Here's another example that leaves me a little bit confused. With max bet of 500 and min 5, this example shows we have 415, on the button and BR1. BR2 has 275 and the book says 140 is the best bet because it says if we both win, I'll win no matter what BR2 bets and if both lose, I still win no matter what BR2 bets also but........ Is the reason why we bet 140 is because both hands WINNING and both hands LOSING have a higher % chance of happening vs. BR2 winning and me losing? Because that's obviously a possibility too... so with this in mind, why woulnd't 135 be the optimal bet in this situation? 415 - 135 = 280, and if he pushes and I lose I'll still win by 5, and if we both win and both lose I'd advance as well so why is 140 the superior bet in this case?
135 is not enough to advance; you will tie on a score of 550 if you both win. And since winning a hand is so much more common than pushing, it is better to have a tie result from the rare lose/push combination than from the much more common win/win combination.
The answer is yes. Because you both play vs the same dealer hand, there is a strong correlation in your results. There is a table in wong's book which gives the probabilities for various combinations of results for different numbers of players. For example, the case you give above: the probability of one player winning while the other loses is only about 12%. With a bet of 140, BR2 could still tie you if you lose while he pushes. That probability is in the table as well and is fairly small (I don't recall the exact number). I like to look at it as a trade off. The value of taking the high outright is close to 43% and you only give up the small probability of losing while BR2 pushes, which is failrly small and you could still beat him in a playoff if this happens.
