Splitting payoffs

Hey there,

I am currently counting cards in a casino and would like to optimise my
strategy on splitten when playing in a group.

Since it's possible to bet a small mount as the first player in the box - lets say 5€ and a lot more, say 500€ in the second place of the same box it can be profitable to split things that usually aren't. For example 3-3 against Ace since only the 5€ of the 1st player have to be added to the bet in order to play on a single 3 with the 500€ (Second player in the box doesn't have to add money for a split if he doesnt want to)...

Now I have a question regarding this table:
/games/blackjack/appendix/9/6ds17r4/

why is the return on splitting for example two 6s against an Ace
(-0.655291) not about twice as much as hitting a 6 (for exampe 2,4) against
an Ace (-0.309156 times 2 i nowhere close to 0.655291)
Are the results of the table correct? Has anyone compared it to other sources or knows a reason for the big difference in the returns stated above?

Would be glad to get some answers =)
cheers
Julian

wizardofodds . com/games/blackjack/appendix/9/6ds17r4/

wasn't allowed to post link :(
...

Firstly if you were allowed to bet $500 behind a $5 then you'd be doing rather well. I once looked at colleagues putting $5 behind the other player's $5 and found you did split certain pairs e.g. 7s vs 9.

Since I assume UK rules so cannot really compare my figures with the US rules. However if I look at whether to split 7's vs 9 [xxx are wizard's finite decks]. ..
Hit 14 has an EV = -0.430930 [438], so the two $5 bets would expect to lose $4.31.
Splitting 7s (assume DDAS) each card has an EV = -0.277879 [289], there would be two bets on the first 7 and only 1 bet on the second, so $15 would expect to lose $4.17. (btw This ignores the advantage that you could re-split 7s on the first hand if you got yet another 7).

If we assume no DDAS after splitting, and halve the EV for the splitting you may notice that the wizard's figures are slightly different for (5,2) -0.285443 (4,3) -0.288453 and (7) -0.295741. I am guessing this is because of the cards gone and therefore this can affect things by up to .01.

With your Aces example I have no idea except if you have 66 the dealer is less likely to have A6 - giving them 17 and you a better chance of you winning - since you already have two of them.

hey, thanks for you answer :)
so you think the calculations in the table are correct?

I only worked out infinite deck numbers using my simple spreadsheet (as I was only after an estimate of strategy for Blackjack and some variants). Sometimes there are small differences between infinite deck and finite deck (e.g double A4v4) but I assume these are due to the cards you've seen actually having a difference.

Sorry but I cannot fully explain the difference you are seeing or make a conclusive judgement as I haven't worked out any finite deck numbers.