MuRRaY said:
If the casino's edge is .66%, does that mean the chance of you actually winning a hand is .34%?
No, it means that if you bet $100, you can expect to get back $99.34 in the long run.
If your bet always pays 1:1 (for example, a craps pass or come bet), the house edge is the difference between the percent you lose and the percent you win. So the pass line, which carries a 1.4% house edge, wins about 49.3% of the time and loses about 50.7% of the time (50.7%-49.3% = 1.4%).
When your bets don't pay 1:1, the situation is more complicated. In blackjack, for example, you get paid 3:2 on blackjacks and sometimes you double down or split (and double your bet). The general formula you use to calculate EV (expected value) is:
EV = [(percent outcome #1)*(amount won on outcome #1) + (percent outcome #2)*(amount won on outcome #2) + ... (percent outcome #n)*(amount won on outcome #n)] / [(percent outcome #1)*(amount bet on outcome #1) + (percent outcome #2)*(amount bet on outcome #2) + ... (percent outcome #n)*(amount bet on outcome #n)]
or simply
EV is the amount you won divided by the amount you bet, corrected for the percent of the time you win and bet.
For blackjack, you have many outcomes: win a blackjack, win a single bet, win a doubled bet, win both split bets, win one split bet and push one split bet, lose a single bet, lose a doubled bet, lose both split bets, lose one split bet and push one split bet, etc. (Note that lost bets should be put in as negative numbers in the formula above)
Overall, the chances of you winning are about 43%, and the chances of you losing are 49%, with the remainder pushing. However, your average win when you win is much greater than 1 (1.25?), while your average loss is only slightly higher than 1 (1.05?). So when you calculate your EV for blackjack, the house edge is much smaller than 6% - it's about 0.3% to 1%.