Any tips would be appreciated. I'm a bit confused with the hands thing. On a typical night at the casino on Tuesday evenings, we deal cards for 28 games (also called hands) of bridge. As each hand has thoroughly shuffled cards we can assume the cards are dealt independently for each hand. What is the probability that in 28 hands of bridge (wherein each hand I am dealt 13 cards), that I have at most 2 hands where I'm dealt no aces and no kings?
I've calculated the probability of getting a hand of no ace and no king first which is : $$ P(no ace\cap noking)=\frac{\binom{4}{0}\binom{4}{0}\binom{44}{13}}{\binom{52}{13}}=0.0817 $$ And then I just know that we want at most two successes out of 28 trials so : $$\binom{28}{0}(0.0817)^0(0.9183)^{28}+\binom{28}{1}(0.0817)^1(0.9183)^{27}+\binom{28}{2}(0.0817)^2(0.9183)^{26} $$