Expeced lose of surrending with two 8's versus dealer 10 or Ace?

I know you're supposed to split two 8's versus a dealer 10 or ace, but I'm curious what the expected loss is for surrendering with two 8's versus a dealer 10 or ace. Can anyone tell me what that expected loss is?

-48.62% for splitting, -50% for surrendering against a ten (-37.07% to -51.04% against an ace, it depends on H17 or not). Not much of a difference. Depending on your betting level, a couple bucks over your lifetime, and a heck of a lot less variance.

Regarding variance, I have a clarifying question. If I'm looking to minimize losses, not necessarily maximize gains, it sounds like it would be a good idea for me to surrrender 8/8 against a ten or ace. Is that correct? My betting level is $10 most of the time, $5 some of the time?

No. Over the long run you will lose less by splitting 8,8 against a ten or ace (in a S17 game). Not that much--it may be worth it to you to pay that cost to keep that extra chip in your stack. But then, why are you gambling at all?

At that level of play, I would always split.

Good point! Thanks for your help.

Appendix 6 says to surrender pair of 8's against an Ace in special case of 2 deck, dealer hits a hard 17.

Appendix 3b says to stand with a pair of 8's against a 10 in special case of 2 deck, dealer stands on 17 and more than 2 cards have been dealt.

When the choice is between losing $200 or losing $50 you should always remember that the calculations done in the cheat sheet are for a freshly shuffled deck where you the first card being dealt is to you. Depending on what has been played, you may in fact splitting may have a lower EV.

They did this scenario on the tv show Entourage once with a huge bet and he keeps getting pairs of 8's.

Quote: pacomartin

Appendix 3b says to stand with a pair of 8's against a 10 in special case of 2 deck, dealer stands on 17 and more than 2 cards have been dealt.

I think what the Wizard means there is how you should generally stand on a 3+ card 16 vs. 10 (S17 or H17 doesn't matter with upcards of 7-10). That would not be applicable to a pair of 8s since that will always be a 2-card hand. Having said that, if you're counting and can't surrender there becomes a point at a really high count where the correct play would shift from splitting to standing. The index from split to surrender (if allowed) is lower and thus more commonly encountered (both of these indices applying since with lots of 10s you become more likely to end up with two 18s losing to a dealer 20).

Now I'll talk about a pair of 8s against an ace; here BS in a shoe game calls for surrendering with H17 if you can, but not with S17. If you're counting, the indices behave differently against an Ace as compared to against a 10; with 88 (as well as a hard 17) you surrender if the count is below a certain point, but not above. The reason being is the post-blackjack-check EV with a dealer Ace goes up with the count (mainly due to being more likely to bust) while it varies inversely with a 10 (since a high count makes it more likely the dealer has a tough-to-beat 20).

Also note that the EV of splitting 88 vs. A is higher with S17 than H17 due to the fact that 18 is the most likely hand you'll get, and that's the total most benefited by S17 (since with H17 a dealer 17 that would make your 18 in S17 win has a good chance of turning into a bigger hand).