Blackjack Switch - A/8 and 8/a vs 6

I ran across a Blackjack Switch game yesterday - my first time playing the game - and had a lot of fun. Afterwards I hit up the Wizard of Odds site to read more about the game. One of the things I don't understand is why you wouldn't switch and then split the following hands:

Dealer - 6

Hand 1 - A-8
Hand 2 - 8-A

It seems like switching the A and 8 to get A-A and 8-8 and then splitting those against the 6 would be the best play. Yet, all the switch calculators I've tried tell me to not switch and their basic strategy is to stand on both hands.

https://wizardofodds.com/games/blackjack/switch/calculator/

http://www.beatingbonuses.com/bjscalc.htm

Seems like following that advice I'm missing out on a couple of good hands.

Any thoughts?

If you look at the expected values of each hand using the table:-

Switch EV's

You will see that A,8 has an ev of 0.3634 against a dealer 6.

So, A,8 & A,8 is worth twice that i.e. 0.7268

A,A has an ev of 0.5018 and 8,8 has an ev of 0.1461.

So, adding these together we get 0.6479. So you are losing around 0.08 ev if you 'Switch' the A,8.


However, if you look at A,7 & A,7 verses a dealer 6 then this would be a 'Switch' because:-

A,7 & A,7 has a combined ev of 0.1779 + 0.1779 = 0.3558

A,A = 0.5018 and 7,7 = - 0.0102 which gives 0.3916.

Hope this helps.

Two 19s vs a 6

is better than

88 and AA vs. 6?

I'm thinking youre forgetting A8 is 19, which is a relatively strong hand in BJ Switch. 18s are significantly weaker though.

Look at Cindy Liu's simple switching strategy in the WOO link to help you out, mrclean.

https://wizardofodds.com/games/blackjack/switch/

If you understand the broad advice to make "Win" "Win" then it's important to realise that A-A is a good hand but has not yet made a definite winning hand, similarly for 8-8. Whereas A-8 is already a likely winner with no risk.

On other occasions it's worth considering 20 or 19 as a better hand than 11, since the stiffs expect to win whereas the 11 is good but not yet bound to win. You probably know the strategy of breaking a Blackjack, that is the same logic to create two winning (or win/draw) stiffs.

Quote: Switch

If you look at the expected values of each hand using the table:-

Switch EV's

You will see that A,8 has an ev of 0.3634 against a dealer 6.

So, A,8 & A,8 is worth twice that i.e. 0.7268

A,A has an ev of 0.5018 and 8,8 has an ev of 0.1461.

So, adding these together we get 0.6479. So you are losing around 0.08 ev if you 'Switch' the A,8.


However, if you look at A,7 & A,7 verses a dealer 6 then this would be a 'Switch' because:-

A,7 & A,7 has a combined ev of 0.1779 + 0.1779 = 0.3558

A,A = 0.5018 and 7,7 = - 0.0102 which gives 0.3916.

Hope this helps.

Thanks for taking the time to reply.

I guess I was thinking more about maximizing the amount I could win on that hand rather than taking the surer money.

Generally it's better to make a good made hand (e.g. 19) than a very good possible hand (e.g. 11 or A-A) as the latter could go wrong. Once I was trying to create a strategy and found out that, against a picture, assuming your second hand was rubbish (i.e. 12-16 or 8 or less) then you made the best first hand 21, 20... before making 11, 10.