
#2
Have you read Katrina Walker's Book, " The Pro's Guide To Spanish 21"? it's worth it since you seem to have some Decent SP21 games close by.
 
#3
daddybo said:
Have you read Katrina Walker's Book, " The Pro's Guide To Spanish 21"? it's worth it since you seem to have some Decent SP21 games close by.
Hi daddybo,

Yes, I did. The book is quite useful, although I need to make some adjustments as some of the rules offered in my local casino are not discussed in the book, especially the Hit-after-Doubling. Anyway, it's a good book for Spanish 21.

Chisanchen
 

FLASH1296

Well-Known Member
#4
I am thinking that Insurance at 5:2 is a good bet with T.C. > +8

It cannot be much more than that. Ordinary 2:1 Insurance requires T.C. >+12
 

k_c

Well-Known Member
#5
Assuming insurance paid 2.5 to 1...

If insurance pays 2.5 to 1, which is the same as 5 to 2, then the perfect insurance count would be to tag tens = +5 and non-tens = -2. (Tags are relative to what remains in shoe as opposed to what has been removed.)

Each spanish 21 deck starts out with 12 tens and 36 non-tens so initial running count would be (5*12-36*2)*(number of decks) = (-12)*(number of decks).

As cards are dealt, adjust initial running count by +2 when a non-ten is dealt and -5 when a ten is dealt. Whenever running count = 0 then insurance @ 5:2 would be an even EV bet. Running count less than 0 would be negative EV. Running count greater than 0 would be positive EV.
 
#6
k_c said:
If insurance pays 2.5 to 1, which is the same as 5 to 2, then the perfect insurance count would be to tag tens = +5 and non-tens = -2. (Tags are relative to what remains in shoe as opposed to what has been removed.)

Each spanish 21 deck starts out with 12 tens and 36 non-tens so initial running count would be (5*12-36*2)*(number of decks) = (-12)*(number of decks).

As cards are dealt, adjust initial running count by +2 when a non-ten is dealt and -5 when a ten is dealt. Whenever running count = 0 then insurance @ 5:2 would be an even EV bet. Running count less than 0 would be negative EV. Running count greater than 0 would be positive EV.
Hi k_c,

Thanks. It would be a little difficult for me if I use the count. I use Stanford Wong's method for regular Blackjack. I'll need to pratice if I want to switch to this new one.

Thanks again for your help, I appreciate it.
 

k_c

Well-Known Member
#7
San said:
Hi k_c,

Thanks. It would be a little difficult for me if I use the count. I use Stanford Wong's method for regular Blackjack. I'll need to pratice if I want to switch to this new one.

Thanks again for your help, I appreciate it.
This count would only be best relative to an insurance bet similar to a perfect insurance count in normal blackjack.

All that is being done in either case is to weight the tags of tens and non-tens to reflect the insurance payoff odds. By doing that the count shows when taking insurance is positive EV.

Other than showing exactly when to insure these insurance counts are not too useful (except maybe for reference.)
 
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