A question for math-minded folks...
If the prior probabilities of a given card landing on the first box, second box, third box are 40%, 15%, 10%, respectively,
what is the posterior probabilities...
That is great data. I am using the KO system (mostly playing 6D, H17, DAS); IRC = -20.
With that said, would it be too much to ask for you to provide similar data as above, but for the KO system...
Yep, I believe this is the correct approach. Therefore, if the Ace didn't land in the first spot, the probabilities increase for the following spots:
25% (instead of 15%) for the second spot, 16.7%...
Here's how I would look at it.
Successes of card landing on spot 1, spot 2, spot 3, eventually somewhere
success(1) = .4
success(2) = .15
success(3) = .1
success(>3) = .35
if trial 1 fails...
Three boxes, A=40%, B=15%, C=10%.
"The 40% event did not happen", remaining 60%.
In this 60%, 15% is in B, 10% is in C.
Therefore, 15/60 will be in B, 10/60 will be in C, and 35/60 went somewhere...
But he wasn't asking for EVs. He was asking for the probability of a card landing in one of the first three boxes and then the probability after one sees that box 1 isn't successful that it will land...
Here's example to show relationship:
Single deck, S17, SPL1, SPLA1, NDAS
Overall EV best strategy: +.0248%
EV given up card A: -36.03%
EV given up card 2: +9.995%
EV given up card 3: +13.59%...
So, a total of 65%.
Did you mean this? Doesn't make sense. Did you mean to say "of the second and third box when the 40% event did not happen?"? If you did, then you have 65% left for those...
In a normally cutted game, the probability of receiving a single given card will never be as high as 40%.
The probability of receiving a 10,J,Q,K may reach 40% and will go up if the next card is a...
You need to get over this paranoia and also learn what kind of variance is typical and can be expected. A true AP does not even begin to think of such things just for losing 80 units (assuming you...