3,3 VS 2 index DD NDAS

[aceside]

Thanks for the update! How would you revise the calculation to account for 6,6?

I don’t understand what Don is talking about, but here is what I would calculate. In this part, get ride of the 6,6 pair contribution to the hard 12 hand.
”Furthermore, the probability that you are dealt a 12 is 0.00375+0.00094+0.00082+0.00094+0.00041 = 0.00686

Hence, the expected true count conditioned on the event that you are dealt a hard 12 is ((0.00375)(1/(101/52))+(0.00094)(2/(101/52))+(0.00082)(2/(101/52))+(0.00094)(2/(101/52))+(0.00041)(3/(101/52)))/0.00686 = 0.779.”

You only count the probability of getting a hard 12, excluding a 6,6 pair. Let me comment a little further. Don has threatened to ban me off from this website a few times!