Please assume that if my first card for a six deck game is a 10, I have (very roughly) a 25% advantage.
And we all know that we can correctly predict a ten 4/13 (~30.8%) times. But what if I can correctly predict a ten 14/37 (37.8%) times?
So, what is my advantage when I predict a 10 will be the next card? In the above example, is it .25 x .378?
Or, perhaps it is .25 x (.378 x .070), where .070 represents .378 - .308.?
Or, is my advantage some other #?
"Please assume that if my first card for a six deck game is a 10, I have (very roughly) a 25% advantage."
Well, I'll do that for the purpose of this exercise, because you asked nicely (!), but you know that, for normal BJ, that number is all wrong, right? The correct number is about 14%.
"And we all know that we can correctly predict a ten 4/13 (~30.8%) times."
It's a funny way to put it, but OK. What you mean is that, whether you predict a ten or any other value, for that matter, a ten will appear, on average 30.8% of the time. Your prediction doesn't alter the probability.
"But what if I can correctly predict a ten 14/37 (37.8%) times?"
That would help! :-)
"So, what is my advantage when I predict a 10 will be the next card? In the above example, is it .25 x .378?"
Certainly not.
"Or, perhaps it is .25 x (.378 x .070), where .070 represents .378 - .308.?"
Nope, not that either.
"Or, is my advantage some other #?"
Some other number. Here's the math. Assume that, overall, your edge in the 6-deck game is, say, -0.5%. (I have to revert to the 14% edge for the ten here, to make the numbers work correctly. If you really had a 25% edge, blackjack wouldn't exist!) If you knew with certainty that the next card was a ten, your edge would be that 14%. If you were just guessing randomly, you would, of course, have no edge at all; you would have a 0.5% disadvantage.
So, we would say that (.308 x .14) + (.692x) = -0.005, where x is your cumulative disadvantage against all the other cards. Solving for x, we get that the global disadvantage against all the other nine cards you could receive is about 0.0695, or 6.95%.
So now, suppose you know with 37.8% accuracy that the next card will be a ten. Either you get that ten or you don't; your edge, once you receive the card (14%) doesn't change. But now, you have a 37.8% chance at that 14% edge and a 62.2% chance of the -6.95% edge. That works out to a positive edge for the player of 0.97%, or, roughly, one percent, which is the answer you seek.
Clear?
Don
Overkill,
To answer your query, we need to know En: your advantage (presumably negative) for a 1st-card non-ten. We can calculate En if we know the overall EV, EVo, of the game, since (if Ex=adv. for 1st-card X, and Px & Pn are the probabilities of X and non-X):
EVo = Ex*Px + En*Pn = Ex*Px + En*(1-Px)
Rearranging this for En gives
En = (EVo - Ex*Px)/(1-Px)
For example, if we use your values of 25% for Ex and 4/13 for Px, and we assume that EVo = -1% (just a w.a.g.), then we get
En = (-1% - 25%*(4/13))/(1 - 4/13) = -12.55...%
Enter Carnac!
Now if Carnac the Magnificent (youngsters, see link below) plays the game, he's able to predict the X's correctly 14/37ths of the time (and thus incorrectly 23/37ths of the time), so Pxc = 14/37 = 37.837...% and Pnc = 100% - Pxc = 62.162...%. Therefore, assuming Ex and En are the same for Carnac as for a random guesser, Carnac's EVc will be
EVc = Ex*Pxc + En*Pnc
So, using the -12.55...% calculated above for En, Carnac's advantage is
EVc = +25%*(14/37) + (-12.55...%)*(23/37) = +1.65...%.
This means that Carnac can transform a -1% game into a +1.65% game... woo hoo!
Using your own value for EVo and the given values for Ex, Px, and Pxc, you can calculate EVc as follows:
EVc = Ex*Pxc + (1 - Pxc)*(EVo - Ex*Px)/(1-Px)
Hope this helps!
Dog Hand
OK, thanks Don and Dog Hand. It looks as though the edge is roughly the same as for card counting.
Keying Aces, not Keying Tens.
