if play 2 boxes, ES10
what is the prob. to lose both 2 boxes simulteanously ???
(neither 1 of them is tie / surr)
thx
thx to Don,
incl. ES10 ?
how about 3 boxes ?
A rough guess would be about half of the 31%.
Don
That in and of itself would be what the push 18% of hands total percent so far of 49% giving you win rate of 51% of the top before counting. That 1% edge right flat bet if you want but still not win enough. If counting counting give you what as much a 3-7%. Spreading 1-6 double deck give you your huge advantage. Right.
"That in and of itself would be what the push 18% of hands total percent so far of 49% giving you win rate of 51% of the top before counting."
No, you misunderstand. You lose more single hands than you win, and you lose more double hands than you win double hands. The distribution isn't symmetrical. So L-L is more common than W-W, and L-P is more common than W-P.
In the end, all the probabilities are preserved, and the dealer edge vs. the BS player cannot be changed by varying the number of simultaneous hands played.
Don
