Blackjackstudent.
The following is a piece from the book, Knock out Black Jack.(p117-118). It is a bit long winded and the author does have a PHD in maths, but after all BJ is about the mathematical edge. That's why we count.
basically he is saying it makes no difference in the long run where you sit or how other people play.
Can the play of another player, no mater how good or bad, affect your chances of winning.Many believe it can. That bad players somehow "bring down" an entire table with their poor plays. NOT SO. In reality, the play of others has no impact on your own fortune. (The only exception to this is if the other players take all the cards forcing a re shuffle.) If there are sufficient cards to finish the round then other players have no overall effect on your expected outcome. Take a mathematical look at why this is true. Say we know that the remaining deck contains, X good (cards 10's-Aces) Y good cards (2's through 7's) ( the ko system includes 7's as + count cards) and Z neutral cards (8 and 9's) At the end of the deck the running count will be +4. So the present running count is 4+x-y. Now let's consider the next hit. The following are the probabilities of getting a good, bad, or neutral card: P(X)=X/(X+Y+Z) P(Y)=Y/(X+Y+Z) P(Z)=Z/(X+Y+Z). Assume for the moment that someone playing ahead of us takes a hit. If the player takes a good card, there are now only (X-1) good cards and (X+Y+Z-1)total cards remaining, so our chances of getting a good card after him would be (X-1)/(X+Y+Z-1). Like-wise if the player gets a bad card, our chances of getting a good card would be X/(X+Y+Z-1). finally if the player gets a neutral card, our chances of getting a good card would again be X/(X+Y+Z-1). To determine our overall chances of getting a good card, we need to multiply each of these conditional chances by their respective probabilities,and add. This yields: P(X)=[X/(x+Y+Z)]x[(X-1)/(X+Y+Z-1)]+[Y/(X+Y+Z)]x[X/(X+Y+Z-1)]+[Z/(X+Y+Z)]x[X/(X+Y+Z-1)] Collecting terms in the numerator give us: P(X)=X(X+Y+Z-1)/[(X+Y+Z)(X+Y+Z-1)] OR SIMPLY PUT, P(X)=X/(x+y+z) and this, as you can see, is precisely the original chance of getting a good card if we had taken the first hit! It makes no difference whether an earlier player takes a hit, two hits, or twenty hits. Overall WE STILL HAVE THE SAME CHANCE OF GETTING THE CARD WE WANT, as long as there is at least one card left after he has done with his hand. Further more, AS LONG AS THERE ARE SUFFICIENT CARDS LEFT IN THE DECK TO FINISH THE ROUND, IT MAKES NO DIFFERENCE HOW OTHER PLAYERS PLAY. In terms of the expected final value of your hand, or the dealers hand, and hence your chance of ultimately winning or loosing.
PERSONALLY I try to sit at 3'rd base so I can see every card as it is dealt and it gives me more time to react and keep the count more accurately (As most people do stop and think about what they are going to do) and also not have to rush to make my decision.
PS If you are a beginner read the book and try the system. I play it and it works.
a5teve.